# 3 Applications Of Derivatives

This book explores different applications of differentiation in order to provide a better understanding of how it can be used to solve problems. It is written for those who have an understanding of basic calculus but want to learn more about how differentiation can be applied in other areas. The author provides detailed examples and explanations which make the material easy to understand. Overall, this book provides a comprehensive look at differentiation and its many applications.

Differentiation is a process that allows us to find the derivative of a function. The derivative can be used to calculate how a function changes at any given point, and it’s an essential tool for understanding how physical systems work. In this book, we’ll discuss what differentiation is and how it works, as well as some of its most common applications.

Differentiation is a fundamental tool of calculus that has countless applications in the real world. Optimization problems are one type of problem where differentiation can be used to great effect. In an optimization problem, the goal is to find the maximum or minimum value of a function. To do this, we take the derivative of the function and set it equal to zero. This will give us the critical points of the function, which are the points where the function is either at a maximum or minimum. From there, we can use some more calculus to determine which of these critical points is actually the point of maximum or minimum for the function. Optimization problems can be applied in many different fields, such as engineering, physics, and economics. In each case, using calculus to find the optimum solution can save time and resources.

One of the most common applications of differentiation is finding the extreme values of a function. Extreme values can tell us things like the highest and lowest points on a graph, or the fastest and slowest points in a motion. To find extreme values, we need to take the derivative of the function and set it equal to zero. This will give us the points where the function is either increasing or decreasing at its fastest rate.

Applications of Differentiation include finding extreme values of a function. The derivative of a function at a point measures the rate of change of the function at that point. So, the derivative can be used to find points where the function is increasing or decreasing at its fastest. These points are known as local extrema, and they can be found by setting the derivative equal to zero and solving. Differentiation can also be used to find points where the function changes from increasing to decreasing (or vice versa). These points are called inflection points, and they can be found by setting the second derivative equal to zero and solving. In addition, differentiation can be used to find absolute extrema, which are the largest and smallest values that a function can take on over a given domain. To find absolute extrema, we need to take into account both the function’s values and its derivatives. So, applications of differentiation are not just limited to finding extreme values; they also allow us to understand the behavior of a function at different points.

The Mean Value Theorem is another important application of differentiation. This theorem states that if a function is continuous on a closed interval, then there exists at least one point in the interval where the function’s derivative is equal to the average rate of change of the function over the interval. In other words, if we take any two points in an interval and find the average rate of change of the function between those points, there will be at least one point in the interval where the function’s derivative is equal to that average rate of change. This theorem has a number of important implications, and we’ll discuss some of them in this chapter.

The Mean Value Theorem is one of the most important results in calculus. And it’s not just because it’s on the AP Calculus exam. The theorem has countless applications in mathematics, science, and engineering. In fact, it’s often referred to as the “workhorse” of calculus. Differentiation is all about finding rates of change. The Mean Value Theorem tells us that for any function that is continuous on an interval, there exists a point where the rate of change is equal to the average rate of change over the entire interval. In other words, the theorem provides a way to find the “average” rate of change. Of course, the theorem is much more than just a simple averaging result. It has profound implications for how we understand and work with functions. In particular, it gives us a way to estimate values and understand relationships between different quantities. As one famous mathematician once said, the Mean Value Theorem is “the most useful single theorem in all of mathematics.” With so many applications, it’s easy to see why!

A function is monotonic if it either always increases or always decreases as we move from left to right on the graph. In other words, a function is monotonic if its derivative is either always positive or always negative. To see why this is true, recall that the derivative at a point measures the rate of change of the function at that point. So, if the derivative is always positive, then the function is always increasing; if the derivative is always negative, then the function is always decreasing. It’s easy to see that a constant function has a derivative of zero, so it is neither increasing nor decreasing.

There are two types of monotonic functions: Increasing functions are those whose values always get larger as we move from left to right. Decreasing functions are those whose values always get smaller as we move from left to right. It’s important to note that a function can be increasing or decreasing without being continuous. In fact, many discontinuous functions are monotonic.

Applications of Differentiation can be broadly classified into two categories: Monotonic Functions and Non-monotonic Functions. A function is said to be monotonic if it either always increases or always decreases as we move along the x-axis. On the other hand, a function is non-monotonic if it changes direction at least once. Examples of monotonic functions include linear functions and exponential functions. On the other hand, examples of non-monotonic functions include polynomial functions and trigonometric functions. Applications of differentiation to monotonic functions are typically concerned with finding the maximum or minimum values of the function, while applications to non-monotonic functions are typically concerned with finding points of inflection.

In our approach, we will dicuss functions that are monotonic with repsect to an interval. In this approach, we see the full power of the derivative.

We say that a function is concave up on an interval if it lies entirely above its tangent lines on that interval. Similarly, we say that a function is concave down on an interval if it lies entirely below its tangent lines on that interval. The points at which a function changes from concave up to concave down (or vice versa) are called inflection points.

Geometrically, concavity corresponds to the second derivative. More precisely, we say that a function is concave up on an interval if its second derivative is positive on that interval, and we say that a function is concave down on an interval if its second derivative is negative on that interval. In other words, the second derivative can be used to determine concavity. This is not too surprising, since the second derivative measures the rate of change of the first derivative. So, if the first derivative is increasing, then the function is concave up; if the first derivative is decreasing, then the function is concave down.

Applications of Differentiation aren’t just limited to finding the slope of a tangent line at a point. In fact, one of the more interesting applications is finding what’s known as an inflection point. An inflection point is a point on a curve where the curve changes from concave up to concave down, or vice versa. To find one, we take the second derivative of a function and set it equal to zero. The second derivative can tell us concavity, so if we set it equal to zero, we’re saying that the concavity changes at that point. And that’s an inflection point! Now that we know how to find them, what are they good for? Well, they’re often used in optimization problems. If we’re trying to maximize or minimize an object’s volume, for example, we might want to know where its inflection points are. That way, we can be sure that we’re not accidentally making the object too big or too small. So next time you see an inflection point, don’t just ignore it - it could be just what you need!

If you’ve ever wondered how your math teacher can tell what a graph is going to look like just by looking at an equation, the answer lies in differentiation. Differentiation is a process of finding the rate of change of a function, and it turns out that this rate of change can be used to sketch the shape of a graph. For example, if a function is increasing at a constant rate, then its graph will be a straight line. If the rate of change is increasing, then the graph will be curved upwards; if the rate of change is decreasing, then the curve will bend downwards. By understanding how differentiation works, we can gain insight into the shape of many mathematical functions.

Differentiation is one of the most powerful tools in mathematics, with applications in everything from physics to engineering. It can also be used to sketch the graphs of functions, a process known as curve sketching. By studying the derivatives of a function, we can learn about its local behavior and how it changes over time. This information can then be used to sketch a rough graph of the function, without having to calculate any points. Curve sketching is a valuable tool for visualizing functions and understanding their behavior. It can also be used to estimate things like turning points and asymptotes, which can be difficult to calculate exactly. In short, curve sketching is a powerful tool that everyone should know how to do.

Differentiation is a powerful tool that can be used to solve a wide variety of optimization problems. Whether you’re trying to find the maximum or minimum value of a function or to optimize some other objective, differentiation can help you find the answer. In this article, we’ll take a look at some of the most common applications of differentiation in optimization. Hopefully, by the end, you’ll see just how powerful this tool can be!

If you’re anything like me, you love nothing more than a good optimization problem. You know, the kind where you have to find the maximum or minimum of some function by differentiating it and setting the derivative equal to zero. But what’s even better than solving those kinds of problems is finding applications for them in the real world. And that’s exactly what applied optimization is all about. By understanding the principles of differentiation, we can figure out how to optimize all sorts of things in the world around us. For example, we can use optimization to find the fastest route between two points, or the least expensive way to produce a product. We can even use it to help design more efficient algorithms. So next time you’re struggling with a difficult optimization problem, just remember that there’s a good chance someone out there has already found an application for it.

An indeterminate form is an expression that takes on different values depending on the values of the variables involved. Indeterminate forms arise in many applications of differentiation, including the determination of maxima and minima. In general, indeterminate forms can be difficult to work with, but there are a few simple rules that can be used to simplify them. One famous example is the expression 0/0, which is known as an indeterminate form because it can take on any value depending on further analysis.

Differentiation is a powerful tool that can be used in a variety of ways to solve mathematical problems. In this book, we learn about several of the most common applications of differentiation. By understanding how to differentiate functions, we can gain insight into their behavior and solve optimization problems more easily. We’ve also seen how indeterminate forms can arise in different situations and learned some simple rules for dealing with them. So if you’re looking to become skilled, it’s important to understand the principles and applications of differentiation inside and out.

## 3.1 Extreme Values

## 3.2 Relative Extreme Values

If a function is defined on an open interval and if at some point in that interval the function reaches a maximum or minimum value (relative to that interval), then we say that the function has a relative extrema on that interval. A maximum or minimum value that occurs at an endpoint is not, by definition, a relative maximum nor a relative minimum. A relative maximum or relative minimum must occur in the interior of an interval.

Now we define relative extrema and state the relative extrema theorem.

**Definition 3.1 **Let \(f\) be a function defined on an open interval \(I\) with \(c\in I\).

- If \(f(c)\geq f(x)\) for all \(x\) in \(I,\) then \(f(c)\) is called a
**relative maximum**of \(f\) on \(I.\) - If \(f(c)\leq f(x)\) for all \(x\) in \(I,\) then \(f(c)\) is called a
**relative minimum**of \(f\) on \(I.\) - If \(c\in I\) and \(f(c)\) is either a relative maximum or a relative minimum then \(f(c)\) is a relative extrema, and we say that \(f(c)\) is a
**relative extreme value**of \(f\).

A relative maximum is sometimes called a **local maximum** .

A relative minimum is sometimes called a **local minimum** .

The following proposition is sometimes called Fermat’s theorem due to acknowledgment that Fermat realized the result first. The following examples show that even when \(f'(c)=0\) there need not be a maximum or minimum at \(c.\) In other words, the converse of Fermat’s Theorem is false in general. Furthermore, there may be an extreme value when \(f'(c)\neq 0\) or when \(f'(c)\) does not exist. We state Fermat’s wonderful observation as the **Relative Extrema Theorem** .

::: {#thm- } [Relative Extrema Theorem] If \(f\) has a relative extremum at \(c\) and \(f'(c)\) exists then \(f'(c)=0.\) :::

*Proof*. Since \(f\) is differentiable at \(c,\) \(f'(c)\) must be positive, zero, or negative. If \[f'(c)=\lim _{x\to c}\frac{f(x)-f(c)}{x-c}>0
\] then there exists an interval \((a,b)\) containing \(c\) such that \(\frac{f(x)-f(c)}{x-c}>0\) for all \(x\neq c\) in \((a,b).\) This produces the following inequalities for \(x\)-values in the interval \((a,b).\) If \(x<c\) then \(f(x)<f(c)\) and so \(f(c)\) is not a relative minimum. If \(x>c\) then \(f(x)>f(c)\) and so \(f(c)\) is not a relative maximum. So the assumption that \(f'(c)>0\) leads to a contradiction. Assuming that \(f'(c)<0\) will also lead to a similar contradiction. Thus it must be the case \(f'(c)=0\) as desired.

**Example 3.1 **Determine if the Relative Extrema Theorem applies and if so find the relative extrema for the function \[
f(x)=
\begin{cases}
2x-3 & x<2/3 \\
3-7x & x\geq 2/3
\end{cases}
\]

*Solution*. The function \(f\) has its maximum value (local and absolute) at \(x=2/3,\) but we can not find this absolute maximum by setting \(f'(x)=0\) because \(f'\) is not defined at \(x=2/3\). Since \(f'(2/3)\) does not exist the Relative Extrema Theorem does not apply.

**Example 3.2 **Determine if the Relative Extrema Theorem applies and if so find the relative extrema for the function \(h(x)=x^3+4\).

*Solution*. Since \(h'(x)=3x^2\) we have \(h'(0)=0.\) However, since \(h\) does not have a relative extremum at \(x=0\) the Relative Extrema Theorem does not hold.

## 3.3 Critical Numbers

In general, the critical numbers divide the domain of a function into intervals on which the sign of the derivative remains the same, either positive or negative. Therefore, if a function is defined on that interval it is either increasing or decreasing on that interval. In particular the graph can not change directions on that interval. This is the crucial idea behind using the derivative to analyze graphs of function.

**Definition 3.2 **A real number \(c\) is called a **critical number** of the function \(f\) provided:

- \(c\) is in the domain of \(f\) and \(f'(c)=0\) or
- \(c\) is in the domain of \(f\) and \(c\) is not in the domain of \(f'\).

Or said differently, a critical number of a function \(f\) is a real number \(c\) in the domain of \(f\) such that \(f'(c)=0\) or \(f'(c)\) does not exist.

**Theorem 3.1 **If \(f\) is a continuous function and has a relative extremum at \(c,\) then \(c\) is a critical number of \(f.\)

**Example 3.3 **Find the critical numbers, if there are any, for the function \[
f(x)=|x+2|-3.
\]

*Solution*. The function \(f(x)=|x+2|-3\) can be rewritten as a piecewise function as \[
f(x)=\left\{
\begin{array}{cc}
x-1 & x\geq -2 \\
-x-5 & x<-2
\end{array}
\right.
\] The graph of the function \(f\) has a sharp corner at \(x=-2\) which can be seen by evaluating the left derivative of \(f\) at \(x=-2\) and the right derivative of \(f\) at \(x=-2\). Therefore \(f'\) is not defined at \(x=-2\). Since \(f'\) is defined everywhere else and \(f'\) is not 0 anywhere the only critical number is \(-2.\)

**Example 3.4 **Find the critical numbers, if there are any, for the function \[
g(x)=x^3+2.
\]

*Solution*. The function \(g\) has derivative \(g'(x)=3x^2\) which is defined for all real numbers. Notice that at \(x=0,\) \(g'(0)=0\) and so 0 is a critical number. Also notice that \(x=0\) is not a relative extrema nor an absolute extrema. In summary, critical numbers allow us to check if there are any extrema at that point, but not conversely.

**Example 3.5 **Find the critical numbers, if there are any, for the function \[
h(x)=\sin x.
\]

*Solution*. The function \(h(x)=\sin x\) has derivative \(h'(x)=\cos x\) which is defined for all real numbers. So to check for extrema we will need to determine where \(h'(x)=0.\) This occurs at \(x=\frac{\pi }{2}+\pi k\) where \(k\) is any integer. In fact, in this case the absolute extrema of \(h\) occurs at these values.

**Example 3.6 **Find the critical numbers, if there are any, for the function \(f(x)=x^{2/3}(1-x)\).

*Solution*. The function \(f\) has derivative, \[f'(x)=\frac{2}{3}x^{-1/3}(1-x)+x^{2/3}(-1)=\frac{2-5 x}{3 x^{1/3}}\] and so we need to check for any \(x\) in the domain of \(f\) such that \(f'(x)=0\) or \(f'(x)\) is undefined. We see that \(f'\) is undefined for \(x=0;\) and \(f'(x)=0\) for \(x=\frac{2}{5}.\) So for this function \(f\) there are two critical numbers namely \(x=0\) and \(x=\frac{2}{5}.\)

**Example 3.7 **Find the critical numbers, if there are any, for the function \[
f(x)=\sqrt{9-x^2}.
\]

*Solution*. The function \(f\) has derivative, \[
f'(x)=\frac{1}{2}\left(9-x^2\right)^{-1/2}(-2x)=-\frac{x}{\sqrt{9-x^2}}
\] and so we need to check for any \(x\) in the domain of \(f\) such that \(f'(x)=0\) or \(f'(x)\) is undefined. We see that \(f'\) is undefined for \(x=\pm 3;\) and \(f'(x)=0\) for \(x=0.\) So there are three critical numbers of \(f\) namely, \(x=0\) and \(x=\pm 3.\)

**Example 3.8 **Find the critical numbers, if there are any, for the function \[
\displaystyle f(x)=\frac{x^2}{x^2-3}.
\]

*Solution*. The function \(f\) has derivative, \[f'(x)=\frac{\left(x^2-3\right)2x-x^2(2x)}{\left(x^2-3\right)^2}=\frac{-6 x}{\left(-3+x^2\right)^2}\] and so we need to check for any \(x\) in the domain of \(f\) such that \(f'(x)=0\) or if \(f'(x)\) is undefined. We see that \(f'\) is undefined for \(x=\pm \sqrt{3};\) and \(f'(x)=0\) for \(x=0.\) So there are three critical numbers of \(f\) namely \(x=0\) and \(x=\pm \sqrt{3}.\)

**Example 3.9 **Find the critical numbers, if there are any, for the function \[
\displaystyle f(x)=\ln \sqrt{x-2}.
\]

*Solution*. The function \(f\) has derivative, \[
f'(x)=\frac{1}{2(-2+x)}
\] and so we need to check for any \(x\) in the domain of \(f\) such that \(f'(x)=0\) or \(f'(x)\) is undefined. We see that \(f'\) is undefined for \(x=2;\) and there are no real numbers with \(f'(x)=0.\) One might be tempted to say that there is one critical numbers of \(x=2.\) However, \(x=2\) is not a critical number because \(x=2\) is not in the domain of \(f(x)=\ln \sqrt{x-2}.\)

## 3.4 Absolute Extrema

Now we turn our attention to absolute extrema which take into account the whole domain of a given \(f\) and not just an open interval in the domain as do relative extrema.

Absolute extrema are defined and a procedure for finding absolute extrema on a given closed bounded interval is given. We give a couple of examples to illustrate this procedure.

**Definition 3.3 **Let \(f\) be a function with domain \(D.\)

- If \(f(c)\geq f(x)\) for all \(x\) in \(D,\) then \(f\) has an
**absolute maximum**at \(c.\) - If \(f(c)\leq f(x)\) for all \(x\) in \(D,\) then \(f\) has an
**absolute minimum**at \(c.\)

- If \(f\) has either an absolute maximum or a absolute minimum at \(c\), then we say \(f\) has an
**absolute extrema**at \(c\), and we say that \(f(c)\) is an**extreme value**.

Sometimes an absolute maximum is called a **global maximum** . Sometimes an absolute minimum is called a **global minimum** .

**Example 3.10 **State whether linear functions, quadratic functions, and the six trigonometric functions have absolute extrema on their domains.

*Solution*. Linear functions \(f(x)=a x+b\) do not have absolute extrema on their natural domain of \(\mathbb{R}\) unless in the trivial case of \(a=0.\) Quadratic functions \(f(x)=a x^2+b x+c\) have an absolute maximum or absolute minimum depending on the sign of \(a.\) The vertex is given by \[
\left(\frac{-b}{2 a},f\left(\frac{-b}{2a}\right)\right)
\] and is always an absolute maximum (if \(a>0\)) or absolute minimum (if \(a<0\)). The trigonometric functions sine and cosine have global maximum and global minimum; and since these functions are periodic they attain these values periodically. However, the functions secant, cosecant, tangent and cotangent do not have any absolute maximum and absolute minimum values on their natural domain. The reader should verify the last statement with either a graph of each of these functions or by analyzing these four functions in terms of sine and cosine.

::: {#thm- } [Extreme Value Theorem] If \(f\) is a continuous function on a closed bounded interval \([a,b],\) then \(f\) must attain an absolute maximum value \(f(s)\) and an absolute minimum value \(f(t)\) at some numbers \(s\) and \(t\) in \([a,b].\) :::

The extreme value theorem is an existence theorem because the theorem tells of the existence of maximum and minimum values but does not show how to find it.

**Example 3.11 **State whether the function has absolute extrema on its domain \[
f(x)=
\begin{cases}
-x^2 & x\neq 0 \\
-1 & x=0
\end{cases}
\]

*Solution*. The function \(f\) does not have a global maximum since it takes on all values less than, but arbitrarily close to 0. However, it never reaches the value of \(0.\) Notice that this function is not continuous on a closed bounded interval containing 0 and so the Extreme Value Theorem does not apply.

**Example 3.12 **State whether the function has absolute extrema on its domain \[
g(x)=
\begin{cases}
x & x> 0 \\
3 & x\leq 0
\end{cases}
\]

*Solution*. The function \(g\) does not have a global minimum since it takes on all values greater than, but arbitrarily close to 0. However, it never reaches the value of \(0.\) Notice that this function is not continuous on a closed bounded interval containing 0 and so the Extreme Value Theorem does not apply.

**Theorem 3.2 **Let \(f\) be a continuous function whose domain contains \([a,b].\) Then to find the absolute extrema of \(f\) on \([a,b]\) perform the following steps.

- Find all critical numbers of \(f\) on \([a,b]\) and evaluate \(f\) for each of these numbers.
- Evaluate \(f\) at the boundary; that is, find \(f(a)\) and \(f(b).\)
- Select the largest and smallest values those listed in (1) and (2).

The largest value is the absolute maximum and the smallest value is the absolute minimum.

**Example 3.13 **Find the absolute extrema of \(f(x)=x^4-4x^2+2\) on the interval \([-3,2]\).

*Solution*. The function \(f\) has derivative, \(f'(x)=4 x^3-8 x\). Since \(f'(x)=0\) only when \(x=0, \pm \sqrt{2}\) and \(f'\) is defined for all \(x\), the only critical numbers are \(x=0, \pm \sqrt{2}\) (see Figure ~\(\ref{fig:graph1}\)). The following table determines the absolute extrema of the function \(f\). \[
\begin{array}{c|c|c|l}
x & f(x) & f'(x) & \text{Conclusion} \\ \hline
-3 & 47 & - & \text{boundary, absolute maximum} \\
-\sqrt{2} & -2 & 0 & \text{critical number, absolute minimum} \\
0 & 2 & 0 & \text{critical number} \\
\sqrt{2} & -2 & 0 & \text{critical number, absolute minimum} \\
2 & 2 & - & \text{boundary}
\end{array}
\]

**Example 3.14 **Find the absolute extrema of \(f(x)=x^{4/5}\) on the interval \([-32,1]\).

*Solution*. The function \(f\) has derivative, \(f'(x)=4/(5 x^{1/5})\). Since \(f'(x)\neq0\) for every \(x\) in the domain of \(f\) and \(f'(x)\) is undefined only when \(x=0\), the only critical numbers are \(x=0\) (see Figure ~\(\ref{fig:graph2}\)). The following table determines the absolute extrema of the function \(f\). \[
\begin{array}{c|c|c|l}
x & f(x) & f'(x) & \text{Conclusion} \\ \hline
-32 & 16 & - & \text{boundary, absolute maximum} \\
0 & 0 & 0 & \text{critical number, absolute minimum} \\
1 & 1 & - & \text{boundary}
\end{array}
\]

**Example 3.15 **Find the absolute extrema of \(\displaystyle f(x)=\frac{x}{x+1}\) on the interval \([1,2]\).

*Solution*. The function \(f\) has derivative, \(f'(x)=\frac{1}{(1+x)^2}\) and so there are no critical numbers because \(x=-1\) is not in our domain of \([1,2]\) (see Figure ~\(\ref{fig:graph3}\)) and \(f'\) is defined for all \(x\) in \([1,2]\). The following table determines the absolute extrema of the function \(f\). \[
\begin{array}{c|c|c|l}
x & f(x) & f'(x) & \text{Conclusion} \\ \hline
1 & 1/2 & - & \text{absolute minimum} \\
2 & 2/3 & - & \text{absolute maximum}
\end{array}
\]

**Example 3.16 **Find the absolute extrema of \(f(x)=\sin x + \cos x\) on the interval \([0,\pi/3]\).

*Solution*. The function \(f\) has derivative, \(f'(x)=\cos x-\sin x\) and so to find our critical number we need to solve \(\sin x=\cos x\) or \(\tan x=1\) which is of course \(x=\frac{\pi }{4}+\pi k\) where \(k\) is any integer. However, for our given domain of \([0,\pi /3]\) the only critical number is \(\pi /4\) (see Figure ~\(\ref{fig:graph4}\)). The following table determines the absolute extrema of the function \(f\). \[
\begin{array}{c|c|c|l}
x & f(x) & f'(x) & \text{Conclusion} \\ \hline
0 & 1 & - & \text{boundary, absolute minimum} \\
\pi /4 & \sqrt{2} & 0 & \text{critical number, absolute maximum} \\
\pi /3 & 1 & - & \text{boundary}
\end{array}
\]

## 3.5 Exercises

**Exercise 3.1 **Find the critical numbers of the given functions.

- \(f(x)=x^2(2x-1)^{2/3}\)
- \(g(x)=\frac{x+1}{x^2+x+1}\)
- \(h(x)=\sqrt[3]{x^2-x}\)

**Exercise 3.2 **Find the absolute maximum and minimum values of the function \(f(x)=x^2-1\) on the interval \(-1\leq x\leq 2\). Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.

**Exercise 3.3 **Find the absolute maximum and minimum values of the function \(f(x)=\frac{-1}{x}\) on the interval \(-2\leq x\leq -1\). Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.

**Exercise 3.4 **Find the absolute maximum and minimum values of the function \(f(x)=\tan x\) on the interval \(\frac{-\pi }{3}\leq x\leq \frac{\pi}{4}\). Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.

**Exercise 3.5 **Find the absolute maximum and minimum values of the function \(f(x)=2-|x|\) on the interval \(-1\leq x\leq 3\). Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.

**Exercise 3.6 **Find the absolute maximum and minimum values of the function \(f(x)=x e^{-x}\) on the interval \(-1\leq x\leq 1\). Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.

**Exercise 3.7 **Find the absolute maximum and minimum values of the function \(f(x)=x^{4/3}\) on the interval \(-1\leq x\leq 8\). Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.

**Exercise 3.8 **Find the absolute maximum and minimum values of the function \(f(x)=3 x^{2/3}\) on the interval \(-27\leq x\leq 8\). Then graph the function. Identify the points on the graph where the absolute extrema occur, and include their coordinates.

**Exercise 3.9 **Find the absolute maximum and absolute minimum values of the given function on the given interval.

- \(h(x)=\frac{x}{x+1}\text{ on }[1,2]\)
- \(i(x)=\cos ^{-1}x \tan ^{-1}x\text{ on }[0,1]\)
- \(j(x)=e^{-x}(\cos x+\sin x)\text{ on }[0,4\pi ]\)
- \(k(x)=\sqrt[3]{x}\sqrt[3]{(x-3)^2}\text{ on }[-1,4]\)

**Exercise 3.10 **Find the extreme values of the function \(y=x^3+x^2-8x+5\) and where they occur.

**Exercise 3.11 **Find the extreme values of the function \(y=\frac{1}{\sqrt{1-x^2}}\) and where they occur.

**Exercise 3.12 **Find the extreme values of the function \(y=\frac{x}{x^2+1}\) and where they occur.

**Exercise 3.13 **Find the extreme values of the function \(y=e^x-e^{-x}\) and where they occur.

**Exercise 3.14 **Find the extreme values of the function \(y=\cos ^{-1} \left(x^2\right)\) and where they occur.

**Exercise 3.15 **Find the derivative at each critical point and determine the local extreme values for the function \(y=x^{2/3}\left(x^2-4\right).\)

**Exercise 3.16 **Find the derivative at each critical point and determine the local extreme values for the function \[y=
\begin{cases}
4-2x & x\leq 1 \\
x+1 & x>1
\end{cases}
\]

**Exercise 3.17 **Find the derivative at each critical point and determine the local extreme values for the function \[y=
\begin{cases}
\frac{-1}{4}x^2-\frac{1}{2}x+\frac{15}{4} & x\leq 1 \\
x^3-6x^2+8x & x>1
\end{cases}
\]

**Exercise 3.18 **Let \(f(x)=\left|x^3-9x\right|.\) Does \(f'(0)\) exist? Does \(f'(3)\) exist? Does \(f'(-3)\) exist? Determine all extrema of \(f.\)

**Exercise 3.19 **What is the largest possible area for a right triangle whose hypotenuse is \(5\text{cm}\) long?

**Exercise 3.20 **The height of a body moving vertically is given by \(s=\frac{-1}{2}g t^2+v_0t+s_0,\) \(g>0\) with \(s\) in meters and \(t\) in seconds. Find the body’s maximum height.

**Exercise 3.21 **Show that \(5\) is a critical number of the function \(g(x)=2+(x-5)^3\) but \(g\) does not have a relative extremum at \(x=5.\)

**Exercise 3.22 **Consider the cubic function \(f(x)=a x^3+b x^2+c x+d\) where \(a\neq 0.\) Show that \(f\) can have zero, one, or two critical numbers and give examples of each.

**Exercise 3.23 **Explain why the function \(f(x)=\frac{8}{\sin x}+\frac{27}{\cos x}\) must attain a minimum in the open interval \(\left(0,\frac{\pi }{2}\right).\)

## 3.6 Mean Value Theorem

## 3.7 Rolle’s Theorem

The Extreme Value Theorem guarantees the existence of a maximum and minimum value of a continuous function on a closed bounded interval. The next theorem is called Rolle’s Theorem and it guarantees the existence of an extreme value on the interior of a closed interval, under certain conditions. Basically Rolle’s theorem states that if a function is differentiable on an open interval, continuous at the endpoints, and if the function values are equal at the endpoints, then it has at least one horizontal tangent. Of course if the function is constant this is automatically true for all points in the interval. So the point is that Rolle’s theorem guarantees us at least one point in the interval where there will be a horizontal tangent. Rolle’s theorem is a special case of the Mean Value Theorem for when the values of the function are the same at the endpoints of the interval.

::: {#thm- } Rolle’s Theorem Let \(f\) be a function that is continuous on \([a,b]\), differentiable on \((a,b)\), and \(f(a)=f(b).\) Then there exists at least one number \(c\) in \((a,b)\) such that \(f'(c)=0.\) :::

*Proof*. If \(f\) is a constant function, then the statement is true; in fact \(f'(c)=0\) for all \(c\) in \((a,b).\) If \(f(x)>f(a)=f(b)\) for some \(x\) in \((a,b),\) then by the Extreme Value Theorem, \(f\) attains its absolute maximum value somewhere in the open interval \((a,b).\) But precisely at this \(c\) we have, \(f'(c)=0.\) If \(f(x)<f(a)=f(b)\) for some \(x\) in \((a,b),\) then by the Extreme Value Theorem, \(f\) attains its absolute minimum value somewhere in the open interval \((a,b).\) But precisely at this \(c\) we have, \(f'(c)=0.\)

**Example 3.17 **Verify Rolle’s theorem for \(f(x)=4-4 x-x^2+x^3\) on \([-2,2].\)

*Solution*. Notice that \(f\) is continuous and differentiable for all real numbers. Also, \(f(-2)\) \(=\text{f(2)}\) \(=0\) and therefore Rolle’s theorem applies and so there is at least one \(c\) in \((-2,2)\) such that \(f'(c)=0.\) We can find it by solving \(f'(c)=-4-2 c+3 c^2=0.\) In fact we find two, namely \(c=\frac{1}{3} \left(1\pm \sqrt{13}\right)\) (see Figure ~\(\ref{fig:graph5}\)).

**Example 3.18 **Show that the equation \(x^3+x-1=0\) has exactly one real root.

*Solution*. Since the function \(f(x)=x^3+x-1\) is a polynomial it is continuous and differentiable for all real numbers. Thus, the Intermediate Value Theorem and Rolle’s Theorem applies. Since \(f(0)=-1<0\) and \(f(1)=1>0,\) by the Intermediate Value Theorem there is a \(c\) in \((0,1)\) such that \(f(x)=0.\) Therefore, the equation has at least one solution. To prove that \(f(x)=0\) for only one \(x,\) we assume that there are two roots namely, \(x_1\) and \(x_2\); and we prove that this can not happen. Thus, assume \(x_1\) and \(x_2\) are solutions, that is \(f\left(x_1\right)=f\left(x_2\right)=0\) with \(x_1<x_2.\) Then by Rolle’s Theorem there exists a \(c\) in \(\left(x_1,x_2\right)\) such that \(f'(c)=0.\) Notice that \(f'(c)=3x^2+1>0\) so that in fact such a \(c\) can not exist. Therefore, there can not be \(x_1<x_2\) and in fact the equation has exactly one real root (see Figure ~\(\ref{fig:graph6}\)).

## 3.8 The Mean Value Theorem

Given a function that is differentiable on an open interval and continuous at the endpoints the Mean Value Theorem states there exists a number in the open interval where the slope of the tangent line at this point on the graph is the same as the slope of the line through the two points on the graph determined by the endpoints of the interval. The **mean** in the Mean Value Theorem is referring to the mean (average) rate of change of \(f\) in the interval.

Next we detail Rolle’s Theorem and the Mean Value Theorem. We provide examples and illustrate why the hypotheses of these two theorems are necessary. We also give applications and detail two other theorems which are consequences of the Mean Value Theorem. We also emphasize that the Mean Value Theorem tells us that between two fixed points of time, the instantaneous velocity is equal to the average velocity.

::: {#thm- } Mean Value Theorem Let \(f\) be a function that is continuous on \([a,b]\) and differentiable on \((a,b).\) Then there exists at least one number \(c\) in \((a,b)\) such that \[ f'(c)=\frac{f(b)-f(a)}{b-a}. \] :::

*Proof*. The equation of the secant line through \((a,f(a))\) and \((b,f(b))\) is \[
y=\left[\frac{f(b)-f(a)}{b-a}\right](x-a)+f(a).
\] Let \(g(x)\) be the difference between \(f(x)\) and \(y.\) Then \[
g(x)=f(x)-\left[\frac{f(b)-f(a)}{b-a}\right](x-a)-f(a).
\] We can see that \(g(a)=g(b)=0.\) Because \(f\) is continuous on \([a,b]\) and differentiable on \((a,b),\) so is \(g.\) By Rolle’s Theorem, there exists a number \(c\) in \((a,b)\) such that \(g'(c)=0,\) which means \[
g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}=0
\] and so \(f'(c)=\frac{f(b)-f(a)}{b-a}\) as desired.

**Example 3.19 **Find all numbers \(c\) in the interval \([a,b]\) such that \[
f'(c)=\frac{f(b)-f(a)}{b-a}
\] for the function \(\displaystyle f(x)=1+\frac{1}{x}\) on \([1,4]\).

*Solution*. Since \(f\) is continuous on \([0,1]\) and differentiable on \((0,1)\) we apply the Mean Value Theorem, \[
f'(c)=-\frac{1}{c^2}=\frac{f(b)-f(a)}{b-a}=\frac{\left(1+\frac{1}{4}\right)-\left(1+\frac{1}{1}\right)}{4-1}=-0.25
\] Solving for \(c\) we obtain, \(c=2\) since \(-2\) is not in \([1,4]\) (see Figure ~\(\ref{fig:graph7}\)).

**Example 3.20 **Find all numbers \(c\) in the interval \([a,b]\) such that \[
f'(c)=\frac{f(b)-f(a)}{b-a}
\] for the function \(f(x)=\tan ^{-1}x\) on \([0,1]\).

*Solution*. Since \(f\) is continuous on \([0,1]\) and differentiable on \((0,1)\) we apply the Mean Value Theorem, \[f'(c)=\frac{1}{1+c^2}=\frac{f(b)-f(a)}{b-a}=\frac{\left(\tan^{-1}1\right)-\left(\tan ^{-1}0\right)}{1-0}=\frac{\pi }{4}\] Solving for \(c\) we obtain, \(c=\sqrt{\frac{4-\pi }{\pi }}\) since \(-\sqrt{\frac{4-\pi }{\pi }}\) is not in \([0,1]\) (see Figure ~\(\ref{fig:graph8}\)).

## 3.9 Using the Mean Value Theorem

**Example 3.21 **Using the Mean Value Theorem evaluate, \[
\lim _{x\to 0}\frac{(1+x)^n-1}{x}
\] where \(n\) is a natural number.

*Solution*. We will use the Mean Value Theorem to find all numbers \(c\) in the interval \([a,b]\) such that \[
f'(c)=\frac{f(b)-f(a)}{b-a}
\] for the function \(f(x)=(1+x)^n\) on \([0,x]\). Since \(f\) is continuous and differentiable on \([0,x]\), we apply the Mean Value Theorem, \[
f'(c)=n(1+c)^{n-1}=\frac{f(b)-f(a)}{b-a}=\frac{(1+x)^n-(1+0)^n}{x-0}=\frac{(1+x)^n-1}{x}
\] Since \(f\) is continuous on \([0,x]\) and differentiable on \((0,x)\) we know this \(c\) must exist. In fact since \(0<c<x\), as \(x\to 0\) we see that \(c\to 0\) because the Mean Value Theorem says that \(c\) is in the open interval \((0,x).\) Thus we can evaluate the limit \[
\lim _{x\to 0}\frac{(1+x)^n-1}{x}=\lim _{c\to 0}n(1+c)^{n-1}=n
\] as desired.

If an object moves in a straight line with position function \(s=f(t),\) then the average velocity between \(t=a\) and \(t=b\) is \[ \frac{f(b)-f(a)}{b-a} \] and the velocity at \(t=c\) is \(f'(c).\) Thus the Mean Value Theorem tells us that at some time \(t=c\) between \(a\) and \(b\) the instantaneous velocity \(f'(c)\) is equal to that average velocity.

**Example 3.22 **Two stationary patrol cars equipped with radar are 1.2 miles apart on a street. As a truck passes the first patrol car, its speed is clocked at 35 miles per hour. One and half minutes later, when the truck passes the second patrol car, its speed is clocked at 30 miles per hour. Prove that the truck must have exceeded the speed limit (of 35 miles per hour) at some time during the one and half minutes.

*Solution*. Let \(t=0\) be the time when the truck passes the first patrol car. The time when it passes the second patrol car is \(1.5/60\) hour. By letting \(s(t)\) represent the distance (in miles) travelled by the truck, we have \(s(0)=0\) and \(s(1.5/60)=1.2.\) So the average velocity is \[
\frac{s\left(\frac{1.5}{60}\right)-s(0)}{\frac{1.5}{60}-0}=\frac{1.2}{\frac{1.5}{60}}=48.0 \text{ mph}
\] Assuming that the position function is differentiable, we can apply the Mean Value Theorem to conclude that the truck must have been traveling at a rate of 48 miles per hour sometime during the one and half minutes.

## 3.10 Exercises

**Exercise 3.24 **Determine the values of the constants \(a,b,c,\) and \(d\) such that the functions \[f(x)=\left\{
\begin{array}{ll}
1 & x=0 \\
a x+b & 0<x\leq 1 \\
x^2+4x+c & 1<x\leq 3
\end{array}
\right.\] \[g(x)=\left\{
\begin{array}{ll}
a & x=-1 \\
2 & 0<x\leq 0 \\
b x^2+c & 0<x\leq 1 \\
d x+4 & 1<x\leq 2
\end{array}
\right.\] satisfies the hypotheses of the Mean Value Theorem on the intervals \([0,3]\) and \([-1,2]\) respectively.

**Exercise 3.25 **Find the value(s) of \(c\) that satisfy the equation \(\frac{f(b)-f(a)}{b-a}=f'(c)\) for \(f(x) =x^{2/3}\) in the conclusion of the Mean Value Theorem on \([0,1].\)

**Exercise 3.26 **Find the value(s) of \(c\) that satisfy the equation \(\frac{f(b)-f(a)}{b-a}=f'(c)\) for \(f(x) =\ln (x-1)\) in the conclusion of the Mean Value Theorem on \([2,4].\)

**Exercise 3.27 **Does the function \(f(x) =x^{2/3}\) satisfy the hypotheses of the Mean Value Theorem on the interval \([-1,8]?\) State why or why not.

**Exercise 3.28 **Does the function \[
f(x) =\left\{
\begin{array}{cc}
\frac{\sin x}{x} & -\pi \leq x\leq 0 \\
0 & x=1
\end{array}
\right.
\] satisfy the hypotheses of the Mean Value Theorem on the interval \([-1,8]?\) State why or why not.

**Exercise 3.29 **Assume \(a_1\neq 0.\) Let \(f(x)=a_1x^2+a_2 x+a_3.\) Prove that for any interval \([a,b]\) the value of \(c\) guaranteed by the Mean Value Theorem for \(f\) is the midpoint of the interval.

**Exercise 3.30 **Show that a cubic can have at most three zeros.

**Exercise 3.31 **Show that the function \(g(t)=\sqrt{t}+\sqrt{1+t}-4\) has exactly one zero in the interval \((0,+\infty).\)

**Exercise 3.32 **Show that the function \(g(t)=2 t-\cos ^2t+\sqrt{2}\) has exactly one zero in the interval \((-\infty ,+\infty ).\)

**Exercise 3.33 **Suppose that \(f(-1)=3\) and that \(f'(x)=0\) for all \(x.\) Must \(f(x)=3\) for all \(x?\) Why or why not?

**Exercise 3.34 **Suppose that \(f'(x)=2x\) for all \(x.\) Find \(f(2)=3\) if \(f(-2)=3.\)

**Exercise 3.35 **Show that the equations \(x^5+10x+3=0\) and \(x^7+5x^3+x-6=0\) each have exactly one real root.

**Exercise 3.36 **Use the Mean Value Theorem to show the following \[\lim _{x\to \pi }\frac{\cos x+1}{x-\pi }=0\] and \[\frac{1}{2x+1}>\frac{1}{5}+\frac{2}{25}(2-x)\] when \(0<x<2.\)

**Exercise 3.37 **Let \(f(x)=\frac{1}{x}\) and \[
g(x)=\left\{
\begin{array}{cc}
\frac{1}{x} & x>0 \\
1+\frac{1}{x} & x<0
\end{array}
\right.
\] Show that \(f'(x)=g'(x)\) for all \(x\) in their domains. Can we conclude that \(f-g\) is constant?

## 3.11 Monotonic Functions

## 3.12 Increasing and Decreasing Functions

To determine where a function \(f\) is increasing or decreasing, we begin by finding the critical numbers. These numbers divide the \(x\)-axis into intervals, and we test the sign of \(f'(x)\) in each of these intervals. This procedure is often called the first derivative test and can be used to determine local extrema and intervals of monotonicity.

**Definition 3.4 **A function \(f\) is called **increasing** on an interval \(I\) if \(f\left(x_1\right)<f\left(x_2\right)\) whenever \(x_1<x_2\) in \(I.\) A function \(f\) is called **decreasing** on an interval \(I\) if \(f\left(x_1\right)>f\left(x_2\right)\) whenever \(x_1<x_2\) in \(I.\)

**Definition 3.5 **A function \(f\) is called **monotonic** on an interval \(I\) if it is either increasing or decreasing on \(I.\)

**Theorem 3.3 **Suppose \(f\) is continuous on \([a,b]\) and differentiable on \((a,b).\)

- If \(f'(x)>0\) for all \(x\) in \((a,b),\) then \(f\) is increasing on \((a,b).\)
- If \(f'(x)<0\) for all \(x\) in \((a,b),\) then \(f\) is decreasing on \((a,b).\)

**Example 3.23 **Find where the function \(f(x)=3x^4-4x^3-12x^2+5\) is increasing and decreasing; that is determine the intervals where \(f\) is monotonic.

*Solution*. The derivative of \(f\) is \[
f'(x)=12x^3-12x^2-24x=12x(x-2)(x+1).
\] Since \(f\) is continuous and differentiable, to test where the function is monotonic we divide the \(x\)-axis according to the sign of \(f'(x)\) which depending on the signs of \(12x,\) \(x-2,\) and \(x+1.\) We put our results into the following table: \[
\begin{array}{c|c|c|c|c|l}
\text{Interval} & 12x & x-2 & x+1 & f'(x)\text{
} & f \\ \hline
x<-1 & - & - & - & - & \text{decreasing on } (-\infty ,-1) \\
-1<x<0 & - & - & + & + & \text{increasing on } (-1,0) \\
0<x<2 & + & - & + & - & \text{decreasing on } (0,2) \\
x>2 & + & + & + & + & \text{increasing on } (2,\infty )
\end{array}
\] (see Figure ~\(\ref{fig:graph9}\)). :::

## The First Derivative Test

::: {#thm- } [First Derivative Test] Suppose that \(c\) is a critical number of a function that is continuous on \([a,b].\) Then the following statements hold:

- If \(f'(x)>0\) for \(a<x<c\) and \(f'(x)<0\) for \(c<x<b,\) then \(f\) has a relative (local) minimum at \(c.\)
- If \(f'(x)<0\) for \(a<x<c\) and \(f'(x)>0\) for \(c<x<b,\) then \(f\) has a relative (local) maximum at \(c.\)
- If neither hold then \(f\) has no relative (local) extremum at \(c.\)

**Example 3.24 **Apply the First Derivative Test to find the local extrema of the function \(f(x)=x(1-x)^{2/5}\) and sketch its graph.

*Solution*. First we find the critical numbers of \(f\) by solving \(f'(x)=0\) and determining where \(f'(x)\) is undefined but \(f(x)\) is defined. The derivative of \(f\) is \[
f'(x)=(1-x)^{2/5}+\frac{2x}{5}(1-x)^{-3/5}(-1)=\frac{5(1-x)-2x}{5(1-x)^{3/5}}=\frac{5-7x}{5(1-x)^{3/5}}.
\] Solving \(f'(x)=0\) we find \(x=5/7.\) Also \(f'(1)\) does not exist but \(f(1)=0;\) and therefore the only critical numbers of \(f\) are \(x=5/7\) and \(x=0.\) We determine the local extrema using the following table \[
\begin{array}{c|c|c|c|l}
\text{Interval} & 5-7x & (1-x)^{3/5} & f'(x) & f(x) \\ \hline
x<\frac{5}{7} & + & + & + & \text{ increasing on} \left(-\infty ,\frac{5}{7}\right) \\
\frac{5}{7}<x<1 & - & + & - & \text{ decreasing on} \left(\frac{5}{7},1\right) \\
x>1 & - & - & + & \text{ increasing on }(1,+\infty )
\end{array}
\] \ Therefore, \(\left(\frac{5}{7},f\left(\frac{5}{7}\right)\right)\) is a local maximum and \((1,f(1))\) is a local minimum. Here is the graph of the function \(f(x)=x(1-x)^{2/5}.\) Notice there is a corner at \((1,f(1))\) because \(f\) is defined there but \(f'\) is not (see Figure ~\(\ref{fig:graph10}\)).

**Example 3.25 **Find the local and absolute extrema values of the function \(f(x)=x^3(x-2)^2\) on the interval \(-1\leq x\leq 3.\) Sketch the graph.

*Solution*. First we notice that \(f\) is a polynomial and so is continuous and differentiable for all real numbers. Next we find the critical numbers of \(f\) by solving \(f'(x)=0\) and determining where \(f'(x)\) is undefined, but \(f(x)\) is defined. The derivative of \(f\) is \[
f'(x)=3x^2(x-2)^2+2x^3(x-2)=x^2(x-2)(5x-6).
\] To find the critical numbers we set \(f'(x)=0\) and obtain \(x=0,2,6/5.\) We determine the local extrema and absolute extrema using the following table: \[
\begin{array}{c|c|c|c|c|l}
\text{Interval} & x^2 & x-2 & 5x-6 & f'(x)\text{ } & f \\ \hline
-1<x<0 & + & - & - & + & \text{ increasing on} (-1,0) \\
0<x<\frac{6}{5} & + & - & - & + & \text{ increasing on } \left(0,\frac{6}{5}\right) \\
\frac{6}{5}<x<2 & + & - & + & - & \text{ decreasing on }\left(\frac{6}{5},2\right) \\
2<x<3 & + & + & + & + & \text{ increasing on } (2,3)
\end{array}
\] \ The function \(f\) does not have a local extrema at \(x=0.\) The local maximum is \[
\left(\frac{6}{5},f\left(\frac{6}{5}\right)\right)
\] and the local minimum is \((2,f(2)).\)

Since \(f\) is a continuous function we can use the extreme value theorem to determine absolute extrema. We compute the functional values at the endpoints, namely \(f(-1)=-9\) and \(f(3)=27.\) Therefore, the absolute maximum is \(f(3)=27\) and the absolute minimum is \(f(-1)=-9\) (see Figure ~\(\ref{fig:graph11}\)).

## 3.13 Exercises

**Exercise 3.38 **For each of the following functions determine the critical points and apply the first derivative test to determine the intervals where the function is increasing or decreasing, and all local extrema.

- \(f(x)=(x-1)^2(x+2)\)
- \(f(x)=(x-1)e^{-x}\)
- \(f'(x)=x^{-1/3}(x+2)\)
- \(f(\theta )=3\theta ^2-4\theta ^3\)
- \(h(r)=(r+7)^3\)
- \(g(x)=x^2\sqrt{5-x}\)
- \(f(x)=\frac{x^3}{3x^2+1}\)
- \(h(x)=x^{1/3}\left(x^2-4\right)\)
- \(f(x)=e^{2x}+e^{-x}\)
- \(f(x)=x \ln x\)
- \(f(x)=(x+1)^2\)
- \(f(x)=-x^2-6x-9\)
- \(k(x)=x^3+3x^2+3x+1\)

**Exercise 3.39 **Sketch the graph of a differentiable function \(y=f(x)\) through the point \((1,1)\) such that

- \(f'(1)=0\)
- \(f'(x)>0\) for \(x<1\)
- \(f'(x)<0\) for \(x>1\)
- \(f'(x)>0\) for \(x\neq 1\)
- \(f'(x)<0\) for \(x\neq 1\)

**Exercise 3.40 **Sketch the graph of a differentiable function \(y=f(x)\) that satisfies all of the following conditions.

- a local minimum at \((1,1)\)
- a local maximum at \((3,3)\)
- local maximum at \((1,1)\)
- a local minimum at \((3,3)\)

**Exercise 3.41 **Sketch the graph of a differentiable function \(y=h(x)\) that satisfies all of the following conditions.

- \(h(0)=0\)
- for all \(x\), \(-2\leq h(x)\leq 2\)
- $h’(x)+$ as \(x\to 0^-\)
- $h’(x)+$ as \(x\to 0^+\)

## 3.14 Concavity and Inflection Points

## 3.15 First and Second Order Critical Points

**Definition 3.6 **Suppose \(c\) is in the domain of a function \(f\). We will call \(c\) a **first-order critical number** of \(f\) when \(f'(c)=0\) or \(f'(c)\) does not exist and a **second-order critical number** of \(f\) when \(f''(c)=0\) or \(f''(c)\) does not exist.

## 3.16 Concavity and Inflection Points

**Definition 3.7 **Suppose a function \(f\) is differentiable on an open interval \(I.\)

- If \(f'\) is increasing on \(I,\) then the graph of \(f\) is called
**concave upward**on \(I.\) - If \(f'\) is decreasing on \(I,\) then the graph of \(f\) is called
**concave downward**on \(I.\)

If the graph of \(f\) lies above all of its tangents on an interval \(I,\) it is concave upward on \(I.\) If the graph of \(f\) lies below all of its tangents, it is concave downward on \(I.\)

**Definition 3.8 **A point \(P(c,f(c))\) on a curve is called an **inflection point** of the graph of \(f\) provided \(f\) has a tangent line at \(c\) and the concavity of \(f\) changes at \(x=c\).

## 3.17 Concavity Test

::: {#thm- } Concavity Test Suppose a function \(f\) is twice differentiable on an interval \(I.\)

- If \(f''(x)>0\) for all \(x\) in \(I,\) then the graph of \(f\) is concave upward on \(I.\)
- If \(f''(x)<0\) for all \(x\) in \(I,\) then the graph of \(f\) is concave downward on \(I.\) :::

**Example 3.26 **Determine where the curve \(y=x^4-4x^3\) is concave upward, where it is concave downward, and where the points of inflection are.

*Solution*. The first and second derivatives of the function \(f\) are \[f'(x)=4x^3-12x^2=4x^2(x-3)\] and \[f''(x)=12x^2-24x=12x(x-2).\] Since \(f''\) is defined for all real numbers and \(f''(x)=0\) only when \(x=0\) and \(x=2\) the second-order critical numbers of \(f\) are \(x=0\) and \(x=2\). We summarize the Concavity Test in the following table: \[
\begin{array}{c|c|c|l}
\text{Interval} & f & f'' & \text{Conclusion} \\ \hline
x<0 & & + & \text{concave up} \\
x=0 & 0 & 0 & \text{inflection point} \\
0<x<2 & & - & \text{concave down} \\
x=2 & -16 & 0 & \text{inflection point} \\
x>2 & & + & \text{concave up}
\end{array}
\] Therefore, \(f\) is concave up on \((-\infty ,0)\cup (2,+\infty )\) and concave down on \((0,2).\) The points \((0,0)\) and \((2,-16)\) are inflection points (see Figure ~\(\ref{fig:graph12}\)).

**Example 3.27 **Determine where the curve \(y=x^3-3x+1\) is concave upward and where it is concave downward. Find all inflection points, local extrema, and sketch the curve.

*Solution*. The first and second derivatives of the function \(f\) are \[f'(x)=3x^2-3=3(x-1)(x+1)\] and \(f''(x)=6x.\) Since \(f'\) and \(f''\) are polynomials they are both defined for all real numbers and since \(f'(x)=0\) and \(f''(x)=0\) only when \(x=\pm 1\) and \(x=0\), respectively, the first-order critical numbers are \(x=\pm 1\) and the second-order critical number is \(x=0\). We summarize the First Derivative Test and the Concavity Test in the following table. \[
\begin{array}{c|c|c|c|l}
\text{Interval} & f & f' & f'' & \text{Conclusion} \\ \hline
x<-1 & & + & & \text{increasing} \\
x=-1 & 3 & 0 & & \text{relative maximium} \\
-1<x<1 & & - & & \text{decreasing} \\
x=1 & -1 & 0 & & \text{relative minimum} \\
x>1 & & + & & \text{increasing} \\
x<0 & & & - & \text{concave down} \\
x=0 & 1 & & 0 & \text{inflection point} \\
x>0 & & & + & \text{concave up}
\end{array}
\] Therefore, the function \(f\) has a local maximum at \((3,0)\) and a local minimum at \((1,-1).\) The function \(f\) is increasing on \((-\infty ,-1)\cup (1,+\infty )\) and decreasing on \((-1,1).\) The point \((0,1)\) is an inflection point because \(f\) is concave up on the interval \((0,+\infty )\) and concave down on \((-\infty ,0).\)

(see Figure ~\(\ref{fig:graph13}\)).

## 3.18 Second Derivative Test

::: {#thm- } Second Derivative Test Suppose \(f''\) in continuous on an open interval that contains \(c\) with \(f'(c)=0.\)

- If \(f''(c)>0,\) then \(f\) has a relative (local) minimum at \(c.\)
- If \(f''(c)<0,\) then \(f\) has a relative (local) maximum at \(c.\) :::

The Second Derivative Test is inconclusive when both \(f'(c)=0\) and \(f''(c)=0.\) For example, if \(f(x)=x^3\) and \(g(x)=x^4,\) both \(f'(0)=g'(0)=0\) and \(f''(0)=g''(0)=0.\) The point \(x=0\) is a minimum for \(g\) but is neither a maximum nor a minimum for \(f.\) For some functions the Second Derivative Test might be a straightforward method for determining whether a point is a local extrema, however for some functions the First Derivative Test is a necessity.

**Example 3.28 **Use the Second Derivative Test to determine whether each first-order critical number of the function

\(f(x)=3x^5-5x^3+2\) corresponds to a relative maximum, a relative minimum, or neither.

*Solution*. Since \(f'(x)=15 x^4-15 x^2=15 (x-1) x^2 (x+1)\) is defined for all real numbers and \(f'(x)=0\) when \(x=0\) and \(x=\pm 1\) the first-order critical numbers are \(x=0\) and \(x=\pm 1\) . Since \[f''(x)=60 x^3-30 x=30 x \left(2 x^2-1\right)\] is continuous for all real numbers we can apply the Second Derivative Test. Since \(f''(-1)<0\) the point \((-1,4)\) is a local maximum and since \(f''(1)>0\) the point \((1,0)\) is a local minimum. Since \(f''(0)=0\) the Second Derivative Test is inconclusive at \(x=0\). (see Figure ~\(\ref{fig:graph14}\)).

**Example 3.29 **Use the Second Derivative Test to determine whether each critical number of the function \(f(x)= x+4/x\) corresponds to a relative maximum, a relative minimum, or neither.

*Solution*. The first derivative of the function \(f\) is \[f'(x)=1-\frac{4}{x^2}=\frac{x^2-4}{x^2}=\frac{(x-2)(x+2)}{x^2}.\] The first derivative is defined for all real numbers except \(x=0\) and \(f'(x)=0\) only when \(x=\pm 2\) the first-order critical numbers of $f $ are \(x= \pm 2\). Note that, even though \(f'(0)\) is undefined, so is \(f(0)\) and so \(0\) is not a critical number. The second derivative of the function \(f\) is \(f''(x)=\frac{4}{x^3}\) and since \(f''\) is continuous for all real numbers except \(x=0\) we can apply the Second Derivative Test. Since \(f''(2)>0\) the point \((2,4)\) is a local minimum and since \(f''(-2)<0\) the point \((-2,-4)\) is a local maximum (see Figure ~\(\ref{fig:graph15}\)).

**Example 3.30 **For the function \(f(x)=x^2e^{-3x}\) find all first-order and second-order critical numbers. Apply the First Derivative Test, Concavity Test, and the Second Derivative Test. Sketch the graph of the function.

*Solution*. The first and second derivatives of the function \(f\) are \[f'(x)=2x e^{-3x}-3x^2 e^{-3x}=e^{-3x}\left(2x-3x^2\right)=x(2-3x)e^{-3x}\] and \[f''(x)=e^{-3x}\left(9x^2-12x+2\right).\] Since both the first and second derivatives are continuous for all real numbers the first-order and second-order critical numbers will be found by solving \(f'(x)=0\) and \(f''(x)=0\), respectively. Solving \(f'(x)=0\) we find the first-order critical numbers to be \(x=0\) and \(x=\frac{2}{3}.\) Solving \(f''(x)=0\) we find the second-order critical numbers to be \(x=\frac{2}{3}\pm \frac{\sqrt{2}}{3}.\) Since \(f\) is continuous we can apply the First Derivative Test as follows: \[
\begin{array}{c|c|c|l}
\text{Interval} & f & f' & \text{Conclusion} \\ \hline
x<0 & & - & \text{decreasing} \\
x=0 & 0 & 0 & \text{relative minimum} \\
0<x<\frac{2}{3} & & + & \text{increasing} \\
x=\frac{2}{3} & \frac{4}{9 e^2} & 0 & \text{relative maximum} \\
x>\frac{2}{3} & & - & \text{decreasing}
\end{array}
\] Since \(f''\) is continuous we can apply the Second Derivative Test as follows: since \(f''(0)>0\) the point \((0,0)\) is a local minimum and since \(f''\left(\frac{2}{3}\right)<0\) the point \(\left(\frac{2}{3},f\left(\frac{2}{3}\right)\right)\) is a local maximum. Applying the Concavity Test: \[
\begin{array}{c|c|c|l}
\text{Interval} & f & f'' & \text{Conclusion} \\ \hline
x<\frac{2}{3}-\frac{\sqrt{2}}{3} & & + & \text{concave up} \\
x=\frac{2}{3}-\frac{\sqrt{2}}{3} & \frac{1}{9} \left(6-4 \sqrt{2}\right) e^{-2+\sqrt{2}} & 0 & \text{inflection point} \\
\frac{2}{3}-\frac{\sqrt{2}}{3}<x<\frac{2}{3}+\frac{\sqrt{2}}{3} & & - & \text{concave down} \\
x=\frac{2}{3}+\frac{\sqrt{2}}{3} & \frac{2}{9} \left(3+2 \sqrt{2}\right) e^{-2-\sqrt{2}} & 0 & \text{inflection point} \\
x>\frac{2}{3}+\frac{\sqrt{2}}{3} & & + & \text{concave up}
\end{array}
\] (see Figure ~\(\ref{fig:graph16}\)).

## 3.19 Exercises

**Exercise 3.42 **For each of the following functions identify the inflection points and local maxima and local minima. Identify the intervals on which the function is concave up and concave down. Sketch the graph showing these specific features.

- \(f(x)=\frac{x^4}{4}-2x^2+4\)
- \(f(x)=\frac{9}{14}x^{1/3}\left(x^2-7\right)\)
- \(f(x)=\tan x-4x\) on \(-\frac{\pi }{2}<x<\frac{\pi }{2}\)
- \(f(x)=2 \cos x-\sqrt{2}x\) on \(-\pi <x<\frac{3\pi }{2}\)
- \(f(x)=x^2-4x+3\)
- \(f(x)=x^4-2x^2\)
- \(f(x)=x-\sin x\) on $0x$
- \(f(x)=\sqrt{|x|}\)
- \(f(x)=e^x-2e^{-x}-3x\)
- \(f(x)=\frac{\ln x}{\sqrt{x}}\)

**Exercise 3.43 **Sketch a smooth curve \(y=f(x)\) with \(f(-2)=8,\) \(f(0)=4,\) \(f(2)=0,\) \(f'(x)>0\) for \(|x|>2,\) \(f'(2)=f'(-2)=0,\) \(f'(x)<0\) for \(|x|<2,\) \(f''(x)<0\) for \(x<0,\) and \(f''(x)>0\) for \(x>0.\)

**Exercise 3.44 **Sketch the graph of the following function \(f(x)=x^{2/3}(x-7) .\) Find all vertical and horizontal asymptotes of the graph of each function. Determine intervals of increasing and decreasing, determine concavity, and locate all critical points and points of inflection. Show all special features such as cusps or vertical tangents.

**Exercise 3.45 **Sketch the graph of the following function \[
h(x)=\frac{3x-2}{\sqrt{2x^2+1}}.
\] Find all vertical and horizontal asymptotes of the graph of each function. Determine intervals of increasing and decreasing, determine concavity, and locate all critical points and points of inflection. Show all special features such as cusps or vertical tangents.

**Exercise 3.46 **Find constants \(a\) and \(b\) that guarantee that the graph of the function defined by \[
f(x)=\frac{a x+5}{3-b x}
\] will have a vertical asymptote at \(x=5\) and a horizontal asymptote at \(y=-3.\) Sketch the graph of the function.

**Exercise 3.47 **Find all vertical tangents and cusps for the function \(f(x)=\sqrt{4-x^2}.\) Justify your work. Sketch the graph of the function.

**Exercise 3.48 **Find all vertical tangents and cusps for the function \[f(x)=\left\{
\begin{array}{cc}
x^{1/3}+3 & x\leq 0 \\
3-x^{1/5} & x\geq 0
\end{array}
\right..\] Justify your work. Sketch the graph of the function.

**Exercise 3.49 **Explain why the function \[
f(x)=\left\{
\begin{array}{cc}
1 & x\leq 0 \\
\frac{1}{x} & x>0
\end{array}
\right.
\] has a vertical asymptote but no vertical tangent. Sketch the graph of the function.

**Exercise 3.50 **Sketch the graph of the curve \[
y^2=\frac{x^3}{2a-x}
\] for [\(0,2a\)). Show all special features such as vertical asymptotes, horizontal asymptotes, cusps, vertical tangents, and intercepts. Sketch the graph of the function.

**Exercise 3.51 **Sketch the graph of the following functions. Find all vertical and horizontal asymptotes of the graph of each function. Determine intervals of increasing and decreasing, determine concavity, and locate all critical points and points of inflection. Show all special features such as cusps or vertical tangents.

- \(f(x)=x^{2/3}(x-7)\)
- \(g(x)=\frac{72-54 x+x^2+2 x^3}{15-17 x+4 x^2}\)
- \(h(x)=\frac{3x-2}{\sqrt{2x^2+1}}\)

## 3.20 Limits Involving Infinity

## 3.21 Infinite Limits

Infinite limits are used to described unbounded behavior of a function near a given real number which is not necessarily in the domain of the function. They are particularly useful for showing the intentions of the graph of a function by drawing dashed lines representing unbounded growth which are called vertical asymptotes.

**Definition 3.9 **Let \(f\) be a function defined on both sides of \(c,\) except possible at \(c\) itself. Then \(\lim_{x\to c}f(x)=+\infty\) means that the values of \(f(x)\) can be made arbitrarily large by taking \(x\) sufficiently close to \(c\) (\(x\neq c\) ).

**Definition 3.10 **Let \(f\) be a function defined on both sides of \(c,\) except possible at \(c\) itself. Then \(\lim_{x\to c}f(x)=-\infty\) means that the values of \(f(x)\) can be made arbitrarily large negative by taking \(x\) sufficiently close to \(c\) (\(x\neq c\) ).

**Example 3.31 **Determine \(\displaystyle \lim_{x\to 0}\frac{1}{x}.\)

*Solution*. Since \(f(x)=\frac{1}{x}\) decreases without bound as \(x\to 0^-\) , we notice \(\lim_{x\to 0^-}f(x)=-\infty\). Also since \(f\) increases without bound as \(x\to 0^+,\) we notice \(\lim_{x\to 0^+}f(x)=+\infty\). Thus \(\lim_{x\to0}f(x)\) does not exist.

Compare the previous example with the following.

**Example 3.32 **Determine \(\displaystyle \lim_{x\to2} \frac{2x^3+x^2-16x+12}{x^2-4}\).

*Solution*. By factoring, \[\begin{align*}
& \lim_{x\to 2}\frac{2x^3+x^2-16x+12}{x^2-4}
=\lim_{x\to 2}\frac{(x-2)\left(2x^2+5x-6\right)}{(x-2)(x+2)} \\
& \qquad =\lim_{x\to 2}\frac{\left(2x^2+5x-6\right)}{(x+2)}
=\frac{\left(2(2)^2+5(2)-6\right)}{((2)+2)} =3.
\end{align*}\]

**Theorem 3.4 **Let \(A\) be a positive real number.

- If \(n\) is a positive even integer, then \(\lim_{x\to c}\frac{A}{(x-c)^n}=+\infty.\)
- If \(n\) is a positive odd integer, then \[ \lim_{x\to c^+}\frac{A}{(x-c)^n}=+\infty \qquad \text{and} \qquad \lim_{x\to c^-}\frac{A}{(x-c)^n}=-\infty. \]

**Example 3.33 **Evaluate \(\displaystyle \lim_{x\to 2^+}\frac{-3}{\sqrt[3]{x-2}}\).

*Solution*. Notice that \(\displaystyle \frac{1}{\sqrt[3]{x-2}}\) increases without bound as \(x\to 2^+\) and therefore, \(\displaystyle \lim_{x\to 2^+}\frac{-3}{\sqrt[3]{x-2}}=-\infty.\)

**Example 3.34 **Evaluate \(\displaystyle \lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{x^2}\right).\)

*Solution*. Note that \[
f(x)=\frac{1}{x}-\frac{1}{x^2}=\frac{-1+x}{x^2}
\]

and that \(\frac{-1+x}{x^2}\) decreases without bound as \(x\to 0^+\) and therefore, \[
\lim_{x\to 0^+}\frac{-1+x}{x^2}=-\infty.
\] This is true because when \(x>0\) is close to \(0\) we know that \(-1+x\) is negative. Similarly, \(\lim_{x\to 0^-}\frac{-1+x}{x^2}=-\infty\) because when \(x<0\) and close to \(0\) we know that \(-1+x\) is negative. Therefore it follows \[
\lim_{x\to 0}\left(\frac{1}{x}-\frac{1}{x^2}\right)=-\infty
\]

## 3.22 Vertical Asymptotes

**Definition 3.11 **The line \(x=c\) is called a **vertical asymptote** of the curve \(y=f(x)\) if at least one of the following statements is true:

- \(\lim_{x\to c}f(x)=+\infty\) \(\lim_{x\to c^-}f(x)=+\infty\) \(\lim_{x\to c^+}f(x)=+\infty\)
- \(\lim_{x\to c}f(x)=-\infty\) \(\lim_{x\to c^-}f(x)=-\infty\) \(\lim_{x\to c^+}f(x)=-\infty\)

The following example demonstrates that not all rational functions have vertical asymptotes.

**Example 3.35 **Determine the vertical asymptotes of the function \[
g(x)=\frac{x^3}{x^2+3x+10}.
\]

*Solution*. The function \(g\) is continuous on its domain which is \(\mathbb{R}\) and therefore there are no vertical asymptotes for this function.

**Example 3.36 **Determine the vertical asymptotes of the function \[
f(x)=\frac{x^3}{x^2+3x-10}.
\]

*Solution*. Since \[
f(x)=\frac{x^3}{x^2+3x-10}=\frac{x^3}{(x-2)(x+5)}
\] and therefore the vertical asymptotes are \(x=2\) and \(x=-5\) because \[
\lim_{x\to 2^+}\frac{x^3}{(x-2)(x+5)}=+\infty
\qquad \text{and} \qquad
\lim_{x\to 5^+}\frac{x^3}{(x-2)(x+5)}=+\infty.
\]

The following example demonstrates that there can be an unlimited number of vertical asymptotes for a function.

**Example 3.37 **Determine the vertical asymptotes of the function \[
h(x)=\tan x-\cot x.
\]

*Solution*. We can rewrite this function as \[
h(x)=\tan x-\cot x=\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}=\frac{\sin^2x-\cos^2x}{\sin x \cos x}=\frac{\cos 2x}{\sin x \cos x}.
\] Therefore the zeros of the sine and cosine functions yield the vertical asymptotes of \(x=\pm \frac{\pi }{2}+k\) for all integers \(k,\) since \(\cos 2x\) is not zero for these values of \(x\) and because, for any \(c=\pm \frac{\pi }{2}+k\) , \(\lim_{x\to c}h(x)\) is either \(\pm \infty\). See Figure \(\eqref{fig:graph1}\).

## 3.23 Limits at Infinity

**Definition 3.12 **Let \(f\) be a function defined on some interval \((a,+\infty ).\)

Then \[\lim_{x\to \infty }f(x)=L\] means that the values of \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently large; or more precisely, for every \(\epsilon>0\) there exists an \(N\) such that if \(x>N\) then \(|f(x)-L|<\epsilon\).

**Definition 3.13 **Let \(f\) be a function defined on some interval \((-\infty ,a).\) Then \[\lim_{x\to -\infty }f(x)=L\] means that the values of \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently large negative; or more precisely, for every\(\epsilon >0\) there exists an \(N\) such that if \(x<N\) then \(|f(x)-L|<\epsilon.\)

The following limit rules are similar to the limit rules used for when \(x\to c\) but instead use \(x\to +\infty ;\) and they are also valid for when \(x\to +\infty\) is replaced by \(x\to -\infty.\)

**Theorem 3.5 **If \(\lim_{x\to \infty }f(x)\) and \(\lim_{x\to \infty }g(x)\) exist, then

- \(\lim_{x\to \infty } k=k\) for any constant \(k\)
- \(\lim_{x\to \infty }k f(x)=k \lim_{x\to \infty }f(x)\)
- \(\lim_{x\to \infty }[f(x)+g(x)]=\lim_{x\to \infty }f(x)+\lim_{x\to \infty }g(x)\)
- \(\lim_{x\to \infty }[f(x)-g(x)]=\lim_{x\to \infty }f(x)-\lim_{x\to \infty }g(x)\)
- \(\lim_{x\to \infty }[f(x)g(x)]=\left( \lim_{x\to \infty }f(x) \right)\left( \lim_{x\to \infty }g(x) \right)\)
- \(\lim_{x\to \infty }[f(x)/g(x)]=\left( \lim_{x\to \infty }f(x) \right)/\left( \lim_{x\to \infty }g(x) \right)\)
- \(\lim_{x\to \infty }[f(x)]^n=\left( \lim_{x\to \infty }f(x) \right){}^n\) where \(n\) is a rational number and whenever the limits exist.

**Theorem 3.6 **Let \(A\) be a real number.

- If \(r>0\) is a rational number, then \[ \lim_{x\to \infty }\frac{A}{x^r}=0. \]
- If \(r>0\) is a rational number with \(x^r\) defined for all \(x,\) then \[ \lim_{x\to -\infty }\frac{A}{x^r}=0. \]

**Example 3.38 **Evaluate \(\displaystyle\lim_{x\to \infty }\frac{3x^2-x-2}{5x^2+4x+1}.\)

*Solution*. To evaluate this limit we divide both the numerator and the denominator by the highest power of \(x\) that occurs. So we have \[
\lim_{x\to \infty }\frac{3x^2-x-2}{5x^2+4x+1}
=\lim_{x\to \infty }\frac{\frac{3x^2}{x^2}-\frac{x}{x^2}-\frac{2}{x^2}}{\frac{5x^2}{x^2}+\frac{4x}{x^2}+\frac{1}{x^2}}
=\lim_{x\to \infty }\frac{3-\frac{1}{x}-\frac{2}{x^2}}{5+\frac{4}{x}+\frac{1}{x^2}}
% =\frac{3-0-0}{5+0+0}
=\frac{3}{5}.
\]

**Example 3.39 **Evaluate \(\displaystyle \lim_{x\to \infty }\left(x-\sqrt{x^2+1}\right).\)

*Solution*. We use the conjugate radical as follows \[\begin{align*}
& \lim_{x\to \infty }\left(x-\sqrt{x^2+1}\right)
=\lim_{x\to \infty }\frac{\left(x-\sqrt{x^2+1}\right)\left(x+\sqrt{x^2+1}\right)}{\left(x+\sqrt{x^2+1}\right)} \\
& \qquad =\lim_{x\to \infty }\frac{x^2-\left(x^2+1\right)}{\left(x+\sqrt{x^2+1}\right)}
=\lim_{x\to \infty }\frac{-1}{x+\sqrt{x^2+1}} \\
& \qquad =\lim_{x\to \infty }\frac{\frac{-1}{x}}{\frac{x}{x}+\sqrt{\frac{x^2+1}{x^2}}}
=\frac{0}{1+\sqrt{1+0}}=0.
\end{align*}\]

## 3.24 Horizontal Asymptotes

In mathematics, the symbol \(\infty\) is not a number, rather it is used to describe the process of unrestricted growth or the result of such a growth.

**Definition 3.14 **The line \(y=L\) is called a **horizontal asymptote** of the curve \(y=f(x)\) if either \(\lim_{x\to \infty }f(x)=L\) or \(\lim_{x\to -\infty} f(x)=L.\)

::: {#thm- } If \(f\) is a rational functions of the form:

\[
f(x)=\frac{a_0+a_1x+a_2x^2+\cdots +a_nx^n}{b_0+b_1x+b_2x^2+\cdots +b_mx^m}
\] where \(n\) is the degree of the polynomial in the numerator and \(m\) is the degree of the polynomial in the denominator, then the horizontal asymptote of the curve \(y=f(x)\) is determined by the following.

- If \(n=m,\) then \(y=\frac{a_n}{b_m}\) is the horizontal asymptote.
- If \(n<m,\) then \(y=0\) is the horizontal asymptote.
- If \(n>m,\) then there is no horizontal asymptote, but rather a slant (oblique) asymptote and can be found be using long division. :::

**Example 3.40 **Find the horizontal asymptote of the graph of the function \[
f(x)=\frac{(1-x)(2+x)}{(1+2x)(2-3x)}.
\]

*Solution*. The degree of the numerator \((1-x)(2+x)=-x^2-x+2\) is 2 and the degree of the denominator \((1+2x)(2-3x)=-6 x^2+x+2\) is 2 we divide both numerator and denominator by \(x^2\) and using the properties of limits, we have \[
%\lim_{x\to \infty} f(x)=
\lim_{x\to \pm\infty} \frac{-x^2-x+2}{-6 x^2+x+2}=
%\lim_{x\to \pm\infty} \frac{\frac{-x^2-x+2}{x^2}}{\frac{-6 x^2+x+2}{x^2}}=
\lim_{x\to \pm\infty} \frac{\frac{-x^2}{x^2}-\frac{x}{x^2}+\frac{2}{x^2}}{\frac{-6 x^2}{x^2}+\frac{x}{x^2}+\frac{2}{x^2}}=
\frac{-1-0+0}{-6+0+0}=6
\] Therefore the only horizontal asymptote of the graph of \(f(x)\) is \(y=6.\)

**Example 3.41 **Find the horizontal asymptote of the graph of the function \[
f(x)=\frac{\sqrt{2x^2+1}}{3x-5}.
\]

*Solution*. Dividing both numerator and denominator by \(x\) and using the properties of limits, we have \[\begin{align*}
\lim_{x\to \infty }\frac{\sqrt{2x^2+1}}{3x-5}
=\lim_{x\to \infty }\frac{\sqrt{2+\frac{1}{x^2}}}{3-\frac{5}{x}}
=\frac{\sqrt{\lim_{x\to \infty }\left(2+\frac{1}{x^2}\right)}}{\lim_{x\to \infty }\left(3-\frac{5}{x}\right)}
=\frac{\sqrt{2+0}}{3-5(0)}=\frac{\sqrt{2}}{3}
\end{align*}\] Therefore the line \(y=\left.\sqrt{2}\right/3\) is a horizontal asymptote. It is also important to realize that \[\begin{align*}
\lim_{x\to -\infty }\frac{\sqrt{2x^2+1}}{3x-5}
& =\lim_{x\to -\infty }\frac{\frac{1}{x}\sqrt{2x^2+1}}{3-\frac{5}{x}} \\
& =\lim_{x\to -\infty }\frac{\left(\frac{-1}{\sqrt{x^2}}\right)\sqrt{2x^2+1}}{3-\frac{5}{x}}
=\lim_{x\to -\infty }\frac{-\sqrt{2+\frac{1}{x^2}}}{3-\frac{5}{x}}
=-\frac{\sqrt{2}}{3}
\end{align*}\] Therefore, the line \(y=\left.-\sqrt{2}\right/3\) is another horizontal asymptote.

**Example 3.42 **Find all horizontal asymptotes of the graph of the function \[
f(x)=\frac{x}{\sqrt[4]{x^4+1}}.
\]

*Solution*. Dividing both numerator and denominator by \(x\) and using the properties of limits, we have \[\begin{align*}
\lim_{x\to \infty }\frac{x}{\sqrt[4]{x^4+1}}
& =\lim_{x\to \infty }\frac{\frac{x}{x}}{\frac{1}{x}\sqrt[4]{x^4+1}}
=\lim_{x\to \infty }\frac{1}{\sqrt[4]{\frac{x^4+1}{x^4}}} \\
& =\lim_{x\to \infty }\frac{1}{\sqrt[4]{1+\frac{1}{x^4}}}
=\frac{1}{\sqrt[4]{\lim_{x\to \infty }\left(1+\frac{1}{x^4}\right)}}
=\frac{1}{\sqrt[4]{1+0}}=1.
\end{align*}\] Therefore, the line \(y=1\) is a horizontal asymptote. In computing the limit \(x\to -\infty\) we must remember that for \(x<0,\) we have \(\sqrt[4]{x^4}=-x,\) so when we divide the numerator by \(x,\) when \(x<0\) we have, \[\begin{align*}
\lim_{x\to -\infty }\frac{x}{\sqrt[4]{x^4+1}}
& =\lim_{x\to -\infty }\frac{\frac{x}{x}}{\frac{1}{x}\sqrt[4]{x^4+1}} \\
& =\lim_{x\to \infty }\frac{1}{\frac{1}{\sqrt[4]{x^4}}\sqrt[4]{\frac{x^4+1}{x^4}}}
=\lim_{x\to \infty }\frac{1}{-\text{ }\sqrt[4]{1+\frac{1}{x^4}}}
=-1.
\end{align*}\] Therefore, the horizontal asymptotes are \(y=\pm 1.\)

## 3.25 Curve Sketching

## 3.26 Curve Sketching as a Procedure

The following steps can be used to graph many commonly used functions.

- If given a function, determine the domain and range.
- If possible algebraically simplify the function.
- Test for symmetry with respect to the \(x\)-axis, \(y\)-axis, and the origin.
- Determine any \(x\)-intercepts and \(y\)-intercepts.
- Determine any vertical asymptotes.
- Determine any horizontal asymptotes.
- Determine the first order critical numbers.
- Apply the First Derivative Test.
- Apply the Second Derivative Test.
- Determine the second order critical numbers.
- Apply the Concavity Test.
- Determine any vertical tangents.
- Determine any cusps.
- Plot Points.
- Sketch the curve.

## 3.27 Vertical Tangents and Cusps

There are four possibilities for unbounded behavior of a derivative \(f'(x)\) around a given real number \(c.\)

**Definition 3.15 **Suppose the function \(f\) is continuous at the point \(P(c,f(c)).\)

The graph of \(f\) has a

**vertical tangent**at \(P\) if one of the following holds.\(\lim _{x\to c^+}f'(x)=\lim _{x\to c^-}f'(x)=+\infty\)

\(\lim _{x\to c^+}f'(x)=\lim _{x\to c^-}f'(x)=-\infty\)

The graph of \(f\) has a

**cusp**at \(P\) if one of the following holds.\(\lim _{x\to c^+}f'(x)=+\infty\) and \(\lim _{x\to c^-}f'(x)=-\infty\)

\(\lim _{x\to c^+}f'(x)=-\infty\) and \(\lim _{x\to c^-}f'(x)=+\infty\)

**Example 3.43 **Sketch the graph of \[
f(x)=3x^{3/5}\left(5-x-4x^2\right)
\] and explain why there is a vertical tangent at \(x=0\).

*Solution*. Using the product rule the derivative of the function \(f\) is \[\begin{align*}
f'(x) & =\frac{9}{5}x^{-2/5}\left(5-x-4x^2\right)+3x^{3/5}(-1-8x) \\
& =\frac{9\left(5-x-4x^2\right)}{5x^{2/5}}+\frac{\left(5x^{2/5}\right)\left(3x^{3/5}\right)(-1-8x)}{5x^{2/5}} \\
& =\frac{9\left(5-x-4x^2\right)}{5x^{2/5}}+\frac{15x(-1-8x)}{5x^{2/5}} \\
& =\frac{45-24 x-156 x^2}{5 x^{2/5}}.
\end{align*}\] To determine any vertical tangents we consider where \(f'(x)\) is undefined. Notice that at \(x=0\) the derivative is undefined but \(f(0)=0.\) Thus, the point \((0,0)\) is a candidate for a being a vertical tangent to the graph of \(f.\) We check the following limits to determine if \((0,0)\) is a vertical tangent. Since \(45-24 x-156 x^2\to 45\) and \(5 x^{2/5}\to -\infty\) as \(x\to 0^-,\) \[\lim _{x\to 0^-}\frac{45-24 x-156 x^2}{5 x^{2/5}}=+\infty.\] Since \(45-24 x-156 x^2\to 45\) and \(5 x^{2/5}\to -\infty\) as \(x\to 0^-,\) \[\lim _{x\to 0^+}\frac{45-24 x-156 x^2}{5 x^{2/5}}=+\infty.\] Therefore the function \(f\) has a vertical tangent at \((0,0)\) which can be seen from the sketch of the graph of \(f.\) (see Figure ~\(\ref{fig:graph17}\)).

**Example 3.44 **Sketch the graph of \(f(x)=x^{2/3}\left(x^2+5x-20\right)\) and explain why there is a cusp at \(x=0\).

*Solution*. Using the product rule we find the derivative as \[\begin{align*}
f'(x) & =\frac{2}{3}x^{-1/3}\left(x^2+5x-20\right)+x^{2/3}(2x+5) \\
& =\frac{2\left(x^2+5x-20\right)}{3x^{1/3}}+\frac{3x^{1/3}x^{2/3}(2x+5)}{3x^{1/3}} \\
& =\frac{2x^2+10x-40}{3x^{1/3}}+\frac{3x(2x+5)}{3x^{1/3}} \\
& =\frac{2x^2+10x-40}{3x^{1/3}}+\frac{6x^2+15x}{3x^{1/3}} \\
& =\frac{8 x^2+25 x-40}{3 \sqrt[3]{x}}.
\end{align*}\] To determine any cusps we consider where \(f'(x)\) is undefined. Notice that at \(x=0\) the derivative is undefined but \(f(0)=0.\) Thus the point \((0,0)\) is a candidate for a being a cusp for the graph of the function \(f\). We check the following limits to determine if \((0,0)\) is a cusp. Since \(8x^2+25x-40\to -40\) and \(3 \sqrt[3]{x}\to -\infty\) as \(x\to 0^-,\)

\[\lim _{x\to 0^-}\frac{8 x^2+25 x-40}{3 \sqrt[3]{x}}=+\infty.\] Since \(8x^2+25x-40\to -40\) and \(3 \sqrt[3]{x}\to +\infty\) as \(x\to 0^+,\) \[\lim _{x\to 0^+}\frac{8 x^2+25 x-40}{3 \sqrt[3]{x}}=-\infty .\] Therefore the function \(f\) has a cusp at \((0,0)\) which can be seen from the sketch of the graph of \(f.\) (see Figure ~\(\ref{fig:graph18}\)).

**Example 3.45 **Determine any vertical tangents and cusps for the function \[
f(x)=x^{2/3} (x - 1)^{1/3}.
\]

*Solution*. To find the vertical tangents and cusps we check where the first derivative is undefined. The derivative of \(f\) is \[f'(x)=\frac{x^{2/3}}{3 (x-1)^{2/3}}+\frac{2 \sqrt[3]{x-1}}{3 \sqrt[3]{x}}=\frac{3 x-2}{3 (x-1)^{2/3} \sqrt[3]{x}}.\] Therefore the points \((0,0)\) and \((1,0)\) are candidates for where a vertical tangent and cusp might occur. The function \(f\) has a vertical asymptote at \(x=1\) since \[
\lim_{x\to 1^-}f'(x)=\lim_{x\to 1^-} \frac{3 x-2}{3 (x-1)^{2/3} \sqrt[3]{x}} = +\infty
%\lim_{x\to 1^+}f'(x)=\lim_{x\to 1^+} \frac{3 x-2}{3 (x-1)^{2/3} \sqrt[3]{x}} = +\infty
\] The function \(f\) has a cusp at \(x=0\) since \[\lim_{x\to 0^-}f'(x)=\lim_{x\to 0^-} \frac{3 x-2}{3 (x-1)^{2/3} \sqrt[3]{x}} = +\infty
% \lim_{x\to 0^+}f'(x)=\lim_{x\to 0^+} \frac{3 x-2}{3 (x-1)^{2/3} \sqrt[3]{x}} = -\infty .
\] (see Figure ~\(\ref{fig:graph19}\)).

**Example 3.46 **Sketch the graph of the rational function \[
f(x)=\frac{2x^2}{x^2-1}
\] showing all special features.

*Solution*. The domain of \(f\) is \[
\left\{x\left|x^2-1\neq 0\right.\right\}
=\{x|\neq \pm 1\}
=(-\infty ,-1)\cup (-1,1)\cup (1,+\infty ).
\] The \(x\) and \(y\) intercepts are both \(0.\) Since \(f(-x)=f(x)\) the function is even and so the curve is symmetric about the \(y\)-axis. Since \[
\lim _{x\to \pm \infty}\frac{2x^2}{x^2-1}
=\lim _{x\to \pm \infty}\frac{2}{1-\frac{1}{x^2}}=2
\] the line \(y=2\) is a horizontal asymptote. Since the denominator is 0 when \(x\pm 1,\) we compute the following limits: \[
\begin{array}{cc}
\displaystyle
\lim _{x\to 1^+}\frac{2x^2}{x^2-1}
=+\infty & \displaystyle \lim _{x\to 1^-}\frac{2x^2}{x^2-1}
=-\infty \\
\displaystyle
\lim _{x\to -1^+}\frac{2x^2}{x^2-1}
=-\infty & \displaystyle \lim _{x\to -1^-}\frac{2x^2}{x^2-1}
=+\infty
\end{array}
\] Therefore, the lines \(x=1\) and \(x=-1\) are vertical asymptotes. Next we find the derivative function. \[ f'(x)=\frac{4x\left(x^2-1\right)-2x^2(2x)}{\left(x^2-1\right)^2}=\frac{-4x}{\left(x^2-1\right)^2}.\] Since \(f'(x)>0\) when \(x<0\) \((x\neq -1)\) and \(f'(x)<0\) when \(x>0\) \((x\neq 1),\) \(f\) is increasing on \((-\infty ,-1)\) and \((-1,0)\) and decreasing on \((0,1)\) and \((1,+\infty ).\) The only critical number is \(x=0.\) Since \(f'\) changes sign from positive to negative at \(0,\) \(f(0)=0\) is a local maximum by the First Derivative Test. Also, \[f''(x)=\frac{-4\left(x^2-1\right)^2+4x(2)\left(x^2-1\right)(2x)}{\left(x^2-1\right)^4}=\frac{12x^2+4}{\left(x^2-1\right)^3}\] Since \(12x^2+4>0\) for all \(x,\) we have \(f''(x)>0 \Longleftrightarrow x^2-1>0 \Longleftrightarrow |x|>1\) and \(f''(x)<0 \Longleftrightarrow |x|<1.\) Thus the curve is concave downward on the intervals \((-\infty ,-1)\) and \((1,+\infty )\) and concave downward on \((-1,1).\) There is no point of inflection since \(1\) and \(-1\) are not in the domain of \(f.\) (see Figure ~\(\ref{fig:graph20}\)).

**Example 3.47 **Sketch the graph of the trigonometric function \[
f(x)=2 \cos x+\sin 2 x
\] showing all special features.

*Solution*. The domain of the function \(f\) is \(\mathbb{R}.\) The \(y\)-intercept is \((0,2)\) since \(f(0)=2.\) The \(x\)-intercepts occur when \(2 \cos x+\sin 2x=2 \cos x+2 \sin x \cos x=2 \cos x(1+\sin x)=0\) which is precisely when \(x=\pi /2\) and \(x=3\pi /2\), because we need only consider \([0,2\pi ]\) since function is periodic via, \[f(x+2\pi )=2 \cos (x+2\pi )+\sin [2(x+2\pi )]=2\cos x+\sin 2x=f(x).\] There are no vertical asymptotes nor horizontal asymptotes. The first derivative of \(f\) is, \[\begin{align*}
f'(x)& =-2 \sin x+2 \cos 2x =-2(2 \sin x-1)(\sin x+1).
\end{align*}\] Thus, \(f'(x)=0\) when \(\sin x=1/2\) or when \(\sin x=-1,\) so in \([0,2\pi ]\) we only consider the critical numbers \(x=\pi /6\), \(x=5\pi /6,\) and \(x=3\pi /2.\) Applying the First Derivative Test we find: \[
\begin{array}{c|c|c|l}
\text{Interval} & f & f' & \text{Conclusion} \\ \hline
0<x<\pi /6 & & + & \text{increasing} \\
x=\pi /6 & 3\left.\sqrt{3}\right/2 & 0 & \text{relative maximum} \\
\pi /6<x<5\pi /6 & & - & \text{decreasing} \\
x=5\pi /6 & -3\left.\sqrt{3}\right/2 & 0 & \text{relative minimum} \\
5\pi /6<x<3\pi /2 & & + & \text{increasing} \\
x=3\pi /2 & 0 & 0 & \text{horizontal tangent} \\
3\pi /2<x<2\pi & & + & \text{increasing}
\end{array}
\] The second derivative of \(f\) is, \[ f''(x)=-2\cos x-4 \sin 2x=-2 \cos x (1+4 \sin x)\] so the second order critical numbers are \(x=\pi /2,3\pi /2, \alpha _1,\) and \(\alpha _2\) where \(\sin \alpha _1=-\frac{1}{4}\) and \(\sin \alpha _2=-\frac{1}{4}.\) Applying the Concavity Test we find, \[
\begin{array}{c|c|c|l}
\text{Interval} & f & f'' & \text{Conclusion} \\ \hline
0<x<\pi /2 & & - & \text{concave down} \\
x=\pi /2 & 0 & 0 & \text{inflection point} \\
\pi /2<x<\alpha _1 & & + & \text{concave up} \\
x=\alpha _1 & f\left(\alpha _1\right) & 0 & \text{inflection point} \\
\alpha _1<x<3\pi /2 & & - & \text{concave down} \\
x=3\pi /2 & 0 & 0 & \text{inflection point} \\
3\pi /2<x<\alpha _2 & & + & \text{concave up} \\
x=\alpha _2 & f\left(\alpha _2\right) & 0 & \text{inflection point} \\
\alpha _2<x<2\pi & & - & \text{concave down}
\end{array}
\]

## 3.28 Exercises

**Exercise 3.52 **Evaluate the following limits.

- \(\lim _{x\to 0^-}\frac{5}{2x}\)

- \(\lim _{x\to -8^+}\frac{2x}{x+8}\)

- \(\lim _{x\to 0}\frac{-1}{x^2(x+1)}\)

- \(\lim _{x\to 0}\frac{-1}{x^2(x-1)}\)

- \(\lim _{x\to 0^+}\frac{2}{x^{1/5}}\)

- \(\lim _{x\to \frac{\pi}{2}^-}\tan x\)

**Exercise 3.53 **Find the limits for the following functions. - \(f(x)=\frac{1}{x^2-4}\) as \(x\to 2^+,\) \(x\to 2^-,\) \(x\to -2^+,\) and \(x\to -2^-.\)

- \(f(x)=\frac{x^2-1}{2x+4}\) as \(x\to -2^+,\) \(x\to -2^-,\) \(x\to 1^+,\) and \(x\to 0^-.\)

- \(f(t)=2-\frac{3}{t^{1/3}}\) as \(t\to 0^+,\) and \(t\to 0^-.\)

- \(f(x)=\left(\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}}\right)\) as \(x\to 0^+,\) \(x\to 0^-,\) \(x\to 1^+,\) and \(x\to 1^-.\)

**Exercise 3.54 **Graph the following rational functions and include the graphs and equations of their asymptotes.

- \(f(x)=\frac{-3}{x-3}\)
- \(f(x)=\frac{x^2}{x-1}\)
- \(f(x)=\frac{x^2-1}{2x+4}\)

- \(f(x)=\frac{x^3-x^2+x-1}{(2x+3)(2x-1)}\)

**Exercise 3.55 **For each of the following sketch the graph of a function \(y=f(x)\) that satisfies the conditions. - \(f(0)=0,\) \(f(1)=2,\) \(f(-1)=-2,\) \(\lim _{x\to -\infty }f(x)=-1,\) and \(\lim _{x\to \infty }f(x)=1\) - \(f(2)=1,\) \(f(-1)=0,\) \(\lim _{x\to \infty }f(x)=0,\) \(\lim _{x\to0^+}f(x)=\infty ,\) \(\lim _{x\to 0^-}f(x)=-\infty ,\) and \(\lim _{x\to -\infty }f(x)=1\)

- \(\lim _{x\to \pm \infty }g(x)=0,\) \(\lim _{x\to 3^-}g(x)=-\infty,\) and \(\lim _{x\to 3^+}g(x)=+\infty\)

**Exercise 3.56 **Sketch the graph of the function \[
f(x)=
\left\{
\begin{array}{cc}
-(x+1)^2+1 & -1\leq x\leq 0 \\
x^2 & 0<x<1 \\ 1 & 1<x<2 \\
2 & x=2 \\ 1 & 2<x\leq 3
\end{array}
\right.
\] and then use the graph to determine which the following statements about the function \(y=f(x)\) are true and which are false?

- \(\lim _{x \to -1^+}f(x)=1\)
- \(\lim _{x\to 2}f(x)\) does not exist
- \(\lim _{x\to 2}f(x)=2\)
- \(\lim _{x\to 1^-}f(x)=2\)
- \(\lim _{x\to 1^+}f(x)=1\)
- \(\lim _{x\to 1}f(x)\) does not exist
- \(\lim _{x\to 0^+}f(x)=\lim _{x\to 0^-}f(x)\)
- \(\lim _{x\to c}f(x)\) exists at every \(c\) in the open interval \((-1,1)\)
- \(\lim _{x\to c}f(x)\) exists at every \(c\) in the open interval \((1,3)\)
- \(\lim _{x\to -1^-}f(x)=0\)
- \(\lim _{x\to 3^+}f(x)\) does not exist

**Exercise 3.57 **Sketch the graph of the function \[
f(x)=\left\{
\begin{array}{cc} 3-x & x<2 \\ 2 & x=2 \\ \frac{x}{2} & x>2\end{array}\right.
\] and then use the graph to determine the following? - \(\text{Find} \lim _{x\to 2^+}f(x),\) \(\lim _{x\to 2^-}f(x),\) and \(f(2).\) - Does \(\lim _{x\to 2}f(x)\) exist? If so, what is it? If not, why not? - Find \(\lim _{x\to -1^-}f(x)\) and \(\lim _{x\to -1^+}f(x).\) - Does \(\lim _{x\to -1}f(x)\) exist? If so, what is it? If not, why not?

**Exercise 3.58 **Let \(g(x)=\sqrt{x}\sin \left(\frac{1}{x}\right).\) Use the graph of \(g\) to determine the following. - Does \(\lim _{x\to 0^+}g(x)\) exist? If so, what is it? If not, why not? - Does \(\lim _{x\to 0^-}g(x)\) exist? If so, what is it? If not, why not? - Does \(\lim _{x\to 0}g(x)\) exist? If so, what is it? If not, why not?

**Exercise 3.59 **Sketch the graph of the function \[
f(x)=\left\{\begin{array}{cc} x^3 & x\neq 1 \\ 0 & x=1\end{array}\right..
\]

Find \(\lim _{x\to 1^-}f(x)\) and \(\lim _{x\to 1^+}f(x).\) Does \(\lim _{x\to 1}f(x)\) exist? If so, what is it? If not, why not?

**Exercise 3.60 **Sketch the graph of the function \[
f(x)=\left\{\begin{array}{cc}
1-x^2 & x\neq 1 \\
2 & x=1\end{array}
\right..
\]

Find \(\lim _{x\to 1^-}f(x)\) and \(\lim _{x\to 1^+}f(x).\) Does \(\lim _{x\to 1}f(x)\) exist? If so, what is it? If not, why not?

**Exercise 3.61 **Sketch the graph of the function

\[
f(x)=\left\{
\begin{array}{ll} x & -1\leq x<0 \quad \text{or} \quad 0<x\leq 1 \\ 1 & x=0 \\ 0 & x<-1 \quad \text{or} \quad x>1.\end{array}\right.\] - What is the domain and range of \(f?\) - At what points \(c,\) if any does \(\lim _{x\to c}f(x)\) exist? - At what points does only the left-hand limit exist? - At what points does only the right-hand limit exist?

**Exercise 3.62 **Find the following one-sided limits algebraically.

- \(\lim _{x\to 2^+} \frac{x(2x+5)}{(x+1)(x^2+x)}\)

- \(\lim _{h\to 0^+}\frac{\sqrt{h^2+4h+5}-\sqrt{5}}{h}\)

- \(\lim _{x\to 1^+}\frac{\sqrt{2x}(x-1)}{|x-1|}\)

**Exercise 3.63 **Find the following two-sided limits.

- \(\lim _{t\to 0}\left(\frac{\sin k t}{t}\right)\) where \(k\) is a constant
- \(\lim _{x\to 0}\left(\frac{\tan 2x}{x}\right)\)
- \(\lim _{x\to 0}\left(6x^2(\cot x)(\csc 2x)\right)\)
- \(\lim _{h\to 0}\left(\frac{\sin (\sin h)}{\sin h}\right)\)
- \(\lim _{\theta \to 0}\left(\frac{\sin \theta }{\sin 2\theta }\right)\)

**Exercise 3.64 **Find the limits for the following functions for both \(x\to +\infty\) and \(x\to -\infty\).

- \(f(x)=\pi -\frac{2}{x^2}\)
- \(f(x)=\frac{-5+\left(\frac{7}{x}\right)}{3-\left(\frac{1}{x}\right)}\)
- \(f(\theta )=\frac{\cos \theta }{3 \theta }\)

- \(f(x)=e^{-x}\sin x\)

- \(f(x)=\frac{2x+3}{5x+7}\)
- \(f(x)=\frac{x+1}{x^2+3}\)

- \(f(x)=\frac{9x^4+x}{2x^4+5x^2-x+6}\)

- \(f(x)=\frac{2\sqrt{x}+x^{-1}}{3x-7}\)

- \(f(x)=\frac{x^{-1}+x^{-4}}{x^{-2}-x^{-3}}\)

## 3.29 Applied Optimization Problems

## 3.30 Optimization Procedures

In this section we give a few examples on how to set up a function to be optimized using its derivative. In general, the first step in solving an application problem is to understand the problem; maybe ask what are the unknowns? and what are the given quantities? Then the next best step is usually to draw a picture, labeling the unknowns and introducing notation. The final step should always be to check the solution to see that it makes sense for the given questions.

An optimization procedure is a step-by-step procedure to solve an application problem using calculus; and in particular, using derivatives. Here is a typical optimization procedure:

- Sketch a graph or draw a diagram.
- Introduce mathematical notation.
- Express information as expressions, equations and functions.
- Formulate the problem mathematically by stating the knowns and unknowns.
- State the knowns and unknowns.
- Identify needed theorems and validate their hypotheses. Apply the theorem.
- Solve the mathematical problem.
- Answer the original problem in terms of the given language.

We will illustrate this optimization procedure in the following examples.

## 3.31 Optimizing with Numbers

**Example 3.48 **Find two nonnegative numbers whose sum is \(8\) and the product of whose squares is as large as possible.

*Solution*. We are looking for two nonnegative numbers, say \(x\) and \(y\) with \(x\geq 0,\) \(y\geq 0,\) and \(x+y=8.\) We want to maximize the function \[
P(x)=x^2y^2=x^2(8-x)^2
\] with \(x\geq 0.\) Since the derivative of \(P\), \[
P'(x)=128 x-48 x^2+4 x^3=4 (-8+x) (-4+x) x
\] is continuous, the only critical numbers are \(x=0, 4, 8.\) We evaluate \(P\) to find \(P(0)=0=P(8)\) and so the largest possible value is \(P(4)=256\) with \(x=y=4.\)

**Example 3.49 **Under the condition that \(2x-5y=18,\) minimize \(x^2y\) when \(x\geq 0\) and \(y\geq 0.\)

*Solution*. We want to minimize \[
P(x)=x^2y=x^2\left(\frac{1}{5}\right) (-18+2 x).
\] Since the derivative of \(P\), \[
P'(x)=\frac{6}{5} \left(-6 x+x^2\right)=\frac{6}{5} (-6+x) x
\] is continuous, the only critical numbers are \(x=0\) and \(x=6\). We evaluate to find \(P(0)=0\) and the smallest possible value to be \(P(6)=-\frac{216}{5}.\) Thus the values are \(x=6\) and \(y=\frac{-6}{5}.\)

## 3.32 Optimizing Volume

**Example 3.50 **Find the dimensions of the right circular cylinder of largest volume that can be inscribed in a sphere of radius \(R.\)

*Solution*. Let \(R\) be the radius of the sphere and \(r\) the radius of the cylinder with height \(2h\) so that \(0\leq h\leq R.\) Since \(r^2+h^2=R^2,\) the volume of the cylinder is given by

\[
V(h)=\pi r^22h=2\pi h\left(R^2-h^2\right)=2\pi \left(R^2h-h^3\right).
\] Using the Extreme Value Theorem, we wish to maximize the continuous function \(V(h)\) on the closed interval \([0,R]\). The derivative of \(V\) is \(V'(h)=2\pi \left(R^2-3h^2\right).\) Since \(V'\) is continuous the only critical numbers are found by solving \(V'(h)=0\) for \(0\leq h\leq R\). Thus the only critical number is \(h=R\left/\sqrt{3}.\right.\) Since \(V(0)=V(R)=0\), \(h=0\) and \(h=R\) give minima, it follows by the Extreme Value Theorem that \(h=R\left/\sqrt{3}\right.\) must be a maximum. After solving for \(r\) we find the dimensions of the cylinder are, height: \(2 R\frac{\sqrt{3}}{3}\) and radius: \(\frac{R}{3}\sqrt{6}.\)

## 3.33 Optimizing with Geometry

**Example 3.51 **Find all points on the circle \(x^2+y^2=a^2\) such that the product of the \(x\)-coordinate and the \(y\)-coordinate is as large as possible.

*Solution*. In the first quadrant we have \(y=\sqrt{a^2-x^2}\) and so we want to maximize \(f(x)=x\sqrt{a^2-x^2}\) subject to \(x>0\). The derivative of \(f\) is, \[
f'(x)=\frac{a^2-2 x^2}{\sqrt{a^2-x^2}}.
\] Thus the critical number is \(x=\frac{a}{\sqrt{2}}.\) The maximum value of the function \(f\) is

\[
f\left(\frac{a}{\sqrt{2}}\right)=\frac{a}{\sqrt{2}}\sqrt{a^2-\frac{a^2}{2}}=\frac{a^2}{2}.
\] Thus the points are \[
\left(\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}}\right) \text{ and } \left(\frac{-a}{\sqrt{2}},\frac{-a}{\sqrt{2}}\right).
\]

## 3.34 Optimizing Area

**Example 3.52 **A woman plans to fence off a rectangular garden whose area is 64 \(\text{ft}^2.\) What should be the dimensions of the garden if she wants to minimize the amount of fencing used?

*Solution*. Let \(x\) and \(y\) be the dimensions of the rectangular plot. The fencing (perimeter) is \(P=2x+2y\) and the area is \(A=x y=64\) with domain \(x>0.\) We want to minimize \(P\) so we write \(P\) as a function of one variable, say \[
P=2x+\frac{2(64)}{x}.
\] Since the derivative of \(P\) is, \[
P'=2-\frac{128}{x^2}=\frac{2x^2-128}{x^2}
\] and is continuous the only critical number is when \(P'=0\) with \(x>0\) which is when \(x=8\). In this example we apply the First Derivative Test to verify that the critical number \(x=8\) is useful as follows \[
\begin{array}{c|c|c|l}
\text{Interval} & f & f' & \text{Conclusion} \\ \hline
0<x<8 & & - & \text{decreasing}\\
x=8 & 8 & 0 & \text{local minimum}\\
x>8 & & + & \text{increasing}\\
\end{array}
\] Since \(P\) is decreasing on \((0,8)\) and increasing on \((0,\infty)\), the function \(P\) has an absolute minimum at \(x=8\); and the dimensions of the garden should be \(8\) ft by \(8\) ft.

::: {#exm- } [Optimizing with Area] Someone with 750 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens? :::

*Solution*. Let \(x\) be the lengths of the 2 sides and let \(y\) be the lengths of the other 5 sides. Since there are three divides making the four pens \(2x+2y+3y=750\). The area of the four pens is \(A=x y,\) thus we can solve \(2x+2y+3y=750\) for \(y\) obtaining \(y=\frac{1}{5} (750-2 x).\) So a function of the area is \[
A(x)=\frac{1}{5}x (750-2 x).
\] The derivative of \(A\) is, \(A'(x)=\frac{1}{5} (750-4 x)\) and so the critical number is \(x=750/4.\) The largest possible area is \(A\left(\frac{750}{4}\right)=14062.5\) square feet.

## 3.35 Optimizing Angles

**Example 3.53 **The bottom of an 8-ft-high mural painted on a vertical wall is 13 ft above the ground. The lens of a camera fixed to a tripod is 4 ft above the ground. How far from the wall should the camera be placed to photograph the mural with the largest possible angle?

*Solution*. Let the horizontal distance from the camera to the wall be \(x.\) Let $$ be the angle of elevation from the camera lens to the top of the mural and let $$ the angle of elevation from the camera to the bottom of the mural. Also, let \(\theta =\alpha -\beta .\) Then

\[
\theta (x)=\tan ^{-1}\frac{17}{x}-\tan ^{-1}\frac{9}{x}.
\] Since the first derivative of \(\theta\) is, \[\begin{align*}
\frac{d\theta }{dx} & = \frac{1}{1+\left(\frac{17}{x}\right)^2}\left(\frac{-17}{x^2}\right)-\frac{1}{1+\left(\frac{9}{x}\right)^2}\left(\frac{-9}{x^2}\right) \\
& =\frac{-17}{x^2+289}+\frac{9}{x^2+81} \\
& =-\frac{8 \left(x^2-153\right)}{\left(x^2+81\right) \left(x^2+289\right)}
\end{align*}\] we see that \(\frac{d\theta }{dx}=0\) and \(x>0\) when \(x=\sqrt{153}.\) Applying the First Derivative Test, the largest possible angle is when \(x=\sqrt{153}\) \(=3\sqrt{17}\) or approximately \(12.4\) feet.

## 3.36 Optimizing Distance

**Example 3.54 **A truck is 250 mi due east of a sports car and is traveling west at a constant speed of 60 mi/h. Meanwhile, the sports car is going north at 80 mi/h. When will the truck and the car be closest to each other? What is the minimum distance between them?

*Solution*. Draw a figure with the car at the origin of a Cartesian coordinate system and the truck at \((250,0).\) At time \(t,\) (in hours) the truck is at position \((250-x,0),\) while the car is at \((0,y).\) Let \(D\) be the distance that separates them. Then \(\frac{dx}{dt}=60\) and \(\frac{dy}{dt}=80\) so that \(x=60 t\) and \(y=80 t.\) We will minimize the square of the distance, \(D^2=(250-x)^2+y^2\), \[ D^2=(250-60 t)^2+(80t)^2=2500 \left(25-12 t+4 t^2\right)\] Since \(\frac{dD^2}{dt}=10,000(2t-3)\) the derivative of the distance squared is 0 when \(t=1.5 \text{hr}.\) Substituting into the equation for \(D^2\) produces the shortest distance: \(x=60(1.5)=90\) and \(y=80(1.5)=120\).

Thus, \(D^2=(250-90)^2+120^2=1600(25)\) and so \(D=40(5)=200\) which is the minimum distance (because there is no maximum distance and the Extreme Value Theorem applies).

## 3.37 Optimizing Time

**Example 3.55 **A jeep is on the desert at a point \(P\) located 40 km from a point \(Q\), which lies on a long straight road. The driver can travel at 45 km/h on the desert and 75 km/h on the road. The driver will win a prize if he arrives at the finish line at point \(F\), 50 km from \(Q\), in 84 minutes or less. What route should he travel to minimize the time of travel? Does he win the prize?

*Solution*. Suppose that the driver heads for a point \(S\) located \(x\) km down the road from \(Q\) towards his destination. We want to minimize the time. We will need to remember the formula \(d=r t,\) or in terms of time \(t=d/r.\) Since the distance between \(P\) and \(S\) is \(\sqrt{x^2+1600}\) and the distance between \(S\) and \(F\) is \(50-x,\) the total time is given by

\[
T(x)=\frac{\sqrt{x^2+1600}}{45}+\frac{50-x}{75} \text{where } 0\leq x\leq 50.
\] Since \[T'(x)=\frac{x}{45 \sqrt{x^2+1600}}-\frac{1}{75}=\frac{5 x-3 \sqrt{1600+x^2}}{225 \sqrt{1600+x^2}}\] we find that \(x=30\) is the only critical number of \(T.\) To find the extreme values we evaluate \(T\) at the endpoints, we find \[
T(30)=\frac{\sqrt{(30)^2+1600}}{45}+\frac{50-30}{75}=\frac{62}{45}=1.37778 \text{ hr}
\] \[
T(0)=\frac{\sqrt{(0)^2+1600}}{45}+\frac{50-0}{75}=\frac{14}{9}=1.55556 \text{ hr}
\] \[
T(50)=\frac{\sqrt{(50)^2+1600}}{45}+\frac{50-50}{75}=\frac{2 \sqrt{41}}{9}=1.42292 \text{ hr}
\] Therefore, the driver can minimize the total driving time by heading for a point that is \(30 \text{km}\) from the point \(Q\) and then traveling on the road to point \(D.\) He wins the prize because the minimal route is only 83 minutes.

## 3.38 Marginal Analysis

Marginal analysis is concerned with the way quantities such as price, cost, revenue, and profit vary with small changes in the level of production. The demand function \(p(x)\) is defined to be the price that consumers will pay for each unit of the commodity when \(x\) units are brought to market. Then \(R(x)=x p(x)\) is the total revenue function derived from the sale of the \(x\) units and \(P(x)=R(x)-C(x)\) is the total profit function where \(C(x)\) is the total cost function for producing \(x\) units.

::: {#exm- } [Optimizing Profits] A toy manufacturer produces an inexpensive doll (Dolly) and an expensive doll (Polly) in units of \(x\) hundred and \(y\) hundred, respectively. Suppose it is possible to produce the dolls in such a way that \(y=\frac{82-10x}{10-x}\) with \(0\leq x\leq 8\) and that the company receives twice as much for selling a Polly doll as for selling a Dolly doll. Find the level of production for both \(x\) and \(y\) for which total revenue derived from selling these dolls is maximized. What vital assumption must be made about sales in the model? :::

::: {#exm- } [Optimizing Revenue] A business manager estimates that when \(p\) dollars are charged for every unit of a product, the sales will be \(x=380-20p\) units. At this level of production, the average cost is modeled by \[ A(x)=5+\frac{x}{30}. \] - Find the total revenue and total cost functions, and express the profit as a function of \(x.\) - What price should the manufacturer charge to maximize profit? - What is the maximum profit? :::

::: {#exm- } [Optimizing Costs] Suppose the total cost (in dollars) of manufacturing \(x\) units of a certain commodity is \(C(x)=3x^2+5x+75\). - At what level of production is the average cost per unit the smallest? - At what level of production is the average cost per unit equal to the marginal cost? - Graph the average cost and the marginal cost on the same set of axes, for \(x>0.\) :::

## 3.39 Exercises

**Exercise 3.65 **You are planning to close off a corner of the first quadrant with a line segment 20 units long running from \((a,0)\) to \((0,b).\) Show that the area of the triangle enclosed by the segment is largest when \(a=b.\)

**Exercise 3.66 **Your iron work has contracted to design and build a \(500 \text{ft}^3,\) square-based, open-top, rectangular steel holding tank for a paper company. The tank is to be made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to find dimensions for the base and height that will make the tank weigh as little as possible. (a) What dimensions do you tell the shop to use? (b) Briefly describe how you took weight into account.

**Exercise 3.67 **The bottom of an 8-ft-high mural painted on a vertical wall is 13 ft above the ground. The lens of a camera fixed to a tripod is 4 ft above the ground. How far from the wall should the camera be placed to photograph the mural with the largest possible angle?

**Exercise 3.68 **A 1125 \(\text{ft}^3\) open-top rectangular tank with a square base \(x\) ft on a side and \(y\) ft deep is to be built with its top flush with the ground to catch runoff water. The costs associated with the tank involve not only the material from which the tank is made but also an excavation charge proportional to the product \(x y.\) (a) If the total cost is \(c=5\left(x^2+ 4 x y\right)+10 x y\) what values of \(x\) and \(y\) will minimize it? (b) Give a possible scenario fro the cost function in part (a).

**Exercise 3.69 **Two sides of a triangle have lengths \(a\) and \(b\), and the angle between then \(\theta .\) What value of $$ will maximize the triangle’s area?

**Exercise 3.70 **The height of an object moving vertically is given by \(s=-16 t^2+96 t+112\) with \(s\) in feet and \(t\) in seconds. Find the object’s velocity when \(t=0.\) Find its maximum height and when it occurs. Also find its velocity when \(s=0.\)

**Exercise 3.71 **Find the dimensions of the right circular cylinder of largest volume that can be inscribed in a sphere of radius \(R.\)

**Exercise 3.72 **Jane is 2 mi offshore in a boat and wishes to reach a coastal village 6 mi down a straight shoreline from the point nearest the boat. She can now row 2 mph can walk 5 mph. Where should she land her boat to reach the village in the least amount of time?

**Exercise 3.73 **The positions of two particles on the \(s\)-axis are \(s_1=\sin t\) and \(s_2=\sin (t+\pi /3)\) with \(s_1\) and \(s_2\) in meters and \(t\) in seconds. (a) At what time(s) in the interval $0t$ do the particles meet? (b) What is the farthest apart that the particles ever get? (c) When in the interval $0t$ is the distance between the particles chaining the fastest?

**Exercise 3.74 **A truck is 250 mi due east of a sports car and is traveling west at a constant speed of 60 mi/h. Meanwhile, the sports car is going north at 80 mi/h. When will the truck and the car be closest to each other? What is the minimum distance between them?

**Exercise 3.75 **A jeep is on the desert at a point \(P\) located 40 km from a point \(Q\), which lies on a long straight road. The driver can travel at 45 km/h on the desert and 75 km//h on the road. The driver will win a prize if he arrives at the finish line at point \(F\), 50 km from \(Q\), in 84 minutes or less. What route should he travel to minimize the time of travel? Does he win the prize?

## 3.40 Indeterminate Forms

We say that \(\lim_{x\to \infty }\frac{3^x-1}{x^3}\) has the intermediate form **intermediate form** \(\frac{\infty }{\infty}\) because \(3^x\to \infty\) and \(x^3\to \infty\) as \(x\to \infty\)

In this section we will consider the following seven indeterminate forms \[ \frac{\infty }{\infty }, \qquad \frac{0}{0}, \qquad \infty-\infty, \qquad 0\cdot \infty, \qquad \infty^0, \qquad 0^0, \qquad 1^{\infty}. \]

## 3.41 l’H^opital’s Rule

Our main investigative tool will be l’H^opital’s rule which says that the limit of a quotient of functions \(f\) and \(g\) is equal to the limit of the quotient of their derivatives \(f'\) and \(g'\) provided some conditions are satisfied. It is especially important to verify the conditions regarding the limits of \(f\) and \(g\) before applying l’H^opital’s rule.

::: {#thm- } l’H^opital’s Rule Suppose \(f\) and \(g\) are differentiable functions and \(g'(x)\neq 0\) on an open interval \(I\) that contains \(c\). Suppose \[ \lim_{x\to c}\frac{f(x)}{g(x)} \] produces an intermediate form \(\frac{0}{0}\) or \(\frac{\infty }{\infty }\) and that \[ \lim_{x\to c}\frac{f'(x)}{g'(x)}\] exists or is infinite, then \[ \lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}. \] :::

To illustrate l’H^opital’s Rule we will evaluate several limits.

## 3.42 Indeterminate Forms \(0/0\) and \(\infty/\infty\)

**Example 3.56 **Evaluate \(\displaystyle \lim_{x\to \infty }\frac{\ln x}{x}\) using l’H^opital’s rule.

*Solution*. The given limit has the indeterminate form \(\frac{\infty }{\infty }\) since \(\lim_{x\to \infty}\ln x=+\infty\) and \(\lim_{x\to +\infty }x=+\infty.\) We apply l’H^opital’s rule to evaluate the limit as follows \[
\lim_{x\to +\infty }\frac{\ln x}{x}=\lim_{x\to +\infty }\frac{\frac{1}{x}}{1}=0.
\]

**Example 3.57 **Evaluate \[
\lim_{x\to -\infty }\frac{x^2}{e^{-x}}
\] using l’H^opital’s rule.

*Solution*. The given limit has the indeterminate form \(\frac{\infty }{\infty }\) since \(\lim_{x\to -\infty }x^2=+\infty\) and \(\lim_{x\to -\infty }e^{-x}=+\infty .\) We apply l’H^opital’s rule to evaluate the limit as follows \[
\lim_{x\to -\infty }\frac{x^2}{e^{-x}}=\lim_{x\to -\infty }\frac{2x}{-e^{-x}}=\lim_{x\to -\infty }\frac{2}{e^{-x}}=0.
\] Thus illustrating that L’Hopital’s rule can be used multiple times.

We are justified in calling \(\frac{\infty}{\infty}\) an indeterminate form since \[ \lim_{x\to +\infty }\frac{\ln x}{x}=0 \neq 1=\lim_{x\to 0 }\frac{\sin x}{x}. \]

**Example 3.58 **Evaluate \[
\lim_{x\to 0 }\frac{\sin x}{x}
\] using l’H^opital’s rule.

*Solution*. The given limit has the indeterminate form \(\frac{0}{0 }\) since \(\lim_{x\to 0 }\sin x=0\) and \(\lim_{x\to 0 }x=0 .\) We apply l’H^opital’s rule to evaluate the limit as follows \[
\lim_{x\to 0 }\frac{\sin x}{x}=\lim_{x\to 0 }\frac{\cos x}{1}=1.
\]

**Example 3.59 **Evaluate \[
\lim_{x\to 0}\frac{e^{2x}-1}{x}
\] using l’H^opital’s rule.

*Solution*. The given limit has the indeterminate form \(\frac{0}{0}\) since \(\lim_{x\to 0}e^{2x}-1=0\) and \(\lim_{x\to 0}x=0.\) We apply l’H^opital’s rule to evaluate the limit as follows \[
\lim_{x\to 0}\frac{e^{2x}-1}{x}=\lim_{x\to 0}\frac{2 e^{2 x}}{1}=2.
\]

We are justified in calling \(\frac{0}{0}\) an indeterminate form since \[ \lim_{x\to 0 }\frac{\sin x}{x}=1 \neq 2=\lim_{x\to 0}\frac{e^{2x}-1}{x}. \]

**Example 3.60 **Evaluate \[
\lim_{x\to 0} \frac{a^x-b^x}{x}
\] using l’H^opital’s rule, where \(a,b>0\).

*Solution*. The given limit has the indeterminate form \(\frac{0}{0}\) since \(\lim_{x\to 0}a^x-b^x=0\) and \(\lim_{x\to 0}x=0.\) We apply l’H^opital’s rule to evaluate the limit as follows \[
\lim_{x\to 0}\frac{a^x-b^x}{x}=\lim_{x\to 0}\frac{(\ln a) a^x-(\ln b) b^x}{1}=\ln a-\ln b=\ln \frac{a}{b}.
\]

## 3.43 Indeterminate Forms \(\infty-\infty\) and \(0\cdot \infty\)

**Example 3.61 **Evaluate \[
\lim_{x\to +\infty }e^{-x}\sqrt{x}
\] using l’H^opital’s rule.

*Solution*. The given limit has indeterminate form \(0\cdot \infty\) since \(\lim_{x\to +\infty }e^{-x}=0\) and \(\lim_{x\to +\infty }\sqrt{x}=+\infty .\) We apply l’H^opital’s rule to evaluate the limit as follows \[
\lim_{x\to +\infty }\frac{\sqrt{x}}{e^x}=\lim_{x\to +\infty }\frac{\frac{1}{2\sqrt{x}}}{e^x}=\lim_{x\to +\infty }\frac{1}{2\sqrt{x}e^x}=0.
\]

**Example 3.62 **Evaluate \[
\lim_{x\to 1^+ } \left(\frac{1}{\ln x}-\frac{1}{x-1}\right)
\] using l’H^opital’s rule.

*Solution*. The given limit has indeterminate form \(\infty -\infty\) since \(\lim_{x\to 1^+ }\frac{1}{\ln x}=+\infty\) and \(\lim_{x\to 1^+ }\frac{1}{x-1}=+\infty .\) We apply l’H^opital’s rule to evaluate the limit as follows \[\begin{align*}
& \lim_{x\to 1^+ } \left(\frac{1}{\ln x}-\frac{1}{x-1}\right)
=\lim_{x\to 1^+ } \frac{x-1-\ln x}{(x-1)\ln x}
=\lim_{x\to 1^+ } \frac{1-\frac{1}{x}}{\ln x+\frac{x-1}{x}} \\
& \qquad =\lim_{x\to 1^+ }\frac{x-1}{x\ln x+x-1}
=\lim_{x\to 1^+ } \frac{1}{\ln x+x\left(\frac{1}{x}\right)+1}
=\frac{1}{2}
\end{align*}\]

## 3.44 Indeterminate Forms \(\infty^0\), \(0^0\), and \(1^{\infty}\)

The limit \[ \lim_{x\to a} f(x)^{g(x)} \] is said to an indeterminate form of the type

- [] \(\infty^0\) if \(\lim_{x\to a} f(x)=\infty\) and \(\lim_{x\to a} g(x)=0\), $$-5pt]
- [] \(0^0\) if \(\lim_{x\to a} f(x)=0\) and \(\lim_{x\to a} g(x)=0\), and $$-5pt]
- [] \(1^{\infty}\) if \(\lim_{x\to a} f(x)=1\) and \(\lim_{x\to a} g(x)=\pm \infty\).

With these indeterminate forms we can apply the identity \(f(x)^{g(x)}=e^{g(x)\ln f(x)}\). Since we know that the exponential function is continuous, we can use \[ \lim_{x\to a} f(x)^{g(x)}=\lim_{x\to a}e^{g(x)\ln f(x)}=e^{\left( \displaystyle \lim_{x\to a} g(x)\ln f(x)\right)} \] and then possibly apply l’H^opital’s rule, of course we must first check whether or not \(\lim_{x\to a} g(x)\ln f(x)\) exists first.

**Example 3.63 **Evaluate \[
\lim_{x\to 0^+}(\sin x)^x
\] using l’H^opital’s rule.

*Solution*. The given limit has the indeterminate form \(0^0\) since \(\lim_{x\to 0^+} \sin x=0\) and \(\lim_{x\to 0^+}x=0.\) Since \(\lim_{x\to 0^+}\ln (\sin x)=+\infty\) and \(\lim_{x\to 0^+}\frac{1}{x}=+\infty\) we have indeterminate form of \(\frac{\infty}{\infty }\) and so we apply l’H^opital’s rule to evaluate the following limit \[
\lim_{x\to 0^+} x \ln \sin x
=\lim_{x\to 0^+}\frac{\ln (\sin x)}{\frac{1}{x}}
=\lim_{x\to 0^+}\frac{\frac{\cos x}{\sin x}}{\frac{-1}{x^2}}
=\lim_{x\to 0^+}\frac{-x^2}{\tan x}
=\lim_{x\to 0^+}\frac{-2x}{\sec ^2x}
=0.
\] Therefore, since the exponential function is continuous, \[
\lim_{x\to 0^+}(\sin x)^x
=\lim_{x\to 0^+} e^{x \ln \sin x}
= e^{\displaystyle \left( \lim_{x\to 0^+} x \ln \sin x\right) }
%=e^{\left(\displaystyle \lim_{x\to 0^+}\frac{\ln (\sin x)}{\frac{1}{x}}\right)}
=e^0
=1.
\]

**Example 3.64 **Evaluate \[
\lim_{x\to +\infty }x^{1/x}
\] using l’H^opital’s rule.

*Solution*. The given limit has the indeterminate form \(\infty ^0\) since \(\lim_{x\to \infty }\frac{1}{x}=0\) and \(\lim_{x\to +\infty }\frac{1}{x}=+\infty .\)

Since \(\lim_{x\to \infty}\ln x=+\infty\) and \(\lim_{x\to \infty}x=+\infty\) the following limit has indeterminate form \(\frac{\infty }{\infty }\) and so we apply l’H^opital’s rule as follows \[
\lim_{x\to +\infty }\frac{1}{x}\ln x
=\lim_{x\to +\infty }\frac{\frac{1}{x}}{1}
=0
\] Since the exponential function is continuous \[
\lim_{x\to +\infty }x^{1/x}
=\lim_{x\to +\infty }e^{\frac{1}{x}\ln x}
=e^{\left(\displaystyle \lim_{x\to +\infty }\frac{1}{x}\ln x \right)}
=e^0
=1.
\]

**Example 3.65 **Evaluate \[
L=\lim_{x\to 0^+}(1+\sin 4x)^{\cot x}
\] using l’H^opital’s rule.

*Solution*. The given limit has the indeterminate form \(1^{\infty }\) since \(\lim_{x\to 0^+}1+\sin 4x=1\) and \(\lim_{x\to 0^+}\cot x=+\infty .\) Notice \(\lim_{x\to 0^+}\frac{\ln (1+\sin 4x)}{\tan x}\) has indeterminate form \(\frac{0}{0}\) since \(\lim_{x\to 0^+}\ln (1+\sin 4x)=0\) and \(\lim_{x\to 0^+}\tan x=0.\) We apply l’H^opital’s rule to find \[
\lim_{x\to 0^+}\frac{\ln (1+\sin 4x)}{\tan x}
=\lim_{x\to 0^+}\frac{\frac{4\cos 4x}{1+\sin 4x}}{\sec ^2x}
=\frac{\frac{4}{1+0}}{1}
=4.
\] Since the exponential function is continuous \[\begin{align*}
\lim_{x\to 0^+}(1+\sin 4x)^{\cot x}
& =\lim_{x\to 0^+}e^{\cot x \ln (1+\sin 4x)} \\
& =e^{\left( \displaystyle \lim_{x\to 0^+} \cot x \ln (1+\sin 4x) \right)}
=e^4.
\end{align*}\]

## 3.45 Change of Variable

In the next example we show how making a change of variable can simplify the process of evaluating a limit.

**Example 3.66 **Evaluate \[
\lim_{x\to +\infty }x^5\left[\sin \left(\frac{1}{x}\right)-\frac{1}{x}+\frac{1}{6x^3}\right]
\] using l’H^opital’s rule.

*Solution*. We make a change of variable to simplify the expression, namely \(u=\frac{1}{x}.\) Since \(u\to 0\) as \(x\to +\infty ,\) we have \[\begin{align*}
\lim_{x\to +\infty }x^5\left[\sin \left(\frac{1}{x}\right)-\frac{1}{x}+\frac{1}{6x^3}\right]
=\lim_{u\to 0}\frac{\sin (u)-u+\frac{1}{6}u^3}{u^5}
\end{align*}\] Now we have indeterminate form \(\frac{0}{0}\) and so applying l’H^opital rule several times yields \[\begin{align*}
& =\lim_{u\to 0}\frac{\sin (u)-u+\frac{1}{6}u^3}{u^5}
=\lim_{u\to 0}\frac{(\cos u)-1+\frac{1}{2}u^2}{5u^4}
=\lim_{u\to 0}\frac{(-\sin u )+u}{20u^3} \\
& =\lim_{u\to 0}\frac{(-\cos u )+1}{60u^2}
=\lim_{u\to 0}\frac{\sin u }{120 u}
=\lim_{u\to 0}\frac{\cos u }{120}
=\frac{1}{120}.
\end{align*}\]

## 3.46 l’H^opital’s Rule Fails

In this final example we illustrate one way in which l’H^opital’s rule can fail even though the value of the limit is finite.

**Example 3.67 **Try to evaluate \[
\lim_{x\to +\infty }\frac{x+\sin x}{x-\cos x}
\] using l’H^opital’s rule.

*Solution*. This limit has indeterminate form since \[
\lim_{x\to +\infty }(x+\sin x)=+\infty \hspace{.5cm}\text{and}\hspace{.5cm} \lim_{x\to +\infty }(x+\cos x)=+\infty .
\] If we try to apply L’Hospitals’s Rule we find, \[ \lim_{x\to +\infty }\frac{x+\sin x}{x-\cos x}=\lim_{x\to \infty }\frac{1+\cos x}{1+\sin x}\] but the limit \(\lim_{x\to \infty }\frac{1+\cos x}{1+\sin x}\) does not exist because of osculating behavior, so we can not use l’H^opital’s rule. To correctly find this limit we divide by \(x\) as follows \[
L=\lim_{x\to +\infty }\frac{x+\sin x}{x-\cos x}
=\lim_{x\to +\infty }\frac{\frac{x}{x}+\frac{\sin x}{x}}{\frac{x}{x}-\frac{\cos x}{x}}
=\frac{1+0}{1-0}
=1.
\]

## 3.47 Exercises

**Exercise 3.76 **Use l’H^opital’s rule to find the following limits.

- \(\lim_{x\to 2}\frac{x-2}{x^2-4}.\)
- \(\lim_{x\to 1}\frac{x^3-1}{4x^3-x-3}.\)
- \(\lim_{x\to -5}\frac{x^2-25}{x+5}.\)
- \(\lim_{x\to 0}\frac{\sin x^2}{x}.\)
- \(\lim_{x\to 0}\frac{\sin x -x}{x^3}.\)
- \(\lim_{x\to \pi /2}\frac{1-\sin x}{1+ \cos 2x}.\)
- \(\lim_{x\to \pi /2}\frac{\ln ( \csc x)}{\left(x-\frac{\pi }{2}\right)^2}.\)
- \(\lim_{x\to \left(\frac{\pi }{2}\right)^-}\left(x-\frac{\pi }{2}\right)\sec x.\)
- \(\lim_{x\to 0}\frac{\left(\frac{1}{2}\right)^x-1}{x}.\)
- \(\lim_{x\to 0^+}\frac{\ln \left(e^x-1\right)}{\ln x}.\)
- \(\lim_{x\to \infty }(\ln 2x-\ln (x+1)).\)
- \(\lim_{x\to 0^+}\left(\frac{3x+1}{x}-\frac{1}{\sin x}\right).\)
- \(\lim_{x\to 0}\frac{\cos x-1}{e^x-x-1}.\)
- \(\lim_{x\to \infty } x^2e^{-x}.\)
- \(\lim_{x\to \infty }(\ln x)^{1/x}.\)
- \(\lim_{x\to \infty }x^{1/ \ln x}.\)
- \(\lim_{x\to \infty }(1+2x)^{1/ 2 \ln x}.\)
- \(\lim_{x\to 0^+}\frac{\sqrt{x}}{\sqrt{\sin x}}.\)

**Exercise 3.77 **Let \[
f(x)=\left\{
\begin{array}{cc}
x+2 & x\neq 0 \\
0 & x=0
\end{array}
\right.
\qquad \text{and} \qquad
g(x)=\left\{
\begin{array}{cc}
x+1 & x\neq 0 \\
0 & x=0
\end{array}
\right.
\] Show that \(\lim_{x\to 0}\frac{f'(x)}{g'(x)}=1\) but \(\lim_{x\to 0}\frac{f(x)}{g(x)}=2\). Explain why this does not contradict l’H^opital’s Rule.

**Exercise 3.78 **Find constant \(a\) and \(b\) so that \(\lim_{x\to 0}\left(\frac{\sin 2x}{x^3}+\frac{a}{x^2+b}\right)=1.\)

**Exercise 3.79 **Find all values of \(a\) and \(b\) so that \(\lim_{x\to 0}\frac{\sin a x+b x}{x^3}=36.\)

**Exercise 3.80 **Find the values of \(a\) so that \(\lim_{x\to 0}\frac{a-\cos b x}{x^2}=2.\)

**Exercise 3.81 **For a certain value of \(a,\) the limit \(\lim_{x\to +\infty }\left(x^4+5x^3+3\right)^a-x\) is finite and nonzero. Find \(a\) and then use l’H^opital’s rule to compute the limit.

**Exercise 3.82 **Evaluate \(\lim_{x\to a}\frac{\sqrt{2a^3x-x^4}-a\sqrt[3]{a^2x}}{a-\sqrt[4]{a x^3}}.\)

**Exercise 3.83 **Determine which values of constants \(a\) and \(b\) is it true that \[
\lim_{x\to 0}\left(x^{-3}\sin 7x+a x^{-2}+b\right)=-2?
\]