Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 5

Applications of the Derivative

Objectives

When you have completed this unit, you should be able to

1. use calculus to sketch the graph of a function in detail.
2. solve optimization problems.
3. apply Newton’s method to solve equations of the form $f(x)=0$.

To sketch the graph of a function, we must be methodical and clear about the information we obtain at each of the steps described in the first seven sections of this unit, and listed on pages 232-234 of the textbook. “Optimization” problems are so named because they involve decisions one can make to optimize the resources available under constraints. We end this unit with a discussion of one more application of the derivative—Newton’s method.

Domain, x-y Intercepts, and Symmetries

In this section, we discuss how to obtain basic information about the graph of a function without using the derivative.

Sketching starts with the domain of a function as defined in Definition 2.6. The domain indicates where the function may have a vertical asymptote, and where the graph is located with respect to the $x$-axis. The functions we work with in this course may have a vertical asymptote in the following two possible cases:

1. points where the function is not defined.
2. the endpoints of the open intervals of the function’s domain.

Therefore, a function has no vertical asymptotes if its domain is ${\displaystyle ℝ}$, all real numbers.

Example 5.1. The domain of the function

${\displaystyle f\left(x\right)=\sqrt{x^2-9}}$

consists of all $x$ such that ${\displaystyle x^2-9\ge 0}$.

If we solve this inequality, we obtain ${\displaystyle x^2\ge 9}$.

Taking the square roots gives us ${\displaystyle \left|x\right|\ge 3\text{.}}$

Therefore, we have two inequalities—$x \leq -3$ and $x \geq 3$—and in interval notation, the domain is $(-\infty,-3] \cup [3,\infty)$.

Hence, the graph of this function does not exist between the vertical lines ${\displaystyle x=-3}$ and ${\displaystyle x=3}$. Since intervals of the domain are closed at the endpoints ${\displaystyle 3}$ and ${\displaystyle -3}$, the function does not have vertical asymptotes.

Example 5.2. The domain of the function

${\displaystyle h\left(x\right)={\displaystyle \frac{4x-5}{4x^2-6x}}}$

consists of all $x$ such that the denominator ${\displaystyle 4x^2-6x}$ is not zero.

If ${\displaystyle 4x^2-6x=0}$, then ${\displaystyle 2x\left(2x-3\right)=0}$ and ${\displaystyle x=0}$ or ${\displaystyle x={\displaystyle \frac{3}{2}}\text{.}}$

The domain is all real numbers except ${\displaystyle 0}$ and ${\displaystyle 3∕2}$. So, this function may have vertical asymptotes at ${\displaystyle x=0}$ or ${\displaystyle x=3∕2}$.

Example 5.3. The domain of the function

${\displaystyle g\left(x\right)={\displaystyle \frac{x^2+1}{x{\left(2x-7\right)}^2}}}$

consists of $x$ such that ${\displaystyle x{\left(2x-7\right)}^2\ne 0}$, hence ${\displaystyle x\ne 0}$ and ${\displaystyle x\ne 7∕2}$

The domain is

${\displaystyle D_g=ℝ-\left\{0,{\displaystyle \frac{7}{2}}\right\}\text{,}}$

and again, the vertical lines ${\displaystyle x=0}$ and ${\displaystyle x=7∕2}$ may be asymptotes of the function.

Example 5.4. The domain of the function

${\displaystyle h\left(x\right)={\displaystyle \frac{x-4}{\sqrt{9x^2-24x+10}}}}$

consists of all $x$ such that ${\displaystyle 9x^2-24x+10>0}$.

Let us solve this inequality:

The domain is the union of the intervals

${\displaystyle \left(-\infty ,{\displaystyle \frac{4-\sqrt{6}}{3}}\right)\text{and}\left({\displaystyle \frac{4+\sqrt{6}}{3}},\infty \right)\text{.}}$

In this case, the vertical lines

${\displaystyle x={\displaystyle \frac{4-\sqrt{6}}{3}}\text{and}{\displaystyle \frac{4+\sqrt{6}}{3}}}$

may be vertical asymptotes of the function, and there is no graph between these two vertical lines.

Example 5.5. In the function,

${\displaystyle u\left(x\right)=\sqrt{1-\text{sin }x}\text{,}}$

we have ${\displaystyle 1-\text{sin }x\ge 0}$, but since ${\displaystyle \left|\text{sin }x\right|\le 1}$, it is the case that ${\displaystyle 1-\text{sin }x\ge 0}$ for any $x$, and the domain of the function is ${\displaystyle ℝ}$. Hence, there are no vertical asymptotes.

Example 5.6. To find the domain of the composition of two functions, we must consider the domain of each function.

In the function ${\displaystyle f\left(x\right)=\text{tan}\left(2x+1\right)}$, the domain of ${\displaystyle \text{tan }u}$ is the union of the intervals of the form

$\left( {\dfrac{{(2k + 1)\pi }}{2},\dfrac{{(2k + 3)\pi }}{2}} \right)$,  for any integer $k$.

Hence, the inside function ${\displaystyle 2x+1}$ must be in these intervals, that is

${\displaystyle {\displaystyle \frac{\left(2k+1\right)\pi }{2}}<2x+1<{\displaystyle \frac{\left(2k+3\right)\pi }{2}}}$

Solving for $x$:

${\displaystyle {\displaystyle \frac{\left(2k+1\right)\pi -2}{4}}<x<{\displaystyle \frac{\left(2k+3\right)\pi -2}{4}}}$

The domain is all intervals of the form

$\left( {\dfrac{{(2k + 1)\pi - 2}}{4},\dfrac{{(2k + 3)\pi - 2}}{4}} \right)$  for any integer $k$.

The function may have vertical asymptotes at

$x = \dfrac{{(2k + 1)\pi - 2}}{4}$   or   $x = \dfrac{{(2k + 3)\pi - 2}}{4}$   for any integer $k$.

The $x$ and $y$ intercepts are the points of the graph that intersect the $x$-axis and the $y$-axis, respectively.

To find the $x$-intercepts, we must find all $a$ such that $f(a) = 0$.

Example 5.7. For the function

${\displaystyle f\left(x\right)=\sqrt{x^2-9}\text{,}}$

we set ${\displaystyle \sqrt{x^2-9}=0}$ and solve for $x$. Thus ${\displaystyle x^2=9}$, if ${\displaystyle x=3}$ or ${\displaystyle x=-3}$, and the $x$ intercepts are $(3,0)$ and $(-3,0)$.

Example 5.8. A quotient

${\displaystyle g\left(x\right)={\displaystyle \frac{x^2+1}{x{\left(2x-7\right)}^2}}}$

is zero if the numerator is zero, but in this case, ${\displaystyle x^2+1\ne 0}$ for any real $x$; hence, the function has no $x$-intercepts.

Example 5.9. For the function

${\displaystyle u\left(x\right)=\sqrt{1-\text{sin }x}\text{,}}$

we set ${\displaystyle \sqrt{1-\text{sin }x}=0}$ and solve for $x$.

This strategy yields ${\displaystyle \text{sin }x=1}$, and we have

$x = \dfrac{{(4k + 1)\pi }}{2}$   for any integer $k$.

The function has infinitely many $x$-intercepts:

$\left( {\dfrac{{(4k + 1)\pi }}{2},0} \right)$   for any integer $k$.

Example 5.10. For the function

${\displaystyle f\left(x\right)=\text{tan}\left(2x+1\right)\text{,}}$

we have ${\displaystyle \text{tan}\left(2x+1\right)=0}$ if ${\displaystyle \text{sin}\left(2x+1\right)=0}$.

Then

${\displaystyle 2x+1=k\pi \text{,}}$

and solving for $x$, we conclude that

$\left({\displaystyle {\displaystyle \frac{k\pi -1}{2}}},0\right)$.

There may be instances where we cannot solve $f(x)=0$ algebraically, and in such cases, we may be able to locate the $x$-intercepts using the Intermediate Value Theorem (see page 67 of the textbook).

Theorem 5.2. The Intermediate Value Theorem

If $f$ is a continuous function on the interval ${\displaystyle \left[a,b\right]}$ and ${\displaystyle f\left(a\right)<M<f\left(b\right)}$, then there is number $c$ in the interval ${\displaystyle \left(a,b\right)}$—that is, ${\displaystyle a<c<b}$—such that ${\displaystyle f\left(c\right)=M}$.

Figure 5.1. Intermediate Value Theorem

In Figure 5.1, above, you can see the interpretation of this theorem.

For ${\displaystyle M=0}$, this theorem says that if ${\displaystyle f\left(a\right)<0<f\left(b\right)}$, then there is a $c$ such that ${\displaystyle f\left(c\right)=0}$.

Example 5.11. To find the $x$-intercept of the function

${\displaystyle g\left(x\right)=x^3+4x^2+x-4\text{,}}$

we must solve for $x$ in the equation ${\displaystyle x^3+4x^2+x-4=0}$. Since it is not easy to do so algebraically, we apply the Intermediate Value Theorem. The function is continuous everywhere, because $g$ is a polynomial. So, we must find two numbers, $a$ and $b$, such that ${\displaystyle g\left(a\right)<0<g\left(b\right)}$.

We observe that ${\displaystyle g\left(0\right)=-4<0}$ and ${\displaystyle g\left(1\right)=2>0}$. Hence, ${\displaystyle g\left(0\right)<0<g\left(1\right)}$, and there must be a number $c$ between ${\displaystyle 0}$ and ${\displaystyle 1}$ such that ${\displaystyle g\left(c\right)=0}$; hence, there is an $x$-intercept in the interval ${\displaystyle \left(0,1\right)}$.

You can check that ${\displaystyle g\left(-1\right)<0<g\left(-2\right)}$; therefore, there is another $x$-intercept on the interval ${\displaystyle (-2,-1)}$.

In the last section of this unit, we consider Newton’s method, which can also be used to locate $x$-intercepts.

Example 5.12. In Examples 5.1 to 5.3, we saw that zero is not in the domain of the functions

${\displaystyle f\left(x\right)=\sqrt{x^2-9}\text{,}h\left(x\right)={\displaystyle \frac{4x-5}{4x^2-6x}}\text{or}g\left(x\right)={\displaystyle \frac{x^2+1}{x{\left(2x-7\right)}^2}}\text{.}}$

Hence, these functions do not have $y$-intercepts—that is, their graphs do not intersect the $y$-axis.

Example 5.13. In Example 5.4, we have the function

${\displaystyle h\left(x\right)={\displaystyle \frac{x-4}{\sqrt{{\left(4-3x\right)}^2-6}}}\text{,}}$

and ${\displaystyle h\left(0\right)=-{\displaystyle \frac{4}{10}}}$. Hence, the $y$-intercept of the function is

${\displaystyle \left(0,-{\displaystyle \frac{4}{\sqrt{10}}}\right)}$

Example 5.14. In Example 5.6, we have the function

${\displaystyle f\left(x\right)=\text{tan}\left(2x+1\right)\text{,}}$

and ${\displaystyle f\left(0\right)=\text{tan}\left(1\right)}$.

Hence, the $y$-intercept of the function is ${\displaystyle \left(0,\text{tan}\left(1\right)\right)\approx \left(0,1\text{.}5574\right)}$.

To sketch the graph of a function, it helps to know when a function has certain symmetries.

Consider the graph in Figure 5.2, below.

Figure 5.2. Basic graph

If the graph in Figure 5.2 were symmetric with respect to the $y$-axis, then the complete graph would be as shown in Figure 5.3:

Figure 5.3. Basic graph symmetrical with respect to the $y$-axis

If the graph in Figure 5.2 were symmetric with respect to the origin (i.e., across the diagonal formed by the identity function), then the complete graph would be as shown in Figure 5.4:

Figure 5.4. Basic graph symmetrical with respect to the origin

If a function is even, then its graph is symmetric with respect to the $y$-axis (see Figure 5.3, above). If the function is odd, then its graph is symmetric with respect to the origin (see Figure 5.4).

The function in Figure 5.5, below, is even.

Figure 5.5. Even function

The function in Figure 5.6 is odd:

Figure 5.6. Odd function

To identify these symmetries, we use the following definition.

Example 5.15.

1. The function ${\displaystyle f\left(x\right)=x^4}$ is even because ${\displaystyle f\left(-x\right)={\left(-x\right)}^4=x^4=f\left(x\right)}$.

   In general, if $n$ is even, then ${\displaystyle E\left(x\right)=x^n}$ is even.

2. The function ${\displaystyle f\left(x\right)=x^3}$ is odd because ${\displaystyle f\left(-x\right)={\left(-x\right)}^3=-x^3=-f\left(x\right)}$.

   In general if $m$ is odd, then ${\displaystyle O\left(x\right)=x^m}$ is odd.

3. From the graphs of the trigonometric functions, we see that sine, tangent, cotangent and cosecant are odd functions; and cosine and secant are even.

   Remember that ${\displaystyle \text{sin}\left(-x\right)=-\text{sin }x}$ and ${\displaystyle \text{cos}\left(-x\right)=\text{cos }x}$

4. A constant function is even. [Why?]
5. The absolute value function is even.

Example 5.16. To determine whether the function ${\displaystyle h\left(x\right)=-2x^3+\text{sin}\left(2x^2\right)}$ is even, odd or neither, we evaluate the function at ${\displaystyle -x\text{.}}$

Hence,

${\displaystyle h\left(-x\right)=-2{\left(-x\right)}^3+\text{sin}\left(2{\left(-x\right)}^2\right)=2x^3+\text{sin}\left(2x^2\right)\text{.}}$

Since

${\displaystyle h\left(-x\right)\ne h\left(x\right)\text{and}h\left(-x\right)\ne -h\left(x\right)\text{,}}$

the function is neither even nor odd.

Example 5.17. The function ${\displaystyle g\left(x\right)={\displaystyle \frac{6x}{\text{cos}\left(4x\right)}}}$ is odd because

${\displaystyle g\left(-x\right)={\displaystyle \frac{6\left(-x\right)}{\text{cos}\left(4\left(-x\right)\right)}}=-{\displaystyle \frac{6x}{\text{cos}\left(4x\right)}}=-g\left(x\right)}$

Example 5.18. Observe that for $f(x)$ and ${\displaystyle g\left(x\right)}$ even, and ${\displaystyle h\left(x\right)}$ and ${\displaystyle p\left(x\right)}$ odd,

1. ${\displaystyle f\left(x\right)+g\left(x\right)}$ is even because ${\displaystyle f\left(-x\right)+g\left(-x\right)=f\left(x\right)+g\left(x\right)}$.

   So the function ${\displaystyle q\left(x\right)=x^4+\text{cos}\left(x\right)}$ is even.

2. ${\displaystyle h\left(x\right)+p\left(x\right)}$ is odd because ${\displaystyle h\left(-x\right)+p\left(-x\right)=-h\left(x\right)-p\left(x\right)=-\left(h\left(x\right)+p\left(x\right)\right)}$.

   So the function ${\displaystyle q\left(x\right)=x+\text{tan }x}$ is odd.

3. ${\displaystyle f\left(x\right)g\left(x\right)}$ is even because ${\displaystyle f\left(-x\right)g\left(-x\right)=f\left(x\right)g\left(x\right)}$.

   So the function ${\displaystyle q\left(x\right)=x^4\text{cos}\left(x\right)}$ is even.

4. ${\displaystyle h\left(x\right)p\left(x\right)}$ is even because ${\displaystyle h\left(-x\right)p\left(-x\right)=\left(-h\left(x\right)\right)\left(-p\left(x\right)\right)=h\left(x\right)p\left(x\right))}$.

   So the function ${\displaystyle q\left(x\right)=x\text{tan }x}$ is even.

5. ${\displaystyle f\left(x\right)h\left(x\right)}$ is odd because ${\displaystyle f\left(-x\right)h\left(-x\right)=f\left(x\right)\left(-h\left(x\right)\right)=-\left(f\left(x\right)h\left(x\right)\right)}$.

   So the function ${\displaystyle q\left(x\right)=x^4\text{tan}\left(x\right)}$ is odd.

6. ${\displaystyle f\left(g\left(x\right)\right)}$ is even because ${\displaystyle f\left(g\left(-x\right)\right)=f\left(g\left(x\right)\right)}$.

   So the function ${\displaystyle q\left(x\right)=\text{cos}\left(x^4\right)}$ is even.

7. ${\displaystyle f\left(h\left(x\right)\right)}$ is even because ${\displaystyle f\left(h\left(-x\right)\right)=f\left(-h\left(x\right)\right)=f\left(h\left(x\right)\right)}$.

   So the function ${\displaystyle q\left(x\right)=\text{cos}\left(x^3\right)}$ is even.

Example 5.19. Observe how we use Example 5.18 to decide if the functions given are even or odd.

1. ${\displaystyle p\left(x\right)=4x^9}$ is odd because the constant function ${\displaystyle 4}$ is even and the function ${\displaystyle x^9}$ is odd—Example 5.18, (e).
2. ${\displaystyle q\left(x\right)=x\text{ tan }x+x^2\text{ cos }x}$ is even—Example 5.18, (c), (d), (a).
3. ${\displaystyle g\left(x\right)=\text{sec}\left(4x\right)\text{ cos }x\text{ tan }x}$ is odd—Example 5.18, (d), (e).