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3.6: Analyzing the Graph of a Function

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

A student will be able to:

• Summarize the properties of function including intercepts, domain, range, continuity, asymptotes, relative extreme, concavity, points of inflection, limits at infinity.
• Apply the First and Second Derivative Tests to sketch graphs.

Introduction

In this lesson we summarize what we have learned about using derivatives to analyze the graphs of functions. We will demonstrate how these various methods can be applied to help us examine a function’s behavior and sketch its graph. Since we have already discussed the various techniques, this lesson will provide examples of using the techniques to analyze the examples of representative functions we introduced in the Lesson on Relations and Functions, particularly rational, polynomial, radical, and trigonometric functions. Before we begin our work on these examples, it may be useful to summarize the kind of information about functions we now can generate based on our previous discussions. Let's summarize our results in a table like the one shown because it provides a useful template with which to organize our findings.

Table Summary
$f(x)$ Analysis
Domain and Range
Intercepts and Zeros
Asymptotes and limits at infinity
Differentiability
Intervals where $f$ is increasing
Intervals where $f$ is decreasing
Relative extrema
Concavity
Inflection points

Example 1: Analyzing Rational Functions

Consider the function $f(x) = \frac{x^2 -4} {x^2 - 2x - 8}.$

General Properties: The function appears to have zeros at $x = \pm 2.$ However, once we factor the expression we see

$\frac{x^2 - 4} {x^2 -2x -8} = \frac{(x + 2)(x - 2)} {(x - 4)(x + 2)} = \frac{x - 2} {x - 4}.$

Hence, the function has a zero at $x = 2,$ there is a hole in the graph at $x = -2,$ the domain is $(-\infty, -2) \cup (-2, 4) \cup (4, +\infty),$ and the $y-$intercept is at $(0, \frac{1} {2}).$

Asymptotes and Limits at Infinity

Given the domain, we note that there is a vertical asymptote at $x = 4.$ To determine other asymptotes, we examine the limit of $f$ as $x \rightarrow \infty$ and $x \rightarrow -\infty$. We have

$\lim_{x \to \infty} \frac{x^2 -4} {x^2 -2x - 8} = \lim_{x \to \infty} \frac{\frac{x^2}{x^2} -\frac{4} {x^2}} {\frac{x^2} {x^2} -\frac{2x} {x^2} - \frac{8} {x^2}} = \lim_{x \to \infty} \frac{1-\frac{4} {x^2}} {1- \frac{2} {x} - \frac{8} {x^2}} = 1.$

Similarly, we see that $\lim_{x \to -\infty} \frac{x^2 -4} {x^2 -2x - 8} =1$. We also note that $y \neq \frac{2}{3}$ since $x \neq -2.$

Hence we have a horizontal asymptote at $y = 1.$

Differentiability

$f'(x) = \frac{-2x^2 -8x - 8} {(x^2 -2x - 8)} = \frac{-2} {(x - 4)^2} <0$. Hence the function is differentiable at every point of its domain, and since $f'(x) < 0$ on its domain, then $f$ is decreasing on its domain, $(-\infty, -2) \cup (-2, 4) \cup (4, +\infty)$.

$f''(x) = \frac{4} {(x - 4)^3}.$

$f''(x) \neq 0$ in the domain of $f.$ Hence there are no relative extrema and no inflection points.

So $f''(x) > 0$ when $x > 4.$ Hence the graph is concave up for $x > 4.$

Similarly, $f''(x) < 0$ when $x < 4.$ Hence the graph is concave down for $x < 4,$ $x \neq -2.$

Let’s summarize our results in the table before we sketch the graph.

Table Summary
$f(x)=\frac{x^2 -4}{x^2-2x-8}$ Analysis
Domain and Range $D=(-\infty,-2)\cup(-2,4)\cup(4,+\infty)$ $R=\big\{\mathrm{all\ reals}\neq 1 \text{ or } \frac {2}{3}\big\}$
Intercepts and Zeros zero at $x = 2,$ $y-$intercept at $\left(0,\frac{1}{2} \right)$
Asymptotes and limits at infinity VA at $x = 4,$ HA at $y = 1,$ hole in the graph at $x = -2$
Differentiability differentiable at every point of its domain
Intervals where $f$ is increasing nowhere
Intervals where $f$ is decreasing $(-\infty,-2)\cup(-2,4)\cup(4,+\infty)$
Relative extrema none
Concavity concave up in $(4,+\infty),$ concave down in $(-\infty,-2)\cup(-2,4)$
Inflection points none

Finally, we sketch the graph as follows:

Let’s look at examples of the other representative functions we introduced in Lesson 1.2.

Example 2:

Analyzing Polynomial Functions

Consider the function $f(x) = x^3 + 2 x^2 - x - 2.$

General Properties

The domain of $f$ is $(-\infty,+\infty)$ and the $y-$intercept at $(0,-2).$

The function can be factored

$f(x) = x^3 + 2 x^2 - x -2 = x^2 (x+2) -1 (x+2) = (x^2 -1) (x + 2) = (x-1) (x + 1) (x + 2)$

and thus has zeros at $x = \pm 1, -2.$

Asymptotes and limits at infinity

Given the domain, we note that there are no vertical asymptotes. We note that $\lim_{x\to \infty} f(x)=+\infty$ and $\lim_{x\to -\infty} f(x)=-\infty.$

Differentiability

$f'(x) = 3 x ^2 + 4 x -1 = 0$ if $x=\frac{-4\pm\sqrt{28}}{6}=\frac{-2\pm\sqrt{7}}{3}$ . These are the critical values. We note that the function is differentiable at every point of its domain.

$f'(x) > 0$ on $\left (-\infty,\frac{-2-\sqrt{7}}{3}\right )$ and $\left (\frac{-2+\sqrt{7}}{3}, +\infty\right )$; hence the function is increasing in these intervals.

Similarly, $f'(x) < 0$ on $\left (\frac{-2-\sqrt{7}}{3},\frac{-2+\sqrt{7}}{3}\right )$ and thus is $f$ decreasing there.

$f''(x) = 6x + 4 = 0$ if $x=-\frac{2}{3},$ where there is an inflection point.

In addition, $f''\left (\frac{-2-\sqrt{7}}{3}\right )<0$. Hence the graph has a relative maximum at $x=\frac{-2-\sqrt{7}}{3}$ and located at the point $(-1.55,0.63).$

We note that $f''(x) < 0$ for $x<-\frac{2}{3}$. The graph is concave down in $\left (-\infty,-\frac{2}{3}\right ).$

And we have $f''\left (\frac{-2+\sqrt{7}}{3}\right )>0$; hence the graph has a relative minimum at $x = \frac{-2+\sqrt{7}}{3}$ and located at the point $(0.22, -2.11).$

We note that $f''(x)> 0$ for $x>-\frac{2}{3}.$ The graph is concave up in $\left (-\frac{2}{3}, +\infty\right ).$

Table Summary
$f(x) = x^3 + 2 x^2 - x - 2$ Analysis
Domain and Range $D=(-\infty,+\infty), R=\left \{ \right . \mathrm{all\ reals}\left . \right \}$
Intercepts and Zeros zeros at $x = \pm 1, -2,$ $y,$ intercept at $(0,-2)$
Asymptotes and limits at infinity no asymptotes
Differentiability differentiable at every point of it’s domain
Intervals where $f$ is increasing $\left (-\infty,\frac{-2-\sqrt{7}}{3}\right )$ and $\left (\frac{-2+\sqrt{7}}{3},+\infty\right )$
Intervals where $f$ is decreasing $\left (\frac{-2-\sqrt{7}}{3},\frac{-2+\sqrt{7}}{3}\right )$
Relative extrema

relative maximum at $x=\frac{-2-\sqrt{7}}{3}$ and located at the point $(-1.55,0.63)$;

relative minimum at $x=\frac{-2+\sqrt{7}}{3}$ and located at the point $(0.22,-2.11).$

Concavity

concave up in $\left (-\frac{2}{3},+\infty\right )$.

concave down in $\left (-\infty,-\frac{2}{3}\right )$ .

Inflection points $x=-\frac{2}{3}$, located at the point $\left (-\frac{2}{3},-.74\right )$

Here is a sketch of the graph:

Consider the function $f(x)=\sqrt{2x-1}.$

General Properties

The domain of $f$ is $\left (\frac{1}{2},+\infty\right )$, and it has a zero at $x=\frac{1}{2}.$

Asymptotes and Limits at Infinity

Given the domain, we note that there are no vertical asymptotes. We note that $\lim_{x\to \infty} f(x)=+\infty$.

Differentiability

$f'(x)=\frac{1}{\sqrt{2x-1}}>0$ for the entire domain of $f.$ Hence $f$ is increasing everywhere in its domain. $f'(x)$ is not defined at $x=\frac{1}{2}$, so $x=\frac{1}{2}$ is a critical value.

$f''(x)=\frac{-1}{\sqrt{(2x-1)^3}}<0$ everywhere in $\left (\frac{1}{2},+\infty\right )$. Hence $f$ is concave down in $\left (\frac{1}{2},+\infty\right ).$ $f'(x)$ is not defined at $x=\frac{1}{2}$, so $x=\frac{1}{2}$ is an absolute minimum.

Table Summary
$f(x)=\sqrt{2x-1}$ Analysis
Domain and Range $D=\left (\frac{1}{2},+\infty\right ),R= \left \{y\ge0 \right \}$
Intercepts and Zeros zeros at $x=\frac{1}{2}$, no $y-$intercept
Asymptotes and limits at infinity no asymptotes
Differentiability differentiable in $\left (\frac{1}{2},+\infty\right )$
Intervals where $f$ is increasing everywhere in $D=\left (\frac{1}{2},+\infty\right )$
Intervals where $f$ is decreasing nowhere
Relative extrema

none

absolute minimum at $x=\frac{1}{2}$, located at $\left (\frac{1}{2},0\right )$

Concavity concave down in $\left (\frac{1}{2},+\infty\right )$
Inflection points none

Here is a sketch of the graph:

Example 4: Analyzing Trigonometric Functions

We will see that while trigonometric functions can be analyzed using what we know about derivatives, they will provide some interesting challenges that we will need to address. Consider the function $f(x) = x -2 \sin x$ on the interval $[-\pi,\pi].$

General Properties

We note that $f$ is a continuous function and thus attains an absolute maximum and minimum in $[-\pi,\pi].$ Its domain is $[-\pi,\pi]$ and its range is $R=\left \{-\pi \le y \le \pi\right \}.$

Differentiability

$f'(x) = 1 - 2 \cos x = 0$ at $x = - \frac{\pi} {3},\frac{\pi}{3}$.

Note that $f'(x) > 0$ on $\left (\frac{\pi} {3}, {\pi}\right )$ and $\left (-\pi, -\frac{\pi} {3}\right )$; therefore the function is increasing in $\left (\frac{\pi} {3}, \pi\right )$ and $\left (-\pi, -\frac{\pi} {3}\right )$.

Note that $f'(x) < 0$ on $\left (-\frac{\pi} {3},\frac{\pi}{3}\right )$; therefore the function is decreasing in $\left (-\frac{\pi} {3},\frac{\pi}{3}\right )$.

$f''(x) = 2 \sin x = 0$ if $x = 0, \pi, -\pi.$ Hence the critical values are at $x = - \pi, -\frac{\pi} {3}, \frac{\pi} {3}, \text { and } \pi.$

$f'' \left (\frac{\pi} {3}\right ) > 0;$ hence there is a relative minimum at $x = \frac{\pi} {3}.$

$f''\left (-\frac{\pi} {3}\right ) < 0$; hence there is a relative maximum at $x = -\frac{\pi} {3}.$

$f''(x) > 0$ on $(0, \pi)$ and $f''(x) < 0$ on $(-\pi, 0).$ Hence the graph is concave up and decreasing on $(0, \pi)$ and concave down on $(-\pi, 0).$ There is an inflection point at $x = 0,$ located at the point $(0, 0).$

Finally, there is absolute minimum at $x = -\pi,$ located at $(-\pi, -\pi),$ and an absolute maximum at $x = \pi,$ located at $(\pi, \pi).$

Table Summary
$f(x) = x - 2\sin x$ Analysis
Domain and Range $D = [-\pi, \pi],$ $R=\left \{-\pi \le y \le \pi\right \}$
Intercepts and Zeros $x = - \frac{\pi} {3}, \frac{\pi} {3}$
Asymptotes and limits at infinity no asymptotes
Differentiability differentiable in $D = [-\pi, \pi]$
Intervals where $f$ is increasing $\left (\frac{\pi} {3}, \pi\right )$ and $(-\pi, -\frac{\pi} {3})$
Intervals where $f$ is decreasing $\left (-\frac{\pi} {3}, \frac{\pi} {3}\right )$
Relative extrema

relative maximum at $x = -\pi/3$

relative minimum at $x = \pi/3$

absolute maximum at $x = \pi$, located at $\left (\pi, \pi \right )$

absolute minimum at $x = -\pi$, located at $\left (-\pi, -\pi \right )$

Concavity concave up in $(0, \pi)$
Inflection points $x = 0,$ located at the point $(0, 0)$

Lesson Summary

1. We summarized the properties of functions, including intercepts, domain, range, continuity, asymptotes, relative extreme, concavity, points of inflection, and limits at infinity.
2. We applied the First and Second Derivative Tests to sketch graphs.

Each of the problems above started with a function and then we analyzed its zeros, derivative, and concavity. Even without the function definition it is possible to sketch the graph if you know some key pieces of information. In the following video the narrator illustrates how to use information about the derivative of a function and given points on the function graph to sketch the function. Khan Academy Graphing with Calculus (9:44).

Another approach to this analysis is to look at a function, its derivative, and its second derivative on the same set of axes. This interactive applet called Curve Analysis allows you to trace function points on a graph and its first and second derivative. You can also enter new functions (including the ones from the examples above) to analyze the functions and their derivatives.

For more information about computing derivatives of higher orders (7.0), see Math Video Tutorials by James Sousa, Higher-Order Derivatives: Part 1 of 2 (7:34)

Review Questions

Summarize each of the following functions by filling out the table. Use the information to sketch a graph of the function.

1. $f(x) = x ^3 + 3 x^2 - x - 3$
$f(x) = x^3 + 3 x^2 - x - 3$ Analysis
Domain and Range
Intercepts and Zeros
Asymptotes and limits at infinity
Differentiability
Intervals where $f$ is increasing
Intervals where $f$ is decreasing
Relative extrema
Concavity
Inflection points
1. $f(x) = -x^4 + 4x^3 - 4x^2$
$f(x) = - x^4 + 4 x^3 - 4x^2$ Analysis
Domain and Range
Intercepts and Zeros
Asymptotes and limits at infinity
Differentiability
Intervals where $f$ is increasing
Intervals where $f$ is decreasing
Relative extrema
Concavity
Inflection points
1. $f(x) = \frac{2x -2} {x^2}$
$f(x) = \frac{2x -2} {x^2}$ Analysis
Domain and Range
Intercepts and Zeros
Asymptotes and limits at infinity
Differentiability
Intervals where $f$ is increasing
Intervals where $f$ is decreasing
Relative extrema
Concavity
Inflection points
1. $f(x) = x - x^\frac{1} {3}$
$f(x) = x - x^\frac{1} {3}$ Analysis
Domain and Range
Intercepts and Zeros
Asymptotes and limits at infinity
Differentiability
Intervals where $f$ is increasing
Intervals where $f$ is decreasing
Relative extrema
Concavity
Inflection points
1. $f(x) = -\sqrt{2x - 6} +3$
$f(x) = - \sqrt{2x - 6} + 3$ Analysis
Domain and Range
Intercepts and Zeros
Asymptotes and limits at infinity
Differentiability
Intervals where $f$ is increasing
Intervals where $f$ is decreasing
Relative extrema
Concavity
Inflection points
1. $f(x) = x^2 -2\sqrt{x}$
$f(x) = x^2 -2\sqrt{x}$ Analysis
Domain and Range
Intercepts and Zeros
Asymptotes and limits at infinity
Differentiability
Intervals where $f$ is increasing
Intervals where $f$ is decreasing
Relative extrema
Concavity
Inflection points
1. $f(x) = 1 + \cos x$ on the interval $[-\pi, \pi]$
$f(x) = 1 + \cos x$ Analysis
Domain and Range
Intercepts and Zeros
Asymptotes and limits at infinity
Differentiability
Intervals where $f$ is increasing
Intervals where $f$ is decreasing
Relative extrema
Concavity
Inflection points

$f(x) = x^3 + 3x^2 - x - 3$ Analysis
Domain and Range $D = (-\infty, +\infty),$ $R =\left \{\mathrm{all\ reals}\right \}$
Intercepts and Zeros zeros at $x = \pm 1, -3,$ $y-$intercept at $(0, -3)$
Asymptotes and limits at infinity no asymptotes
Differentiability differentiable at every point of its domain
Intervals where $f$ is increasing $\left (-\infty, \frac{-3-2\sqrt{3}} {3}\right )$ and $\left (\frac{-3 + 2\sqrt{3}} {3} , +\infty\right )$
Intervals where $f$ is decreasing $\left(\frac{-3-2\sqrt{3}} {3} , \frac{-3 + 2\sqrt{3}} {3}\right)$
Relative extrema

relative maximum at $x = \frac{-3-2\sqrt{3}} {3},$ located at the point $(-2.15, 3.07)$;

relative minimum at $x = \frac{-3+2\sqrt{3}} {3},$ located at the point $(0.15, -3.07)$

Concavity

concave up in $(-1, +\infty)$

concave down in $(-\infty, -1)$

Inflection points $x = -1,$ located at the point $(-1, 0)$
$f(x) = -x^4 + 4x^3 - 4x^2$ Analysis
Domain and Range $D = (-\infty, +\infty),$ $R = \left \{y \le 0\right \}$
Intercepts and Zeros zeros at $x = 0, 2,$ $y-$intercept at $(0, 0)$
Asymptotes and limits at infinity no asymptotes
Differentiability differentiable at every point of its domain
Intervals where $f$ is increasing $(-\infty, 0)$ and $(1, 2)$
Intervals where $f$ is decreasing $(0, 1)$ and $(2, +\infty)$
Relative extrema

relative maximum at $x = 0$, located at the point $(0, 0)$; and at at $x = 2$ located at the point $(2, 0)$

relative minimum at $x = 1$, located at the point $(1, -1)$

Concavity

concave up in $\left (1- \frac{\sqrt{3}} {3}, 1+ \frac{\sqrt{3}} {3}\right )$

concave down in $\left (-\infty, 1- \frac{\sqrt{3}} {3}\right )$ and $\left (1+ \frac{\sqrt{3}} {3}, +\infty\right )$

Inflection points $x=1+ \frac{\sqrt{3}} {3}, 1-\frac {\sqrt{3}} {3}$, located at the points $\left (1.577, -0.444\right )$ and $\left (0.423, -0.444\right )$
$f(x) = \frac{2x - 2} {x^2}$ Analysis
Domain and Range $D = (-\infty, 0) \cup (0, +\infty),$ $R = \left \{y \neq 0 \right \}$
Intercepts and Zeros zeros at $x = 1$, no $y-$intercept
Asymptotes and limits at infinity HA $y = 0$
Differentiability differentiable at every point of its domain
Intervals where $f$ is increasing $(0, 2)$
Intervals where $f$ is decreasing $(-\infty, 0)$ and $(2, +\infty)$
Relative extrema relative maximum at $x = 2,$ located at the point $(2, 0.5)$
Concavity

concave up in $(3, +\infty)$

concave down in $(-\infty, 0)$ and $(0, 3)$

Inflection points $x = 3,$ located at the point $(3, \frac{4} {9})$
$f(x) = x - x^\frac{1} {3}$ Analysis
Domain and Range $D = (-\infty, +\infty),$ $R = \left \{\mathrm{all\ reals} \right \}$
Intercepts and Zeros zeros at $x = \pm 1, 0,$ $y-$intercept at $(0, 0)$
Asymptotes and limits at infinity no asymptotes
Differentiability differentiable in $(\infty, 0) \cup (0, +\infty)$
Intervals where $f$ is increasing $\left (-\infty, \frac{-\sqrt{3}} {9}\right )$ and $\left (\frac{\sqrt{3}} {9}, +\infty\right )$
Intervals where $f$ is decreasing $\left (\frac{-\sqrt{3}} {9}, \frac{\sqrt{3}} {9}\right )$
Relative extrema

relative maximum at $\mathit x = \frac{-\sqrt{3}} {9},$ located at the point $\left (\frac{-\sqrt{3}} {9}, 0.384\right )$

relative minimum at $x = \frac{\sqrt{3}} {9},$ located at the point $\left (\frac{\sqrt{3}} {9}, - 0.384\right )$

Concavity

concave up in $(0, +\infty)$

concave down in $(-\infty, 0)$

Inflection points $x = 0,$ located at the point $(0, 0)$
$f(x) = - \sqrt{2x - 6} + 3$ Analysis
Domain and Range $D = (3, +\infty),$ $R = \left \{y \le 3 \right \}$
Intercepts and Zeros zero at $x = \frac{15} {2},$ no $y-$intercept
Asymptotes and limits at infinity no asymptotes
Differentiability differentiable in $(3, +\infty)$
Intervals where $f$ is increasing nowhere
Intervals where $f$ is decreasing everywhere in $D = (3, + \infty)$
Relative extrema none absolute maximum at $x = 3,$ located at $(3, 3)$
Concavity concave up in $(3, +\infty)$
Inflection points none
$f(x) = x^2 - 2\sqrt{x}$ Analysis
Domain and Range $D = [0, +\infty),$ $R = \left \{\text {all reals } \ge - 1.19 \right \}$
Intercepts and Zeros zeros at $x=0 \text { and } x = \sqrt[3]{4},$ $y-$intercept at $(0, 0)$
Asymptotes and limits at infinity no asymptotes
Differentiability differentiable in $(0, +\infty)$
Intervals where $f$ is increasing $\left (\frac{\sqrt[3]{16}} {4}, +\infty\right )$
Intervals where $f$ is decreasing $\left (0, \frac{\sqrt[3]{16}} {4}\right )$
Relative extrema relative minimum at $x = \frac{\sqrt[3]{16}} {4},$ located at the point $x = \left (\frac{\sqrt[3]{16}} {4}, -1.19\right )$
Concavity concave up in $(0, +\infty)$
Inflection points none
$f(x) = 1 + \cos x$ Analysis
Domain and Range $D = [- \pi, \pi],$ $R = \left \{0 \le y \le 2 \right \}$
Intercepts and Zeros zeros at $x = - \pi, \pi,$ $y-$intercept at $(0, 2)$
Asymptotes and limits at infinity no asymptotes; $\lim_{x \leftrightarrow \infty} f(x)$ does not exist
Differentiability differentiable at every point of its domain
Intervals where $f$ is increasing $(- \pi, 0)$
Intervals where $f$ is decreasing $(0, \pi)$
Relative extrema

absolute max at $x = 0,$ located at the point $(0, 2)$

absolute minimums at $x = \pm \pi,$ located at the points $(-\pi, 0)$ and $(\pi, 0)$

Concavity concave down in $(-\frac{\pi}{2}, \frac{\pi}{2})$, concave up in $(-\pi, -\frac{\pi}{2})$ and $(\frac{\pi}{2}, \pi)$
Inflection points $x = \pm \frac{\pi} {2}$, located at the points $\left (-\frac{\pi} {2}, 1\right )$ and $\left (\frac{\pi} {2}, 1\right )$

Feb 23, 2012

Dec 03, 2014