4.3: Graphing Functions

Learning Objectives

• Use concavity and inflection points to explain how the sign of the second derivative affects the shape of a function’s graph.
• Explain the concavity test for a function over an open interval.
• Explain the relationship between a function and its first and second derivatives.
• State the second derivative test for local extrema.

In this section, we see how the first and second derivatives provide information about the extrema and shape of a graph by describing whether the graph of a function curves upward or curves downward. Now that we have developed the tools we need to determine where a function is increasing and decreasing, as well as acquired an understanding of the basic shape of the graph, we can provide accurate graphs of a large variety of functions.

Guidelines for Drawing the Graph of a Function

We now have enough analytical tools to draw graphs of a wide variety of algebraic and transcendental functions. Before showing how to graph specific functions, let’s look at a general strategy to use when graphing any function.

Problem-Solving Strategy: Drawing the Graph of a Function

Given a function $f$, use the following steps to sketch a graph of $f$:

1. Determine the domain of the function.
2. Locate the $x$- and $y$-intercepts.
3. Evaluate $\displaystyle \lim_{x→∞}f(x)$ and $\displaystyle \lim_{x→−∞}f(x)$ to determine the end behavior. If either of these limits is a finite number $L$, then $y=L$ is a horizontal asymptote. If either of these limits is $∞$ or $−∞$, determine whether $f$ has an oblique asymptote. If $i$s a rational function such that $f(x)=\dfrac{p(x)}{q(x)}$, where the degree of the numerator is greater than the degree of the denominator, then $f$ can be written as $$f(x)=\frac{p(x)}{q(x)}=g(x)+\frac{r(x)}{q(x),}$$ where the degree of $r(x)$ is less than the degree of $q(x)$. The values of $f(x)$ approach the values of $g(x)$ as $x→±∞$. If $g(x)$ is a linear function, it is known as an oblique asymptote.
4. Determine whether $f$ has any vertical asymptotes.
5. Calculate $f′.$ Find all critical points and determine the intervals where $f$ is increasing and where $f$ is decreasing. Determine whether $f$ has any local extrema.
6. Calculate $f''.$ Determine the intervals where $f$ is concave up and where $f$ is concave down. Use this information to determine whether $f$ has any inflection points. The second derivative can also be used as an alternate means to determine or verify that $f$ has a local extremum at a critical point.

Now let’s use this strategy to graph several different functions. We start by graphing a polynomial function.

Example $\PageIndex{1}$: Sketching a Graph of a Polynomial

Sketch a graph of $f(x)=(x−1)^2(x+2).$

Solution

Step 1: Since $f$ is a polynomial, the domain is the set of all real numbers.

Step 2: When $x=0,\; f(x)=2.$ Therefore, the $y$-intercept is $(0,2)$. To find the $x$-intercepts, we need to solve the equation $(x−1)^2(x+2)=0$, gives us the $x$-intercepts $(1,0)$ and $(−2,0)$

Step 3: We need to evaluate the end behavior of $f.$ As $x→∞, \;(x−1)^2→∞$ and $(x+2)→∞$. Therefore, $\displaystyle \lim_{x→∞}f(x)=∞$.

As $x→−∞, \;(x−1)^2→∞$ and $(x+2)→−∞$. Therefore, $\displaystyle \lim_{x→∞}f(x)=−∞$.

To get even more information about the end behavior of $f$, we can multiply the factors of $f$. When doing so, we see that

$$f(x)=(x−1)^2(x+2)=x^3−3x+2. \nonumber$$

Since the leading term of $f$ is $x^3$, we conclude that $f$ behaves like $y=x^3$ as $x→±∞.$

Step 4: Since $f$ is a polynomial function, it does not have any vertical asymptotes.

Step 5: The first derivative of $f$ is

$$f′(x)=3x^2−3. \nonumber$$

Therefore, $f$ has two critical points: $x=1,−1.$ Divide the interval $(−∞,∞)$ into the three smaller intervals: $(−∞,−1), \;(−1,1)$, and $(1,∞)$. Then, choose test points $x=−2, x=0$, and $x=2$ from these intervals and evaluate the sign of $f′(x)$ at each of these test points, as shown in the following table.

Interval	Test point	Sign of Derivative $f'(x)=3x^2−3=3(x−1)(x+1)$	Conclusion
$(−∞,−1)$	$x=−2$	$(+)(−)(−)=+$	$f$ is increasing
$(−1,1)$	$x=0$	$(+)(−)(+)=−$	$f$ decreasing
$(1,∞)$	$x=2$	$(+)(+)(+)=+$	$f$ is increasing

From the table, we see that $f$ has a local maximum at $x=−1$ and a local minimum at $x=1$. Evaluating $f(x)$ at those two points, we find that the local maximum value is $f(−1)=4$ and the local minimum value is $f(1)=0.$

Step 6: The second derivative of $f$ is

$$f''(x)=6x. \nonumber$$

The second derivative is zero at $x=0.$ Therefore, to determine the concavity of $f$, divide the interval $(−∞,∞)$ into the smaller intervals $(−∞,0)$ and $(0,∞)$, and choose test points $x=−1$ and $x=1$ to determine the concavity of $f$ on each of these smaller intervals as shown in the following table.

Interval	Test Point	Sign of $''(x)=6x$	Conclusion
$(−∞,0)$	$x=−1$	$−$	$f$ is concave down..
$(0,∞)$	$x=1$	$+$	$f$ is concave up.

We note that the information in the preceding table confirms the fact, found in step $5$, that f has a local maximum at $x=−1$ and a local minimum at $x=1$. In addition, the information found in step $5$—namely, $f$ has a local maximum at $x=−1$ and a local minimum at $x=1$, and $f′(x)=0$ at those points—combined with the fact that $f''$ changes sign only at $x=0$ confirms the results found in step $6$ on the concavity of $f$.

Combining this information, we arrive at the graph of $f(x)=(x−1)^2(x+2)$ shown in the following graph.

Exercise $\PageIndex{1}$

Sketch a graph of $f(x)=(x−1)^3(x+2).$

Hint

$f$ is a fourth-degree polynomial.

Answer

Example $\PageIndex{2}$: Sketching a Rational Function

Sketch the graph of $f(x)=\dfrac{x^2}{1−x^2}$.

Solution

Step 1: The function $f$ is defined as long as the denominator is not zero. Therefore, the domain is the set of all real numbers $x$ except $x=±1.$

Step 2: Find the intercepts. If $x=0,$ then $f(x)=0$, so $0$ is an intercept. If $y=0$, then $\dfrac{x^2}{1−x^2}=0,$ which implies $x=0$. Therefore, $(0,0)$ is the only intercept.

Step 3: Evaluate the limits at infinity. Since $f$is a rational function, divide the numerator and denominator by the highest power in the denominator: $x^2$.We obtain

$\displaystyle \lim_{x→±∞}\frac{x^2}{1−x^2}=\lim_{x→±∞}\frac{1}{\frac{1}{x^2}−1}=−1.$

Therefore, $f$ has a horizontal asymptote of $y=−1$ as $x→∞$ and $x→−∞.$

Step 4: To determine whether $f$ has any vertical asymptotes, first check to see whether the denominator has any zeroes. We find the denominator is zero when $x=±1$. To determine whether the lines $x=1$ or $x=−1$ are vertical asymptotes of $f$, evaluate $\displaystyle \lim_{x→1}f(x)$ and $\displaystyle \lim_{x→−1}f(x)$. By looking at each one-sided limit as $x→1,$ we see that

$\displaystyle \lim_{x→1^+}\frac{x^2}{1−x^2}=−∞$ and $\displaystyle \lim_{x→1^−}\frac{x^2}{1−x^2}=∞.$

In addition, by looking at each one-sided limit as $x→−1,$ we find that

$\displaystyle \lim_{x→−1^+}\frac{x^2}{1−x^2}=∞$ and $\displaystyle \lim_{x→−1^−}\frac{x^2}{1−x^2}=−∞.$

Step 5: Calculate the first derivative:

$f′(x)=\dfrac{(1−x^2)(2x)−x^2(−2x)}{\Big(1−x^2\Big)^2}=\dfrac{2x}{\Big(1−x^2\Big)^2}$.

Critical points occur at points $x$ where $f′(x)=0$ or $f′(x)$ is undefined. We see that $f′(x)=0$ when $x=0.$ The derivative $f′$ is not undefined at any point in the domain of $f$. However, $x=±1$ are not in the domain of $f$. Therefore, to determine where $f$ is increasing and where $f$ is decreasing, divide the interval $(−∞,∞)$ into four smaller intervals: $(−∞,−1), (−1,0), (0,1),$ and $(1,∞)$, and choose a test point in each interval to determine the sign of $f′(x)$ in each of these intervals. The values $x=−2,\; x=−\frac{1}{2}, \;x=\frac{1}{2}$, and $x=2$ are good choices for test points as shown in the following table.

Interval	Test point	Sign of $f′(x)=\frac{2x}{(1−x^2)^2}$	Conclusion
$(−∞,−1)$	$x=−2$	$−/+=−$	$f$ is decreasing.
$(−1,0)$	$x=−/2$	$−/+=−$	$f$ is decreasing.
$(0,1)$	$x=1/2$	$+/+=+$	$f$ is increasing.
$(1,∞)$	$x=2$	$+/+=+$	$f$ is increasing.

From this analysis, we conclude that $f$ has a local minimum at $x=0$ but no local maximum.

Step 6: Calculate the second derivative:

$$\begin{align*} f''(x)&=\frac{(1−x^2)^2(2)−2x(2(1−x^2)(−2x))}{(1−x^2)^4}\\[4pt] &=\frac{(1−x^2)[2(1−x^2)+8x^2]}{\Big(1−x^2\Big)^4}\\[4pt] &=\frac{2(1−x^2)+8x^2}{\Big(1−x^2\Big)^3}\\[4pt] &=\frac{6x^2+2}{\Big(1−x^2\Big)^3}. \end{align*}$$

To determine the intervals where $f$ is concave up and where $f$ is concave down, we first need to find all points $x$ where $f''(x)=0$ or $f''(x)$ is undefined. Since the numerator $6x^2+2≠0$ for any $x, f''(x)$ is never zero. Furthermore, $f''$ is not undefined for any $x$ in the domain of $f$. However, as discussed earlier, $x=±1$ are not in the domain of $f$. Therefore, to determine the concavity of $f$, we divide the interval $(−∞,∞)$ into the three smaller intervals $(−∞,−1), \, (−1,−1)$, and $(1,∞)$, and choose a test point in each of these intervals to evaluate the sign of $f''(x)$. in each of these intervals. The values $x=−2, \;x=0$, and $x=2$ are possible test points as shown in the following table.

Interval	Test Point	Sign of $f''(x)=\frac{6x^2+2}{(1−x^2)^3}$	Conclusion
$(−∞,−1)$	$x=−2$	$+/−=−$	$f$ is concave down.
$(−1,−1)$	$x=0$	$+/+=+$	$f$ is concave up
$(1,∞)$	$x=2$	$+/−=−$	$f$ is concave down.

Combining all this information, we arrive at the graph of $f$ shown below. Note that, although $f$ changes concavity at $x=−1$ and $x=1$, there are no inflection points at either of these places because $f$ is not continuous at $x=−1$ or $x=1.$

Exercise $\PageIndex{2}$

Sketch a graph of $f(x)=\dfrac{3x+5}{8+4x}.$

Hint

A line $y=L$ is a horizontal asymptote of $f$ if the limit as $x→∞$ or the limit as $x→−∞$ of $f(x)$ is $L$. A line $x=a$ is a vertical asymptote if at least one of the one-sided limits of $f$ as $x→a$ is $∞$ or $−∞.$

Answer

Example $\PageIndex{3}$: Sketching a Rational Function with an Oblique Asymptote

Sketch the graph of $f(x)=\dfrac{x^2}{x−1}$

Solution

Step 1: The domain of $f$ is the set of all real numbers $x$ except $x=1.$

Step 2: Find the intercepts. We can see that when $x=0, \,f(x)=0,$ so $(0,0)$ is the only intercept.

Step 3: Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of the denominator, $f$ must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials to write

$f(x)=\dfrac{x^2}{x−1}=x+1+\dfrac{1}{x−1}$.

Since $\dfrac{1}{x−1}→0$ as $x→±∞, f(x)$ approaches the line $y=x+1$ as $x→±∞$. The line $y=x+1$ is an oblique asymptote for $f$.

Step 4: To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at $x=1.$ Looking at both one-sided limits as $x→1,$ we find

$\displaystyle \lim_{x→1^+}\frac{x^2}{x−1}=∞$ and $\displaystyle \lim_{x→1^−}\frac{x^2}{x−1}=−∞.$

Therefore, $x=1$ is a vertical asymptote, and we have determined the behavior of $f$ as $x$ approaches $1$ from the right and the left.

Step 5: Calculate the first derivative:

$f′(x)=\dfrac{(x−1)(2x)−x^2(1)}{(x−1)^2}=\dfrac{x^2−2x}{(x−1)^2}.$

We have $f′(x)=0$ when $x^2−2x=x(x−2)=0$. Therefore, $x=0$ and $x=2$ are critical points. Since $f$ is undefined at $x=1$, we need to divide the interval $(−∞,∞)$ into the smaller intervals $(−∞,0), (0,1), (1,2),$ and $(2,∞)$, and choose a test point from each interval to evaluate the sign of $f′(x)$ in each of these smaller intervals. For example, let $x=−1, x=\frac{1}{2}, x=\frac{3}{2}$, and $x=3$ be the test points as shown in the following table.

Interval	Test Point	Sign of $f'(x)=\dfrac{x^2−2x}{(x−1)^2}$	Conclusion
$(−∞,0)$	$x=−1$	(−)(−)/+=+	$f$ is increasing.
$(0,1)$	$x=1/2$	(+)(−)/+=−	$f$ is decreasing.
$(1,2)$	$x=3/2$	(+)(−)/+=−	$f$ is decreasing.
$(2,∞)$	$x=3$	(+)(+)/+=+	$f$ is increasing.

From this table, we see that $f$ has a local maximum at $x=0$ and a local minimum at $x=2$. The value of $f$ at the local maximum is $f(0)=0$ and the value of $f$ at the local minimum is $f(2)=4$. Therefore, $(0,0)$ and $(2,4)$ are important points on the graph.

Step 6. Calculate the second derivative:

$$\begin{align*} f''(x) &= \frac{(x−1)^2(2x−2)−2(x−1)(x^2−2x)}{(x−1)^4}\\[4pt] &=\frac{2(x−1)[(x−1)^2−(x^2−2x)]}{(x−1)^4}\\[4pt] &=\frac{2[x^2-2x+1−x^2+2x]}{(x−1)^3}\\[4pt] &=\frac{2}{(x−1)^3}. \end{align*}$$

We see that $f''(x)$ is never zero or undefined for $x$ in the domain of $f$. Since $f$ is undefined at $x=1$, to check concavity we just divide the interval $(−∞,∞)$ into the two smaller intervals $(−∞,1)$ and $(1,∞)$, and choose a test point from each interval to evaluate the sign of $f''(x)$ in each of these intervals. The values $x=0$ and $x=2$ are possible test points as shown in the following table.

Interval	Test Point	Sign of $f''(x)=\dfrac{2}{(x−1)^3}$	Conclusion
$(−∞,1)$	$x=0$	$+/−=−$	$f$ is concave down.
$(1,∞)$	$x=2$	$+/+=+$	$f$ is concave up

From the information gathered, we arrive at the following graph for $f.$

Exercise $\PageIndex{3}$

Find the oblique asymptote for $f(x)=\dfrac{3x^3−2x+1}{2x^2−4}$.

Hint

Use long division of polynomials.

Answer

$y=\frac{3}{2}x$

Example $\PageIndex{4}$: Sketching the Graph of a Function with a Cusp

Sketch a graph of $f(x)=(x−1)^{2/3}$

Solution

Step 1: Since the cube-root function is defined for all real numbers $x$ and $(x−1)^{2/3}=(\sqrt[3]{x−1})^2$, the domain of $f$ is all real numbers.

Step 2: To find the $y$-intercept, evaluate $f(0)$. Since $f(0)=1,$ the $y$-intercept is $(0,1)$. To find the $x$-intercept, solve $(x−1)^{2/3}=0$. The solution of this equation is $x=1$, so the $x$-intercept is $(1,0).$

Step 3: Since $\displaystyle \lim_{x→±∞}(x−1)^{2/3}=∞,$ the function continues to grow without bound as $x→∞$ and $x→−∞.$

Step 4: The function has no vertical asymptotes.

Step 5: To determine where $f$ is increasing or decreasing, calculate $f′.$ We find

$$f′(x)=\frac{2}{3}(x−1)^{−1/3}=\frac{2}{3(x−1)^{1/3}}$$

This function is not zero anywhere, but it is undefined when $x=1.$ Therefore, the only critical point is $x=1.$ Divide the interval $(−∞,∞)$ into the smaller intervals $(−∞,1)$ and $(1,∞)$, and choose test points in each of these intervals to determine the sign of $f′(x)$ in each of these smaller intervals. Let $x=0$ and $x=2$ be the test points as shown in the following table.

Interval	Test Point	Sign of $f′(x)=\frac{2}{3(x−1)^{1/3}}$	Conclusion
$(−∞,1)$	$x=0$	$+/−=−$	$f$ is decreasing
$(1,∞)$	$x=2$	$+/+=+$	$f$ is increasing

We conclude that $f$ has a local minimum at $x=1$. Evaluating $f$ at $x=1$, we find that the value of $f$ at the local minimum is zero. Note that $f′(1)$ is undefined, so to determine the behavior of the function at this critical point, we need to examine $\displaystyle \lim_{x→1}f′(x).$ Looking at the one-sided limits, we have

$$\lim_{x→1^+}\frac{2}{3(x−1)^{1/3}}=∞\text{ and } \lim_{x→1^−}\frac{2}{3(x−1)^{1/3}}=−∞.\nonumber$$

Therefore, $f$ has a cusp at $x=1.$

Step 6: To determine concavity, we calculate the second derivative of $f:$

$$f''(x)=−\dfrac{2}{9}(x−1)^{−4/3}=\dfrac{−2}{9(x−1)^{4/3}}.$$

We find that $f''(x)$ is defined for all $x$, but is undefined when $x=1$. Therefore, divide the interval $(−∞,∞)$ into the smaller intervals $(−∞,1)$ and $(1,∞)$, and choose test points to evaluate the sign of $f''(x)$ in each of these intervals. As we did earlier, let $x=0$ and $x=2$ be test points as shown in the following table.

Interval	Test Point	Sign of $f''(x)=\dfrac{−2}{9(x−1)^{4/3}}$	Conclusion
$(−∞,1)$	$x=0$	$−/+=−$	$f$ is concave down
$(1,∞)$	$x=2$	$−/+=−$	$f$ is concave down

From this table, we conclude that $f$ is concave down everywhere. Combining all of this information, we arrive at the following graph for $f$.

Exercise $\PageIndex{4}$

Consider the function $f(x)=5−x^{2/3}$. Determine the point on the graph where a cusp is located. Determine the end behavior of $f$.

Hint

A function $f$ has a cusp at a point a if $f(a)$ exists, $f'(a)$ is undefined, one of the one-sided limits as $x→a$ of $f'(x)$ is $+∞$, and the other one-sided limit is $−∞.$

Answer

The function $f$ has a cusp at $(0,5)$, since $\displaystyle \lim_{x→0^−}f′(x)=∞$ and $\displaystyle \lim_{x→0^+}f′(x)=−∞$. For end behavior, $\displaystyle \lim_{x→±∞}f(x)=−∞.$

Key Concepts

• If $c$ is a critical point of $f$ and $f'(x)>0$ for $x<c$ and $f'(x)<0$ for $x>c$, then $f$ has a local maximum at $c$.
• If $c$ is a critical point of $f$ and $f'(x)<0$ for $x<c$ and $f'(x)>0$ for $x>c,$ then $f$ has a local minimum at $c$.
• If $f''(x)>0$ over an interval $I$, then $f$ is concave up over $I$.
• If $f''(x)<0$ over an interval $I$, then $f$ is concave down over $I$.
• If $f'(c)=0$ and $f''(c)>0$, then $f$ has a local minimum at $c$.
• If $f'(c)=0$ and $f''(c)<0$, then $f$ has a local maximum at $c$.
• If $f'(c)=0$ and $f''(c)=0$, then evaluate $f'(x)$ at a test point $x$ to the left of $c$ and a test point $x$ to the right of $c$, to determine whether $f$ has a local extremum at $c$.

Glossary

concave down

if $f$ is differentiable over an interval $I$ and $f'$ is decreasing over $I$, then $f$ is concave down over $I$

concave up

if $f$ is differentiable over an interval $I$ and $f'$ is increasing over $I$, then $f$ is concave up over $I$

concavity

the upward or downward curve of the graph of a function

concavity test

suppose $f$ is twice differentiable over an interval $I$; if $f''>0$ over $I$, then $f$ is concave up over $I$; if $f''<$ over $I$, then $f$ is concave down over $I$

first derivative test

let $f$ be a continuous function over an interval $I$ containing a critical point $c$ such that $f$ is differentiable over $I$ except possibly at $c$; if $f'$ changes sign from positive to negative as $x$ increases through $c$, then $f$ has a local maximum at $c$; if $f'$ changes sign from negative to positive as $x$ increases through $c$, then $f$ has a local minimum at $c$; if $f'$ does not change sign as $x$ increases through $c$, then $f$ does not have a local extremum at $c$

inflection point

if $f$ is continuous at $c$ and $f$ changes concavity at $c$, the point $(c,f(c))$ is an inflection point of $f$

second derivative test

suppose $f'(c)=0$ and $f'$' is continuous over an interval containing $c$; if $f''(c)>0$, then $f$ has a local minimum at $c$; if $f''(c)<0$, then $f$ has a local maximum at $c$; if $f''(c)=0$, then the test is inconclusive

Contributors and Attributions

• Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.