OBJECTIVES

When you finish this section, you should be able to:

1. Graph a function using calculus (p. 308)

Steps for Graphing a Function \({y=f(x)}\)

Step 1 Find the domain of \(f\) and any intercepts.

Step 2 Identify any asymptotes, and examine the end behavior.

Step 3 Find \(y^\prime =f^\prime (x)\) and use it to find any critical numbers of \(f.\) Determine the slope of the tangent line at the critical numbers. Also find \(f^{\prime \prime} .\)

Step 4 Use the Increasing/Decreasing Function Test to identify intervals on which \(f\) is increasing \((f^\prime >0)\) and the intervals on which \(f\) is decreasing \((f^\prime <0)\).

Step 5 Use either the First Derivative Test or the Second Derivative Test (if applicable) to identify local maximum values and local minimum values.

Step 6 Use the Test for Concavity to determine intervals where the function is concave up \((f^{\prime \prime} >0)\) and concave down \((f^{\prime \prime} <0)\). Identify any inflection points.

Step 7 Graph the function using the information gathered. Plot additional points as needed. It may be helpful to find the value of the derivative at such points.

Using Calculus to Graph a Polynomial Function

Graph \(f(x)=3x^{4}-8x^{3}\).

Solution We follow the steps for graphing a function.

Step 1 \(f\) is a fourth-degree polynomial; its domain is all real numbers. \(f( 0) =0\), so the \(y\)-intercept is \(0\). We find the \(x\)-intercepts by solving \(f(x) =0\). \[ \begin{eqnarray*} 3x^{4}-8x^{3} &=&0 \\ x^{3}( 3x-8) &=&0 \\ x &=&0 \qquad\hbox{or}\qquad x=\dfrac{8}{3} \end{eqnarray*} \]

There are two \(x\)-intercepts: \(0\) and \(\dfrac{8}{3}\). We plot the intercepts \((0,0)\) and \(\left( \dfrac{8}{3},0\right)\)

Step 2 Polynomials have no asymptotes, but the end behavior of \(f\) resembles the power function \(y=3x^{4}\).

Step 3 \(f^\prime (x) =12x^{3}-24x^{2}=12x^{2}(x-2)\qquad f^{\prime \prime} (x) =36x^{2}-48x=12x( 3x-4) \)

For polynomials, the critical numbers occur when \(f^\prime (x) =0\). \[ \begin{eqnarray*} 12x^{2}( x-2) &=&0 \\ x &=&0 \qquad \hbox{or}\qquad x=2 \end{eqnarray*} \]

The critical numbers are \(0\) and \(2\). At the points \((0,0)\) and \(( 2,-16)\) , the tangent lines are horizontal. We plot these points.

Step 4 To apply the Increasing/Decreasing Function Test, we use the critical numbers \(0\) and \(2\) to form three intervals.

Step 5 We use the First Derivative Test. Based on the table above, \(f( 0) =0\) is neither a local maximum value nor a local minimum value, and \(f(2) =-16\) is a local minimum value.

Step 6 \(f^{\prime \prime} (x) =36x^{2}-48x=12x( 3x-4) ;\) the zeros of \(f^{\prime \prime}\) are \(x=0\) and \(x=\dfrac{4}{3}\). To apply the Test for Concavity, we use the numbers \(0\) and \(\dfrac{4}{3}\) to form three intervals.

The concavity of \(f\) changes at \(0\) and \(\dfrac{4}{3}\), so the points \( (0,0)\) and \(\left( \dfrac{4}{3},-\dfrac{256}{27}\right)\) are inflection points. We plot the inflection points.

Step 7 Using the points plotted and the information about the shape of the graph, we graph the function. See Figure 46.

NOW WORK

Problem 1.

Using Calculus to Graph a Rational Function

Graph \(f(x)=\dfrac{6x^{2}-6}{x^{3}}\).

Solution We follow the steps for graphing a function.

Step 1 \(f\) is a rational function; the domain of \(f\) is \(\{ x|x\neq 0\} \). The \(x\)-intercepts are \(-1\) and \(1\). There is no \(y\)-intercept. We plot the intercepts \((-1,0)\) and \((1,0) .\)

Step 2 The degree of the numerator is less than the degree of the denominator, so \(f\) has a horizontal asymptote. We find the horizontal asymptotes by finding the limits at infinity. Using L’Hôpital’s Rule, we get \[ \lim\limits_{x\rightarrow \infty }\frac{6x^{2}-6}{x^{3}}=\lim\limits_{x \rightarrow \infty }\dfrac{12x}{3x^{2}}=\lim\limits_{x\rightarrow \infty } \frac{12}{6x}=0 \]

The line \(y=0\) is a horizontal asymptote as \(x\rightarrow -\infty\) and as \(x\rightarrow \infty \).

We identify vertical asymptotes by checking for infinite limits. [\(\lim\limits_{x\rightarrow 0}f(x) \) is the only one to check–do you see why?] \[ \lim\limits_{x\rightarrow 0^{-}}{\frac{6x^{2}-6}{x^{3}}}=\infty \qquad \lim\limits_{x\rightarrow 0^{+}}{\frac{6x^{2}-6}{x^{3}}} =-\infty \]

The line \(x=0\) is a vertical asymptote. We draw the asymptotes on the graph.

Step 3 \[ \begin{array}{l} f'(x) &=& \frac{d}{{dx}}\left( {\frac{{6x^2 - 6}}{{x^3 }}} \right) = \frac{{12x(x^3 ) - (6x^2 - 6)(3x^2 )}}{{x^6 }} \\ &=& \frac{{6x^2 (3 - x^2 )}}{{x^6 }} = \frac{{6(3 - x^2 )}}{{x^4 }} \\ f''(x) &=& \frac{d}{{dx}}\left[ {\frac{{6(3 - x^2 )}}{{x^4 }}} \right] = 6 \cdot \frac{{( - 2x)x^4 - (3 - x^2 )(4x^3 )}}{{x^8 }} \\ &=& \frac{{12x^3 (x^2 - 6)}}{{x^8 }} = \frac{{12(x^2 - 6)}}{{x^5 }} \\ \end{array} \]

Critical numbers occur where \(f^\prime (x) =0\) or where \(f^\prime (x)\) does not exist. Since \(f^\prime (x) =0\) when \(x=\pm \sqrt{3}\), there are two critical numbers, \(\sqrt{3}\) and \(-\sqrt{3}\). (The derivative does not exist at \(0\), but, since \(0\) is not in the domain of \(f\), it is not a critical number.)

Step 4 To apply the Increasing/Decreasing Function Test, we use the numbers \(-\sqrt{3},\) \(0,\) and \(\sqrt{3}\) to form four intervals.

Step 5 We use the First Derivative Test. From the table in Step 4, we conclude

1. \(f( -\sqrt{3}) =\) \(-\dfrac{4}{\sqrt{3 }}\approx -2.31\) is a local minimum value.
2. \(f(\sqrt{3}) =\) \(\dfrac{4}{\sqrt{3}}\approx 2.31\) is a local maximum value

Plot the local extreme points.

Step 6 We use the Test for Concavity. \(f^{\prime \prime} (x) =\dfrac{12( x^{2}-6) }{x^{5}}=0\) when \(x=\pm \sqrt{6}\). Now use the numbers \(-\sqrt{6},\) \(0,\) and \(\sqrt{6}\) to form four intervals.

The concavity of the function \(f\) changes at \(-\sqrt{6},\) \(0,\) and \(\sqrt{6}\). So, theeak points \(\left( -\sqrt{6},-\dfrac{5\sqrt{6}}{6}\right)\) and \(\left( \sqrt{6},\dfrac{5\sqrt{6}}{6}\right)\) are inflection points. Since \(0\) is not in the domain of \(f,\) there is no inflection point at \(0.\) We plot the inflectioneak points.

Step 7 We now use the information about where the function increases/decreases and the concavity to complete the graph. See Figure 47. Notice the apparent symmetry of the graph with respect to the origin. We can verify this symmetry by showing \(f( -x) =-f(x)\) and concluding that \(f\) is an odd function.

Figure 47 \(f(x)=\dfrac{6x^2-6}{x^3}\)

NOW WORK

Problem 9.

Using Calculus to Graph a Rational Function

Graph \(f(x) =\dfrac{x^{2}-2x+6}{x-3}\).

Solution

Step 1 \(f\) is a rational function; the domain of \(f\) is \(\{ x|x\neq 3\}\). There are no \(x\)-intercepts, since \( x^{2}-2x+6=0\) has no real solutions. (Its discriminant is negative.) The \(y\) -intercept is \(f( 0) =-2\). Plot the intercept \(( 0,-2).\)

Step 2 We identify any vertical asymptotes by checking for infinite limits (\(3\) is the only number to check). Since \(\lim\limits_{x\rightarrow 3^{-}}\left( \dfrac{x^{2}-2x+6}{x-3}\right) =-\infty \) and \(\lim\limits_{x\rightarrow 3^{+}}\left( \dfrac{x^{2}-2x+6}{x-3 }\right) =\infty \), the line \(x=3\) is a vertical asymptote.

The degree of the numerator of \(f\) is 1 more than the degree of the denominator, so \(f\) will have an oblique asymptote. We divide \(x^{2}-2x+6\) by \( x-3\) to find the line \(y=mx+b\). \[ f(x) =\frac{x^{2}-2x+6}{x-3}=x+1+\frac{9}{x-3} \]

Since \( \lim\limits_{x\rightarrow \infty }[ f(x) -( x+1) ] =\lim\limits_{x\rightarrow \infty }\dfrac{9}{x-3}=0,\) then \(y=x+1\) is an oblique asymptote of the graph of \(f\). Draw the asymptotes on the graph.

Step 3\[ \begin{eqnarray*} f^\prime (x) &=&\dfrac{d}{\textit{dx}}\left( \dfrac{x^{2}-2x+6}{x-3}\right) =\dfrac{( 2x-2) (x-3) -( x^{2}-2x+6) ( 1) }{(x-3) ^{2}}\\ &=&\dfrac{ x^{2}-6x}{(x-3)^{2}}=\dfrac{x( x-6) }{(x-3) ^{2}}\\ f^{\prime \prime} (x) &=&\dfrac{d}{\textit{dx}}\left[ \dfrac{x^{2}-6x}{ (x-3) ^{2}}\right] =\dfrac{( 2x-6) (x-3) ^{2}-( x^{2}-6x) [ 2(x-3) ] }{(x-3) ^{4}} \\ &=&(x-3) \dfrac{( 2x^{2}-12x+18) -( 2x^{2}-12x) }{(x-3) ^{4}}=\dfrac{18}{(x-3) ^{3} } \end{eqnarray*} \]

\(f^\prime (x) =0\) at \(x=0\) and at \(x=6\). So, \(0\) and \(6\) are critical numbers. The tangent lines are horizontal at \(0\) and at \(6.\) Since \(3\) is not in the domain of \(f\), \(3\) is not a critical number.

Step 4 To apply the Increasing/Decreasing Function Test, we use the numbers \(0,\) \(3,\) and \(6\) to form four intervals.

Step 5 Using the First Derivative Test, we find that \(f( 0) =-2\) is a local maximum value and \(f( 6) =10\) is a local minimum value. Plot the points \(\left( 0,-2\right)\) and \(\left( 6,10\right)\).

Step 6 \(f^{\prime \prime} (x) =\dfrac{18}{(x-3) ^{3}}\). If \(x<3\), \(f^{\prime \prime} (x) <0\), and if \(x>3\) , \(f^{\prime \prime} (x) >0\). From the Test for Concavity, we conclude \(f\) is concave down on the interval \(\left( -\infty ,3\right)\) and \(f\) is concave up on the interval \((3,\infty) \). The concavity changes at \(3\), but \(3\) is not in the domain of \(f\). So, the graph of \(f\) has no inflection point.

Step 7 We use the information to complete the graph of the function. See Figure 48.

Figure 48 \(f(x)=\dfrac{x^2-2x+6}{x-3}\)

NOW WORK

Problem 11.

Using Calculus to Graph a Function

Graph \(\ f(x)=\dfrac{x}{\sqrt{x^{2}+4}}\).

Solution

Step 1 The domain of \(f\) is all real numbers. The only intercept is \((0,0)\). So, we plot the point \((0,0) .\)

Step 2 We check for horizontal asymptotes. Since \[ \begin{eqnarray*} \lim\limits_{x\rightarrow \infty }f(x) =\lim\limits_{x\rightarrow \infty }\dfrac{x}{\sqrt{x^{2}+4}}=\sqrt{ \lim\limits_{x\rightarrow \infty }\dfrac{x^{2}}{x^{2}+4}} \underset{\underset{{\color{#0066A7}{\hbox{L’Hôpital’s Rule}}}} {\color{#0066A7}{{{\left\uparrow{\vphantom{width0pc height9pt depth0pt}}\right.}}}}}{=} \sqrt{\lim\limits_{x\rightarrow \infty }\dfrac{2x}{2x}}=1\\[-12.9pt] \end{eqnarray*} \]

the line \(y=1\) is a horizontal asymptote as \(x\rightarrow \infty .\) Since \(\lim\limits_{x\rightarrow -\infty }f(x) =-1,\) the line \(y=-1\) is also a horizontal asymptote as \(x\rightarrow -\infty .\) We draw the horizontal asymptotes on the graph.

Since \(f\) is defined for all real numbers, there are no vertical asymptotes.

Step 3\[ \begin{array}{@{\hspace*{-2.6pc}}rcl} \!\!f^\prime (x) &=&\dfrac{d}{\textit{dx}}\left( \dfrac{x }{\sqrt{x^{2}+4}}\right) =\dfrac{1\cdot \sqrt{x^{2}+4}-x\left[ \dfrac{1}{2} ( x^{2}+4) ^{-1/2}\cdot 2x\right] }{x^{2}+4}\\[5pt] &=&\dfrac{\sqrt{x^{2}+4}-\dfrac{x^{2}}{\sqrt{x^{2}+4}}}{x^{2}+4} =\dfrac{x^{2}+4-x^{2}}{( x^{2}+4) \sqrt{x^{2}+4}}=\dfrac{4}{(x^{2}+4)^{3/2}}\\ f^{\prime \prime} (x) &=& \dfrac{d}{\textit{dx}} \left[ \dfrac{4}{( x^{2}+4) ^{3/2}}\right] =4\left[ -\dfrac{3}{2} ( x^{2}+4) ^{-5/2}\right] \cdot 2x=\dfrac{-12x}{( x^{2}+4) ^{5/2}} \end{array} \]

Since \(f^\prime (x)\) is never \(0\) and \(f^\prime\) exists for all real numbers, there are no critical numbers.

Step 4 Since \(f^\prime (x) >0\) for all \(x\), \(f\) is increasing on \(( -\infty ,\infty)\) .

Step 5 Because there are no critical numbers, there are no local extreme values.

Step 6 We test for concavity. Since \(f^{\prime \prime} (x) >0\) on the interval \((-\infty ,0) \), \(f\) is concave up on \((-\infty ,0)\); and since \(f^{\prime \prime} (x) <0\) on the interval \(( 0,\infty ) \), \(f\) is concave down on \(( 0,\infty ) \). The concavity changes at \(0,\) and \(0\) is in the domain of \(f\), so the point \((0,0)\) is an inflection point of \(f\). Plot the inflection point.

Step 7 Figure 49 shows the graph of \(f\).

NOW WORK

Problem 25.

Using Calculus to Graph a Function

Graph \(f(x)=4x^{1/3}-x^{4/3}\).

Solution

Step 1 The domain of \(f\) is all real numbers. Since \(f(0) = 0\), the \(y\)-intercept is \(0\). Now \(f(x) = 0\) when \(4x^{1/3}-x^{4/3}=x^{1/3}\left( 4-x\right) =0\) or when \(x=0\) or \(x=4.\) So, the \(x\)-intercepts are \(0\) and \(4\). Plot the intercepts \((0,0)\) and \(\left( 4,0\right).\)

Step 2 Since \(\lim\limits_{x\rightarrow \infty }f(x) =\lim\limits_{x\rightarrow \infty }[ x^{1/3}(4-x)] =-\infty\), there is no horizontal asymptote. Since the domain of \(f\) is all real numbers, there is no vertical asymptote.

Step 3 \[ \begin{eqnarray*} f^\prime (x) &=&\dfrac{d}{\textit{dx}}( 4x^{1/3}-x^{4/3}) =\dfrac{4}{3}x^{-2/3}-\dfrac{4}{3}x^{1/3} = \dfrac{4}{3}\!\left( \dfrac{1}{x^{2/3}}-x^{1/3}\right)\\[5pt] &=&\dfrac{4}{3}\!\left( \dfrac{1-x}{ x^{2/3}}\right) \\[5pt] f^{\prime \prime} (x) &=&\dfrac{d}{\textit{dx}}\!\left( \dfrac{4}{ 3}x^{-2/3}-\dfrac{4}{3}x^{1/3}\right) =-\dfrac{8}{9}x^{-5/3}-\dfrac{4}{9} x^{-2/3}=-\dfrac{4}{9}\!\left( \dfrac{2}{x^{5/3}}+\dfrac{1}{x^{2/3}}\right)\\[5pt] &=&-\dfrac{4}{9}\cdot \dfrac{2+x}{x^{5/3}} \end{eqnarray*} \]

Since \(f^\prime (x) =\dfrac{4}{3}\!\left( \dfrac{1-x}{ x^{2/3}}\right) =0\) when \(x=1\) and \(f^\prime (x)\) does not exist at \(x = 0,\) the critical numbers are \(0\) and \(1\). At the point \((1,3)\), the tangent line to the graph is horizontal; at the point \((0,0)\), the tangent line is vertical. Plot these points.

Step 4 To apply the Increasing/Decreasing Function Test, we use the critical numbers \(0\) and \(1\) to form three intervals.

Step 5 By the First Derivative Test, \(f( 1) =3\) is a local maximum value and \(f( 0) =0\) is not a local extreme value.

Step 6 We now test for concavity by using the numbers \(-2\) and \( 0 \) to form three intervals.

The concavity changes at \(-2\) and at \(0.\) Since \[ f( -2) =4( -2) ^{1/3}- ( -2) ^{4/3}= 4\sqrt[3]{-2}-\sqrt[3]{16}\approx -7.56, \]

the inflection points are \(( -2,-7.56)\) and \((0,0)\) . Plot the inflection points.

Step 7 The graph of \(f\) is given in Figure 50.

NOW WORK

Problem 19.

Using Calculus to Graph a Trigonometric Function

Graph \(f(x)=\sin x-\cos ^{2}x,\) \(0\leq x\leq 2\pi\) .

Solution

Step 1 The domain of \(f\) is \(\{ x|0\leq x\leq 2\pi \} \). Since \(f( 0) =\sin 0-\cos ^{2}0=-1\), the \(y\)-intercept is \(-1\). The \(x\)-intercepts satisfy the equation \[ \sin x-\cos ^{2}x=\sin x-( 1-\!\sin ^{2}x) =\sin ^{2}x+\sin x-1=0 \]

This trigonometric equation is quadratic in form, so we use the quadratic formula. \[ \sin x=\dfrac{-1\pm \sqrt{1-4( 1) (-1) }}{2}=\dfrac{ -1\pm \sqrt{5}}{2} \]

Since \(\dfrac{-1-\sqrt{5}}{2}<-1\) and \(-1\leq \sin x\leq 1\), the \(x\) -intercepts occur at \(x=\sin ^{-1}\left( \dfrac{-1+\sqrt{5}}{2}\right) \approx 0.67\) and at \(x=\pi -\sin ^{-1}\left( \dfrac{-1+\sqrt{5}}{2}\right) \approx 2.48.\) Plot the intercepts.

Step 2 The function \(f\) has no asymptotes.

Step 3 \[ \begin{array}{@{}rcl} f' (x)&=&\dfrac{d}{\textit{dx}}( \sin x-\cos ^{2} x) =\cos x+2\, \cos\, x\, \sin x=\cos x(1+2\sin x)\\ f'' (x)& =&\dfrac{d}{\textit{dx}}[ {\cos x(1+2\sin x)}] =2\, \cos ^{2}x-\!\sin x(1+2\sin x)\\ &=&2\cos ^{2}x-\!\sin x-2\sin ^{2}x=-4\sin ^{2}x-\!\sin x+2 \end{array} \]

The critical numbers occur where \(f^\prime (x) =0\). That is, where \[ \cos x=0 \qquad\hbox{or}\qquad 1+2\sin x=0 \]

In the interval \([0,2\pi ] \): \(\cos x=0\) if \(x=\dfrac{\pi }{2}\) or if \(x=\dfrac{3\pi }{2}\); and \(1+2\sin x=0\) when \(\sin x=-\dfrac{1}{2}\), that is, when \(x=\dfrac{7\pi }{6}\) or \(x=\dfrac{11\pi }{6}\). So, the critical numbers are \(\dfrac{\pi }{2}\), \(\dfrac{7\pi }{6}\), \(\dfrac{3\pi }{2}\), and \(\dfrac{11\pi }{6}\). At each of these numbers, the tangent lines are horizontal.

Step 4 To apply the Increasing/Decreasing Function Test, we use the numbers \(0,\) \(\dfrac{\pi }{2}\), \(\dfrac{7\pi }{6}\), \(\dfrac{3\pi }{2}\), \( \dfrac{11\pi }{6},\) and \(2\pi\) to form five intervals.

Step 5 By the First Derivative Test, \(f\left( \dfrac{\pi }{2} \right) =1\) and \(f\left( \dfrac{3\pi }{2}\right) =-1\) are local maximum values, and \(f\left(\dfrac{7\pi}{6}\right) =-\dfrac{5}{4}\) and \(f\left(\dfrac{11\pi}{6}\right) =-\dfrac{5}{4}\) are local minimum values. Plot these points.

Figure 51 \(f(x)=\sin x - \cos^{2}x\), \(0\le x\le 2\pi\)

Step 6 We apply the Test for Concavity. To solve \(f^{\prime \prime} (x) >0\) and \(f^{\prime \prime} (x) <0\), we first solve the equation \(f^{\prime \prime} (x) =-4\sin ^{2}x-\!\sin x+2=0,\) or equivalently, \[ \begin{eqnarray*} 4\sin ^{2}x+\sin x-2 &=&0 \qquad 0\leq x\leq 2\pi \\[3pt] \sin x &=&\dfrac{-1\pm \sqrt{1+32}}{8} \\[3pt] \sin x &\approx &0.593\qquad \hbox{or}\qquad \sin x\approx -0.843 \\ x &\approx &0.63 \qquad x\approx 2.51 \qquad x\approx 4.14 \qquad x\approx 5.28 \end{eqnarray*} \]

We use these numbers to form five subintervals of \([0,2\pi ]\).

The inflection points are \(( 0.63,-0.06) \), \((2.51,-0.06) \), \(( 4.14,-1.13) \), and \(( 5.28,-1.13) .\) Plot the inflection points.

Step 7 The graph of \(f\) is given in Figure 51.

NOW WORK

Problem 33.

Using Calculus to Graph a Function

Graph \(f(x)=e^{x}( x^{2}-3) \).

Solution

Step 1 The domain of \(f\) is all real numbers. Since \(f( 0) =-3\), the \(y\)-intercept is \(-3\). The \(x\) -intercepts occur where \[ \begin{eqnarray*} f(x) &=&e^{x}( x^{2}-3) =0 \\[3pt] x &=&\sqrt{3}\qquad\hbox{or}\qquad x=-\sqrt{3} \end{eqnarray*} \]

Plot the intercepts.

Step 2 Since the domain of \(f\) is all real numbers, there are no vertical asymptotes. To determine if there is a horizontal asymptote, we find the limits at infinity. \[ \lim\limits_{x\rightarrow -\infty }f(x) =\lim\limits_{x\rightarrow -\infty } [ e^{x} ( x^{2}-3 ) ] \]

Since \(e^{x} ( x^{2}-3 )\) is an indeterminate form at \(-\infty\) of the type \(0\cdot \infty \), we write \(e^{x} ( x^{2}-3 ) =\dfrac{x^{2}-3}{e^{-x}}\) and use L’Hôpital’s Rule.

The line \(y=0\) is a horizontal asymptote as \(x\rightarrow -\infty\).

Since \( \lim\limits_{x\rightarrow \infty }[ e^{x}( x^{2}-3) ] =\infty ,\) there is no horizontal asymptote as \(x\rightarrow \infty .\) Draw the asymptote on the graph.

Step 3 \[ \begin{array}{@{}rcl} f^\prime (x) &=&\dfrac{d}{\textit{dx}}[e^{x}( x^{2}-3) ] =e^{x}(2x) +e^{x}(x^{2}-3) =e^{x}( x^{2}+2x-3)\\[8pt] &=&e^{x}( x+3) ( x-1)\\[11pt] f^{\prime \prime} (x) &=&\dfrac{d}{\textit{dx}}[e^{x}( x^{2}+2x-3) ] =e^{x}( 2x+2) +e^{x}(x^{2}+2x-3)\\[8pt] &=&e^{x}( x^{2}+4x-1) \end{array} \]

Solving \(f^\prime (x) =0,\) we find that the critical numbers are \(-3\) and \(1\).

Step 4 To apply the Increasing/Decreasing Function Test, we use the critical numbers \(-3\) and \(1\) to form three intervals.

Step 5 We use the First Derivative Test to identify the local extrema. From the table in Step 4, there is a local maximum at \(-3\) and a local minimum at \(1.\) Then \(f(-3) =6e^{-3}\approx 0.30\) is a local maximum value and \(f(1) =-2e\approx -5.44\) is a local minimum value. Plot the local extrema.

Figure 52 \(f(x)=e^x(x^2-3)\)

Step 6 To determine the concavity of \(f,\) first we solve \( f^{\prime \prime} (x) =0\). We find \(x=-2\pm \sqrt{5}\). Now we use the numbers \(-2-\sqrt{5}\approx -4.24\) and \(-2+\sqrt{5}\approx 0.24\) to form three intervals and apply the Test for Concavity.

The inflection points are \(( -4.24,0.22) \) and \(( 0.24,-3.73) .\) Plot the inflection points.

Step 7 The graph of \(f\) is given in Figure 52.

NOW WORK

Problem 39.