Using Calculus to Graph a Function
Graph \(f(x)=e^{x}( x^{2}-3) \).
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.
| Interval | Sign of f\(^\prime\) | Conclusion |
|---|---|---|
| \(( -\infty ,-3)\) | positive | \(f\) is increasing on the interval \(( -\infty ,-3)\) |
| \((-3,1)\) | negative | \(f\) is decreasing on the interval \(( -3,1)\) |
| \((1,\infty )\) | positive | \(f\) is increasing on the interval \( (1,\infty )\) |
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.
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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.
| Interval | Sign of \({f''}\) | Conclusion |
|---|---|---|
| \(( -\infty ,-2-\sqrt{5})\) | positive | \(f\) is concave up on the interval \((-\infty ,-2 - \sqrt{5})\) |
| \((-2-\sqrt{5},-2+\sqrt{5})\) | negative | \(f\) is concave down on the interval \(( -2 -\sqrt{5}, -2+\sqrt{5})\) |
| \((-2+\sqrt{5},\infty)\) | positive | \(f\) is concave up on the interval \(( -2+\sqrt{5},\infty )\) |
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.