Sketch the following curve.
\[f\left(x\right)=xe^{x}\]
- Find the critical points \[\mbox{Take the derivative using the product rule}\] \[f'\left(x\right)=\frac{d}{dx}\left[x\right]e^{x}+x\frac{d}{dx}\left[e^{x}\right]=e^{x}+xe^{x}\] \[\mbox{Set the derivative equal to 0 and solve}\] \[e^{x}+xe^{x}=0\] \[xe^{x}=-e^{x}\] \[\mbox{Since \(e^x\) is never equal to 0, we can divide it out}\] \[x=-1\] Since \(f'(-1)\) exists, \(x=-1\) is the only critical point.\[\]
- Find the inflection points \[\mbox{Take the second derivative using the product rule}\] \[f''\left(x\right)=\frac{d}{dx}\left[e^{x}\right]+\frac{d}{dx}\left[xe^{x}\right]=\] \[\mbox{Substitute what we found for \(\frac{d}{dx}\left[xe^{x}\right]\)}\] \[e^{x}+e^{x}+xe^{x}=2e^{x}+xe^{x}\] \[\mbox{Set the derivative equal to 0 and solve}\] \[2e^{x}+xe^{x}=0\] \[xe^{x}=-2e^{x}\] \[\mbox{Since \(e^x\) is never equal to 0, we can divide it out}\] \[x=-2\] Once again, since the functions \(x\) and \(e^x\), the second derivative is also never discontinuous; therefore, \(x=-2\) is the only inflection point. \[\]
- Draw a super sign diagram \begin{array} {|r|r|}\hline x & x<-2 &="" -2="" -2<x<-1="" -1="" x="">-1 \\ \hline \mbox{sign of }f'(x) & f'(x<-2) 0="" &="" &="" f'(-2<x<-1)="" f'(x="">-1) \\ \hline \mbox{sign of }f''(x) & f''(x<-2) 0="" &="" f''(-2<x<-1)="" &="" f''(x="">-1) \\ \hline f(x) & & f(-2) & & f(-1) & \\ \hline \end{array} We can ignore the blank spaces. \[\] To fill in the table we can pick a value in the each specific interval. \begin{array} {|r|r|}\hline x & x<-2 &="" -2="" -2<x<-1="" -1="" x="">-1 \\ \hline \mbox{sign of }f'(x) & f'(-3) & & f'(-\frac{3}{2}) & 0 & f'(0) \\ \hline \mbox{sign of }f''(x) & f''(-3) & 0 & f''(-\frac{3}{2}) & & f''(0) \\ \hline f(x) & &f(-2)& & f(-1)& \\ \hline \end{array} Compute the chosen values. Remember \(e^{x}\) is never negative; therefore, the value multiplied by it will determine whether it is negative or positive. \begin{array} {|r|r|}\hline f'\left(-3\right)=e^{-3}-3e^{-3}=-2e^{-3} & f'\left(-\frac{3}{2}\right)=e^{-\frac{3}{2}}-\frac{3}{2}e^{-\frac{3}{2}}=-\frac{1}{2}e^{-\frac{3}{2}} \\ \hline f''\left(-3\right)=2e^{-3}-3e^{-3}=-e^{-3} & f''\left(-\frac{3}{2}\right)=2e^{-\frac{3}{2}}-\frac{3}{2}e^{-\frac{3}{2}}=\frac{1}{2}e^{-\frac{3}{2}} \\ \hline \end{array} \begin{array} {|r|r|}\hline f'\left(0\right)=e^{0}+0e^{0}=1 \\ \hline f''\left(0\right)=2e^{0}+0e^{0}=2 \\ \hline \end{array} We can now fill in the sign diagram. \begin{array} {|r|r|}\hline x & x<-2 &="" -2="" -2<x<-1="" -1="" x="">-1 \\ \hline \mbox{sign of }f'(x) & - & & - & 0 & + \\ \hline \mbox{sign of }f''(x) & - & 0 & + & & + \\ \hline f(x) & &f(-2)& & f(-1)& \\ \hline \end{array} Next, find the value of \(f(-2)\) and \(f(-1)\). \[f\left(-2\right)=-2e^{-2}=-2\frac{1}{e^{2}}\] \[\mbox{Since \(e\approx3\), we can substitute it in}\] \[\approx-2\frac{1}{\left(3\right)^{2}}=-\frac{2}{9}\] Since we are just sketching the curve, we do not need exact values. \[f\left(-1\right)=-e^{-1}=-\frac{1}{e}\approx-\frac{1}{3}\] Lastly, find the \(y\)-intercept. \[f\left(0\right)=0\cdot e^{0}=0\cdot1=0\]</-2></-2></-2)></-2)></-2>
- Check the end behavior \[\] \begin{array} {|r|r|}\hline \mathop {\lim }\limits_{x \to -\infty} xe^{x} & \mathop {\lim }\limits_{x \to \infty} xe^{x} \\ \hline \end{array} \[\] \[\mathop {\lim }\limits_{x \to -\infty} xe^{x}=\mathop {\lim }\limits_{x \to -\infty} x \cdot \mathop {\lim }\limits_{x \to -\infty} e^{x}=\] \[-\infty\cdot e^{-\infty}=-\infty\cdot\frac{1}{e^{\infty}}=-\infty\cdot\frac{1}{\infty}=-\infty \cdot 0\] Since we get an indeterminate product we can use L'Hôpital's Rule. \[\mbox{Use the formula: \(\mathop {\lim }\limits_{x \to a} f\left(x\right)g\left(x\right) =\mathop {\lim }\limits_{x \to a} \frac{f\left(x\right)}{\frac{1}{g\left(x\right)}}\)}\] \[\mathop {\lim }\limits_{x \to -\infty} xe^{x}= \mathop {\lim }\limits_{x \to -\infty} \frac{x}{\frac{1}{e^{x}}} = \mathop {\lim }\limits_{x \to -\infty} \frac{x}{e^{-x}}\] \[\mbox{Now use L'Hôpital's Rule}\] \[\mathop {\lim }\limits_{x \to -\infty} \frac{x}{e^{-x}}=\mathop {\lim }\limits_{x \to -\infty} \frac{\frac{d}{dx}\left[x\right]}{\frac{d}{dx}\left[e^{-x}\right]}=\] \[\mathop {\lim }\limits_{x \to -\infty} \frac{1}{-e^{-x}}=\frac{1}{-e^{\infty}}=-\frac{1}{\infty}=0\] \[\] \[\mathop {\lim }\limits_{x \to \infty} xe^{x}= \mathop {\lim }\limits_{x \to \infty} x \cdot \mathop {\lim }\limits_{x \to \infty} e^{x}=\] \[\infty \cdot e^{\infty} =\infty \cdot \infty = \infty\]
- Using the sign diagram, sketch the curve
First, plot the points found in sign diagram.
Next, graph where the curve is increasing and decreasing. The curve is decreasing when \(x<-1\) and="" is="" increasing="" when="" \(x="">-1\). </-1\)>
Apply the concavity and the proper end points. The curve is concave down when \(x<-2\) and="" is="" concave="" up="" \(x="">-2\).</-2\)>
As we can see from the graph, the points we estimated do not directly line up with the graph; however, they are still fairly close and are sufficient when sketching the curve by hand.
Your curve should be similar to the one above.
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