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Q. 77

Expert-verifiedFound in: Page 276

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Sketch careful, labeled graphs of each function *f* in Exercises *63–82* by hand, without consulting a calculator or graphing utility. As part of your work, make sign charts for the signs, roots, and undefined points of $f,f\text{'},andf\text{'}\text{'},$ and examine any relevant limits so that you can describe all key points and behaviors of *f*.

$f\left(x\right)=\frac{{e}^{x}}{x}$

The sign chart is

The sketch of the graph is

The given function is $f\left(x\right)=\frac{{e}^{x}}{x}.$

To find the roots we will put the given function equal to zero.

So,

$f\left(x\right)=\frac{{e}^{x}}{x}\phantom{\rule{0ex}{0ex}}0=\frac{{e}^{x}}{x}$

Therefore, the given function is undefined at $x=0.$

Now, let's test the sign for $f\text{'}andf\text{'}\text{'}.$

Let's differentiate the equation to find $f\text{'}.$

So,

$f\text{'}\left(x\right)=\frac{\left(x-1\right){e}^{x}}{{x}^{2}}\phantom{\rule{0ex}{0ex}}0=\frac{\left(x-1\right){e}^{x}}{{x}^{2}}\phantom{\rule{0ex}{0ex}}0=\left(x-1\right){e}^{x}\phantom{\rule{0ex}{0ex}}x=1$

Thus, $f\text{'}$ has a local minimum at $x=1.$ It is positive on the interval $\left(1,\infty \right)$ and negative elsewhere. Hence the graph of *f *will be increasing during the positive interval and decrease during the negative interval.

Let's differentiate again.

So,

$f\text{'}\text{'}\left(x\right)=\frac{\left({x}^{2}-2x+2\right){e}^{x}}{{x}^{3}}$

Thus, $f\text{'}\text{'}$ is positive on the interval $\left(0,\infty \right)$ and negative elsewhere. Hence, the graph of *f *will be concave up on positive interval and concave down elsewhere.

The sign chart is

Let's examine the limits.

$\underset{x\to \infty}{\mathrm{lim}}f\left(x\right)=\infty \phantom{\rule{0ex}{0ex}}\underset{x\to -\infty}{\mathrm{lim}}f\left(x\right)=0$

The graph of the function is

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