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Q. 76

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*.

$f\left(x\right)=\mathrm{ln}({x}^{2}+1)$

The sign chart is

The sketch of the graph is

The given function is $f\left(x\right)=\mathrm{ln}\left({x}^{2}+1\right).$

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

So,

$f\left(x\right)=\mathrm{ln}\left({x}^{2}+1\right)\phantom{\rule{0ex}{0ex}}0=\mathrm{ln}\left({x}^{2}+1\right)\phantom{\rule{0ex}{0ex}}x=0$

Therefore, the given function has roots at $x=0.$

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

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

So,

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

Thus, $f\text{'}$ has a local minimum at $x=0.$ It is positive on the interval $\left(0,\infty \right)$ 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,

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

Thus, $f\text{'}\text{'}$ has an inflection points at $x=1,-1.$ It is positive on the interval $\left(-1,1\right)$ 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)=\infty $

The graph of the function is

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