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Q. 76

Expert-verified
Found in: Page 276

### Calculus

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

The sign chart is

The sketch of the graph is

See the step by step solution

## Step 1. Given Information.

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

## Step 2. Finding the roots.

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 $x=0.$

## Step 3. Testing the signs.

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,

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

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.

## Step 4. Sketch the sign chart.

The sign chart is

## Step 5. Examine the relevant limit.

Let's examine the limits.

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

## Step 6. Sketch the graph of function f.

The graph of the function is