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Answers without the blur. Sign up and see all textbooks for free! Q. 33

Expert-verified Found in: Page 261 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Use a sign chart for ${\mathbit{f}}^{\mathbf{\text{'}}}$ to determine the intervals on which each function $\mathbit{f}$ is increasing or decreasing. Then verify your algebraic answers with graphs from a calculator or graphing utility.$\mathbit{l}\mathbit{n}\mathbf{\left(}{\mathbit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{\right)}$

Ans:

After inserting the root values we can find the increasing and decreasing intervals of the given function.

Intervals of the given function are

Increasing at $\left(0,\infty \right)$

Decreasing at$\left(-\infty ,0\right)$

See the step by step solution

## Step 1. Given information:

$\mathbit{l}\mathbit{n}\mathbf{\left(}{\mathbit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{\right)}$

## Step 2. Finding the derivative of the function:

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

## Step 3. Inserting the root points on the number line(Sign chart):

After inserting the root values we can find the increasing and decreasing intervals of the given function.

Intervals of the given function are

Increasing at $\mathbf{\left(}\mathbf{0}\mathbf{,}\mathbf{\infty }\mathbf{\right)}$

Decreasing at$\mathbf{\left(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{0}\mathbf{\right)}$ ## Step 4. Verifying algebraic answers with graphs :  ### Want to see more solutions like these? 