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Q. 11

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Found in: Page 260

### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

# Use the definitions of increasing and decreasing to argue that $f\left(x\right)={x}^{4}$ is decreasing on $\left(-\infty ,0\right]$ and increasing on $\left[0,\infty \right)$. Then use derivatives to argue the same thing.

The statement has been proven.

See the step by step solution

## Step 1. Given information

We have been given a function $f\left(x\right)={x}^{4}$.

We have to use the definitions of increasing and decreasing to argue that this function is decreasing on $\left(-\infty ,0\right]$ and increasing on $\left[0,\infty \right)$.

## Step 2. Using the definition

For a and b in the interval $\left(-\infty ,0\right]$

Now if $a

Then,

role="math" localid="1648442267515" ${a}^{4}>{b}^{4}$

Thus the function is decreasing in the interval $\left(-\infty ,0\right]$

Also,

For a and b in the interval role="math" localid="1648442301380" $\left[0,\infty \right)$

Now if $0 then,

${a}^{4}<{b}^{4}$

Thus the function is increasing in the interval $\left[0,\infty \right)$

## Step 3. Using the derivative

The derivative of the function is given by :

${f}^{\mathrm{\prime }}\left(x\right)=4{x}^{3}$

The function ${f}^{\mathrm{\prime }}\left(x\right)=4{x}^{3}$ is always negative for x<0

The function ${f}^{\mathrm{\prime }}\left(x\right)=4{x}^{3}$ is always positive for x>0

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