Chapter 1.7, Problem 54E

a.

To determine

To calculate: the open intervals on which the graph transformation is increasing and decreasing.

Answer to Problem 54E

  $y=f\left(-x\right)$ is increasing over $\left(-2,1\right)$ and decreasing over $\left(-\infty ,-2\right)\&\left(1,\infty \right)$

Explanation of Solution

Given information:

  $y=f\left(-x\right)$

  

Calculation:

The graph of $f\left(-x\right)$ will be mirror image of given graph with respect to y-axis. So now increasing intervals will become decreasing and vice-versa.

In this graph increasing intervals are $\left(-\infty ,-1\right)\&\left(2,\infty \right)$ .So now decreasing intervals will be $\left(1,\infty \right)\&\left(-\infty ,-2\right)$

In this graph decreasing interval is .So now increasing interval is $\left(-2,1\right)$

b.

To determine

To Find: the open intervals on which the graph transformation is increasing and decreasing using Graph.

Answer to Problem 54E

  $y=-f\left(x\right)$ is increasing over $\left(-1,2\right)$ and decreasing over $\left(-\infty ,-1\right)\&\left(2,\infty \right)$

Explanation of Solution

Given information:

  $y=-f\left(x\right)$

  

Calculation:

The graph of $-f\left(x\right)$ will be mirror image of given graph with respect to x -axis. So now increasing intervals will become decreasing and vice-versa.

In this graph increasing intervals are $\left(-\infty ,-1\right)\&\left(2,\infty \right)$ .So now decreasing intervals will be $\left(2,\infty \right)\&\left(-\infty ,-1\right)$

In this graph decreasing interval is $\left(-1,2\right)$ So now increasing interval is $\left(-1,2\right)$

c.

To determine

To calculate: the open intervals on which the graph transformation is increasing and decreasing.

Answer to Problem 54E

Function is increasing in $\left(-\infty ,-1\right)\&\left(2,\infty \right)$ and function is decreasing in $\left(-1,2\right)$

Explanation of Solution

Given information:

  $y=\frac{1}{2}f\left(x\right)$

  

Calculation:

If a function $y=f\left(x\right)$ changes to $y=tf\left(x\right)$ then intervals remain same, just vertical y-axis is multiplied by t.

So here also the new function will shrink vertically by $\frac{1}{2}$ and intervals will remain same.

So function is increasing in $\left(-\infty ,-1\right)\&\left(2,\infty \right)$ and function is decreasing in $\left(-1,2\right)$

d.

To determine

To calculate: the open intervals on which the graph transformation is increasing and decreasing.

Answer to Problem 54E

Function is decreasing in $\left(-\infty ,0\right)\&\left(3,\infty \right)$ and function is decreasing in $\left(0,3\right)$

Explanation of Solution

Given information:

  $y=-f\left(x-1\right)$

  

Calculation:

The graph of $-f\left(x\right)$ will be mirror image of given graph with respect to x-axis and the graph of $f\left(x-1\right)$ will shift the interval right by 1. So graph of $-f\left(x-1\right)$ will reflection over x-axis and shift the interval right by 1.

In this graph increasing intervals are (-8,-1) and (2,8). So now decreasing intervals will be $\left(3,\infty \right)\&\left(-\infty ,0\right)$ .

In this graph decreasing interval is $\left(-1,2\right)$ . So now increasing interval is $\left(0,3\right)$ .

e.

To determine

To calculate: the open intervals on which the graph transformation is increasing and decreasing.

Answer to Problem 54E

The function is increasing in $\left(-\infty ,1\right)\&\left(2,\infty \right)$ and the function is decreasing in $\left(1,4\right)$ .

Explanation of Solution

Given information:

  $y=f\left(x-2\right)+1$

  

Calculation:

The graph of $y=f\left(x-2\right)+1$ will shift the interval right by 2. So,

In this graph increasing intervals are $\left(-\infty ,1\right)\&\left(2,\infty \right)$ . So now increasing intervals will be $\left(4,\infty \right)$ and $\left(-\infty ,1\right)$ .

In this graph decreasing interval is $\left(-1,2\right)$ . So now decreasing interval is $\left(1,4\right)$