Chapter 0, Problem 60RE

(a)

To determine

To graph : The graph of $y=f\left(-x\right)$ .

Explanation of Solution

Given information :

The graph of $f$ .

  

Graph:

First find the equation of this curve.

Now, equation of straight line passing through the points $\left(0,0\right)$ and $\left(1,-1\right)$ is

  $\begin{array}{l}\frac{y-0}{x-0}=\frac{0+1}{0-1}\\ \frac{y}{x}=-1\\ y=-x\end{array}$

Now, equation of straight line which passing through the points $\left(1,-1\right)$ and $\left(3,2\right)$ is

  $\begin{array}{l}\frac{y+1}{x-1}=\frac{-1-2}{1-3}\\ \Rightarrow \frac{y+1}{x-1}=\frac{-3}{-2}\\ \Rightarrow 2y+2=3x-3\\ \Rightarrow 2y=3x-5\\ \Rightarrow y=\frac{3x-5}{2}\end{array}$

Therefore, equation of this curve is

  $\begin{array}{l}f\left(x\right)=-x;x\in \left[0,1\right]\\ =\frac{3x-5}{2}\text{; }x\in \left[1,3\right]\end{array}$

  $y=f\left(-x\right)$

So,

  $\begin{array}{l}y=f\left(-x\right)=-\left(-x\right)\text{; }x\in \left[0,1\right]\\ =\frac{3\left(-x\right)-5}{2}\text{; }x\in \left[1,3\right]\end{array}$

Or,

  $\begin{array}{l}y=f\left(-x\right)=x\text{; }x\in \left[0,1\right]\\ =\frac{-3x-5}{2}\text{; }x\in \left[1,3\right]\end{array}$

  

Interpretation:

  $\begin{array}{l}y=f\left(-x\right)=x,x\in \left[0,1\right]\\ y=f\left(-x\right)=\frac{-3x-5}{2},x\in \left[1,3\right]\end{array}$

(b)

To determine

To graph : The graph of $y=-f\left(x\right)$ .

Explanation of Solution

Given information :

The graph of $f$ .

  

Graph:

First find the equation of this curve.

Now, equation of straight line passing through the points $\left(0,0\right)$ and $\left(1,-1\right)$ is

  $\begin{array}{l}\frac{y-0}{x-0}=\frac{0+1}{0-1}\\ \frac{y}{x}=-1\\ y=-x\end{array}$

Now, equation of straight line which passing through the points $\left(1,-1\right)$ and $\left(3,2\right)$ is

  $\begin{array}{l}\frac{y+1}{x-1}=\frac{-1-2}{1-3}\\ \Rightarrow \frac{y+1}{x-1}=\frac{-3}{-2}\\ \Rightarrow 2y+2=3x-3\\ \Rightarrow 2y=3x-5\\ \Rightarrow y=\frac{3x-5}{2}\end{array}$

Therefore, equation of this curve is

  $\begin{array}{l}f\left(x\right)=-x;x\in \left[0,1\right]\\ =\frac{3x-5}{2}\text{; }x\in \left[1,3\right]\end{array}$

  $y=-f\left(x\right)$

So,

  $\begin{array}{l}y=-f\left(x\right)=-\left(-x\right)\text{; }x\in \left[0,1\right]\\ =-\frac{3x-5}{2}\text{; }x\in \left[1,3\right]\end{array}$

Or,

  $\begin{array}{l}y=-f\left(x\right)=x\text{; }x\in \left[0,1\right]\\ =\frac{-3x+5}{2}\text{; }x\in \left[1,3\right]\end{array}$

  

Interpretation:

  $\begin{array}{l}y=f\left(-x\right)=x,x\in \left[0,1\right]\\ y=f\left(-x\right)=-\left(\frac{3x-5}{2}\right),x\in \left[1,3\right]\end{array}$

(c)

To determine

To graph : The graph of $y=-2f\left(x+1\right)$ .

Explanation of Solution

Given information :

The graph of $f$ .

  

Graph:

First find the equation of this curve.

Now, equation of straight line passing through the points $\left(0,0\right)$ and $\left(1,-1\right)$ is

  $\begin{array}{l}\frac{y-0}{x-0}=\frac{0+1}{0-1}\\ \frac{y}{x}=-1\\ y=-x\end{array}$

Now, equation of straight line which passing through the points $\left(1,-1\right)$ and $\left(3,2\right)$ is

  $\begin{array}{l}\frac{y+1}{x-1}=\frac{-1-2}{1-3}\\ \Rightarrow \frac{y+1}{x-1}=\frac{-3}{-2}\\ \Rightarrow 2y+2=3x-3\\ \Rightarrow 2y=3x-5\\ \Rightarrow y=\frac{3x-5}{2}\end{array}$

Therefore, equation of this curve is

  $\begin{array}{l}f\left(x\right)=-x;x\in \left[0,1\right]\\ =\frac{3x-5}{2}\text{; }x\in \left[1,3\right]\end{array}$

  $y=-2f\left(x+1\right)+1$

So,

  $\begin{array}{l}y=-2f\left(x+1\right)+1=-2\left(-\left(x+1\right)\right)\text{+1; }x\in \left[0,1\right]\\ =-\frac{3\left(x+1\right)-5}{2}\left(x-2\right)\text{; }x\in \left[1,3\right]\end{array}$

Or,

  $\begin{array}{l}y=-2f\left(x+1\right)+1=2x+3\text{; }x\in \left[0,1\right]\\ =-3x+3\text{; }x\in \left[1,3\right]\end{array}$

  

Interpretation:

  $\begin{array}{l}y=-2f\left(x+1\right)+1=2x+3,x\in \left[0,1\right]\\ y=-2f\left(x+1\right)=-3x+3,x\in \left[1,3\right]\end{array}$

(d)

To determine

To graph : The graph of $y=3f\left(x-2\right)-2$ .

Explanation of Solution

Given information :

The graph of $f$ .

  

Graph:

First find the equation of this curve.

Now, equation of straight line passing through the points $\left(0,0\right)$ and $\left(1,-1\right)$ is

  $\begin{array}{l}\frac{y-0}{x-0}=\frac{0+1}{0-1}\\ \frac{y}{x}=-1\\ y=-x\end{array}$

Now, equation of straight line which passing through the points $\left(1,-1\right)$ and $\left(3,2\right)$ is

  $\begin{array}{l}\frac{y+1}{x-1}=\frac{-1-2}{1-3}\\ \Rightarrow \frac{y+1}{x-1}=\frac{-3}{-2}\\ \Rightarrow 2y+2=3x-3\\ \Rightarrow 2y=3x-5\\ \Rightarrow y=\frac{3x-5}{2}\end{array}$

Therefore, equation of this curve is

  $\begin{array}{l}f\left(x\right)=-x;x\in \left[0,1\right]\\ =\frac{3x-5}{2}\text{; }x\in \left[1,3\right]\end{array}$

  $y=3f\left(x-2\right)-2$

So,

  $\begin{array}{l}y=3f\left(x-2\right)-2=\left\{3\times -\left(x-2\right)\right\}\text{-2; }x\in \left[0,1\right]\\ =3\times \frac{3\left(x-2\right)-5}{2}-2\text{; }x\in \left[1,3\right]\end{array}$

Or,

  $\begin{array}{l}y=3f\left(x-2\right)-2=-3x+4\text{; }x\in \left[0,1\right]\\ =\frac{9x-37}{2}\text{; }x\in \left[1,3\right]\end{array}$

  

Interpretation:

  $\begin{array}{l}y=3f\left(x-2\right)-2=-3x+4,x\in \left[0,1\right]\\ y=3f\left(x-2\right)-2=\frac{9x-37}{2},x\in \left[1,3\right]\end{array}$