Chapter 3.7, Problem 18PPS

To determine

To sketch: the given function and to state the domain and range.

Answer to Problem 18PPS

The final range of the function is $\left(-3,\infty \right)$ .

Domain of the total function is $\left(-\infty ,\infty \right)$ .

Explanation of Solution

Given information:

Given function,

  $f\left(x\right)=\{\begin{array}{l}2x+3\text{if}x\ge -3\\ -\frac{1}{3}x+1\text{if}x<-3\end{array}$

Calculation:

Graph:

  

Given the $x\ge -3$ for the first equation $2x+3$ .

Therefore the domain of the first function is $\left[-3,\infty \right)$ .

For the second equation the $f\left(x\right)=-\frac{1}{3}x+1$ where, $x<-3$ .

Therefore the domain of the second function is $\left(-\infty ,-3\right)$ .

So these two inequalities $x\ge -3$ and $x<-3$ covers the entire number line.

Therefore, the combined final domain of the total function is $\left(-\infty ,\infty \right)$ .

The range of the first function $f\left(x\right)=2x+3$ starts with $-3$ as the boundary ( $-3$ is included) when $x>-3$ the value of $f\left(x\right)$ raises above $-3$ .

Therefore, the range of the first function is $\left[-3,\infty \right)$ .

For the second equation $f\left(x\right)=-\frac{1}{3}x+1,x<-3$ when $x=-3,f\left(x\right)=2$

  $f\left(x\right)=-\frac{1}{3}x+1$

When $x=-6,f\left(x\right)=3$

Therefore, the range of the second function starts from 2 and raises above 2.

Therefore the range of the second function is $\left[2,\infty \right)$

So the final range of the function is $\left(-3,\infty \right)$ .