Chapter 3.7, Problem 20PPS

To determine

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

Answer to Problem 20PPS

The final range of the function is $\left[-2.5,\infty \right)$ .

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

Explanation of Solution

Given information:

Given function,

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

Calculation:

Graph:

  

Given the $x>-1$ for the first equation $3x+2$ .

Therefore the domain of the first function is $\left(-1,\infty \right)$ .

For the second equation the $f\left(x\right)=-\frac{1}{2}x-3$ where, $x\le -1$ .

Therefore the domain of the second function is $\left(-\infty ,-1\right]$ .

So these two inequalities $x>-1$ and $x\le -1$ 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 $f\left(x\right)=3x+2$ starts with $-1$ as the boundary ( $-1$ is included) when $x>-1$ the value of $f\left(x\right)$ raises above $-1$ .

Therefore, the range of the first function is $\left(-1,\infty \right)$ .

For the second equation $f\left(x\right)=-\frac{1}{2}x-3,x\le -1$ when $x=-1,f\left(x\right)=-2.5$

  $f\left(x\right)=-\frac{1}{3}x+1,$

When $x=-8,f\left(x\right)=1$

Therefore, the range of the second function starts from $-2.5$ and raises above 2.

Therefore the range of the second function is $\left[-2.5,\infty \right)$

So the final range of the function is $\left[-2.5,\infty \right)$ .