Chapter 14.4, Problem 45STP

To determine

To Find:Domain and range $f(x)=\left|4x+1\right|-8$

Answer to Problem 45STP

  $D=\left\{allrealNumbers\right\},R=\{y|y\ge -8\}$

Explanation of Solution

Given information:

We know that, the given function is $f(x)=\left|4x+1\right|-8$

Calculation:

  $f(x)=\left|4x+1\right|-8$

  $\because$ Modular function is always define.

  $\left|4x+1\right|$ is always define

  $\therefore$ Domain of $f(x)=\left|4x+1\right|-8$ will be all real number

  $\therefore$ Domain $\in R.$

  $\begin{array}{l}\because f(x)=\left|4x+1\right|-8\\ \because \left|4x+1\right|\ge 0\end{array}$

at $x=-\frac{1}{4}\left|4x+1\right|=0$

  $\therefore$ Range "or" the value of $f(x)$ will be $f(x)\ge -8$

  $\therefore$ Range : $(-8,\infty )$