Chapter 3.10, Problem 22WE

To determine

To check:whether the functions f and g are equal or not.

Answer to Problem 22WE

The functions f and g are equal on the domain $D=\left\{x\text{ :}\text{}x\ge -1\right\}$ .

Explanation of Solution

Given information:

Two functions $f\left(x\right)\to x^2+x\text{ and }g\left(x\right)\to x\left|1+x\right|$ .

The domain and range of the functions f and g are same.

The domain of the functions $D=\left\{x\text{ :}\text{}x\ge -1\right\}$

Calculation:

It is given that the function $f\left(x\right)=x^2+x\text{ and }g\left(x\right)=x\left|1+x\right|$ .

Note that, the modulus of x defines as $\left|x\right|=\left\{\begin{array}{l}x\text{ if }x\ge 0\\ -x\text{ if }x<0\end{array}\right\}$

The modulus of $x+1$ can be written as follows,

  $\begin{array}{l}\left|x+1\right|=\left\{\begin{array}{l}x+1\text{ if }x+1\ge 0\\ -\left(x+1\right)\text{ if }x+1<0\end{array}\right\}\\ \left|x+1\right|=\left\{\begin{array}{l}x+1\text{ if }x\ge -1\\ -\left(x+1\right)\text{ if }x<-1\end{array}\right\}\end{array}$

That is, $\left|x+1\right|=x+1$ for $x\ge -1$ .

Since the domain of the both functions is $D=\left\{x\text{ :}\text{}x\ge -1\right\}$ , substitute $\left|x+1\right|=x+1$ in $g\left(x\right)=x\left|1+x\right|$ ,

  $\begin{array}{c}g\left(x\right)=x\left(1+x\right)\\ =x+x^2\\ =x^2+x\end{array}$

The function $g\left(x\right)=x^2+x$ on $D=\left\{x\text{ :}\text{}x\ge -1\right\}$ which is equal to the function $f\left(x\right)=x^2+x$ .

Therefore, the functions f and g are equal on the domain $D=\left\{x\text{ :}\text{}x\ge -1\right\}$ .