Chapter 1.2, Problem 8CFU

To determine

To find: domain of $\left[f\circ g\right]\left[x\right]$

Answer to Problem 8CFU

  $\left\{x|x\ne -2\right\}$

Explanation of Solution

Given information: The functions given are as follows:

  $\begin{array}{l}f\left(x\right)=\frac{1}{{\left(x-1\right)}^2}\\ g\left(x\right)=x+3\end{array}$

Calculation:

The denominator of $f\left(x\right)$ should not be zero. So, put the denominator equal to zero and find the excluded value as follows:

  $\begin{array}{l}{\left(x-1\right)}^2=0\\ x-1=0\left[\text{taking square root both sides}\right]\\ x=1\end{array}$

Thus, the domain of $f\left(x\right)$ is all real numbers except 1.

So, for $x$ to be in the domain of $\left[f\circ g\right]\left[x\right]$ , $g\left(x\right)\ne 1$ .

Substitute the value of $g\left(x\right)$ in $g\left(x\right)\ne 1$ to get the excluded values.

  $\begin{array}{c}g\left(x\right)\ne 1\\ x+3\ne 1\\ x\ne 1-3\\ x\ne -2\end{array}$

Therefore, the domain of $\left[f\circ g\right]\left[x\right]$ is all real numbers except $-2$ .

For $g\left(x\right)=x^3$ , x is equal to all real numbers. Thus, the domain of $g\left(x\right)$ is all real numbers.

Use $f\left(x\right)\text{ and }g\left(x\right)$ to evaluate $\left[f\circ g\right]\left[x\right]$ as follows:

  $\begin{array}{c}\left[f\circ g\right]\left[x\right]=f\left(g\left(x\right)\right)\\ =5\left(x^3\right)\\ =5x^3\end{array}$

Therefore, the domain of $\left[f\circ g\right]\left[x\right]$ is all real numbers.