Chapter 5, Problem 10CR

To determine

The composite function $\left(f\circ g\right)\left(x\right)$ , its domain, and the value of $\left(f\circ g\right)\left(5\right)$ , where $f\left(x\right)=x^2+2,g\left(x\right)=\frac{2}{x-3}$ .

Answer to Problem 10CR

Solution:

The composite function $\left(f\circ g\right)\left(x\right)$ is $\underset{\_}{\left(f\circ g\right)\left(x\right)=\frac{4}{{\left(x-3\right)}^2}+2}$ .

The domain of $\left(f\circ g\right)\left(x\right)$ is $\underset{\_}{\left(-\infty ,3\right)\cup \left(3,\infty \right)}$ .

The value of $\left(f\circ g\right)\left(5\right)$ is $3$ .

Explanation of Solution

Given information:

The functions $f\left(x\right)=x^2+2,g\left(x\right)=\frac{2}{x-3}$ .

The definition of composite function is $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ ,

Plug in $g\left(x\right)=\frac{2}{x-3}$ ,

  $\Rightarrow \left(f\circ g\right)\left(x\right)=f\left(\frac{2}{x-3}\right)$

As $f\left(x\right)=x^2+2$ , then $f\left(\frac{2}{x-3}\right)={\left(\frac{2}{x-3}\right)}^2+2$ ,

  $\Rightarrow \left(f\circ g\right)\left(x\right)={\left(\frac{2}{x-3}\right)}^2+2$

  $\Rightarrow \left(f\circ g\right)\left(x\right)=\frac{2^2}{{\left(x-3\right)}^2}+2$

  $\Rightarrow \left(f\circ g\right)\left(x\right)=\frac{4}{{\left(x-3\right)}^2}+2$

Therefore, the composite function is $\left(f\circ g\right)\left(x\right)=\frac{4}{{\left(x-3\right)}^2}+2$ .

As $f\left(x\right)=x^2+2$ is a polynomial function and defined for all real numbers, then the domain of $f\left(x\right)$ is $\left(-\infty ,\infty \right)$ .

Now, $g\left(x\right)=\frac{2}{x-3}$ is a rational function, which is not defined if the denominator is zero.

Therefore, $g\left(x\right)$ is not defined when $x-3=0\Rightarrow x=3$ . So, exclude $x=3$ from the domain of $g\left(x\right)$ .

Hence, the domain of $g\left(x\right)=\frac{2}{x-3}$ is $\left\{x\in R|x\ne 3\right\}$ .

Since $x=3$ is not in domain of $g\left(x\right)$ , it will not be in domain of $\left(f\circ g\right)\left(x\right)$ also.

So, domain of $\left(f\circ g\right)\left(x\right)$ would be all real numbers except $3$ .

Therefore, the domain of composite function $\left(f\circ g\right)\left(x\right)$ is $\left(-\infty ,3\right)\cup \left(3,\infty \right)$ .

To find $\left(f\circ g\right)\left(5\right)$ , substitute $x=5$ in $\left(f\circ g\right)\left(x\right)=\frac{4}{{\left(x-3\right)}^2}+2$ ,

  $\Rightarrow \left(f\circ g\right)\left(5\right)=\frac{4}{{\left(5-3\right)}^2}+2$

  $\Rightarrow \left(f\circ g\right)\left(5\right)=\frac{4}{2^2}+2$

  $\Rightarrow \left(f\circ g\right)\left(5\right)=\frac{4}{4}+2$

  $\Rightarrow \left(f\circ g\right)\left(5\right)=1+2$

  $\Rightarrow \left(f\circ g\right)\left(5\right)=3$

Therefore, $\left(f\circ g\right)\left(5\right)=3$ .