Chapter 5, Problem 10RE

To determine

To find : The expression for $f\circ g$ , $f\circ f$ , $g\circ f$ , $g\circ g$ and state the domain of each composite function.

Explanation of Solution

Given: The functions are $f\left(x\right)=\sqrt{3x}$ and $g\left(x\right)=1+x+x^2$ .

The value of $\left(f\circ g\right)\left(x\right)$ is calculated as,

  $\begin{array}{c}\left(f\circ g\right)\left(x\right)=\sqrt{3\left(1+x+x^2\right)}\\ =\sqrt{3+3x+3x^2}\end{array}$

The domain of $g$ is the set of all real numbers.

So, the domain of $f\circ g$ is the set of all real numbers.

The value of $\left(g\circ f\right)$ is calculated as,

  $\begin{array}{c}\left(g\circ f\right)\left(x\right)=1+\sqrt{3x}+{\left(\sqrt{3x}\right)}^2\\ =1+\sqrt{3x}+3x\end{array}$

The domain of $f$ is $\left[0,\infty \right)$

So, the domain of $g\circ f$ is $\left[0,\infty \right)$ .

The value of $\left(g\circ g\right)\left(x\right)$ is calculated as,

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

The domain of $g$ is the set of all real numbers.

So, the domain of $\left(g\circ g\right)\left(x\right)$ is the set of all real numbers

The value of $\left(f\circ f\right)\left(x\right)$ is calculated as,

  $\left(f\circ f\right)\left(x\right)=\sqrt{3\sqrt{3x}}$

The domain of $f$ is $\left[0,\infty \right)$

So, the domain of $\left(f\circ f\right)\left(x\right)$ is $\left[0,\infty \right)$ .