Chapter 3, Problem 11CLT

a .

To determine

To find: $\left(f\circ g\right)\left(x\right)$ where $f\left(x\right)=2x^2$ and $g\left(x\right)=\sqrt{x+6}$

Answer to Problem 11CLT

  $\left(f\circ g\right)\left(x\right)=2x+12$

Domain of $f\circ g$ : All real numbers x such that $x\ge -6$

Explanation of Solution

Consider,

  $\begin{array}{c}\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)\\ =f\left(\sqrt{x+6}\right)\\ =2{\left(\sqrt{x+6}\right)}^2\end{array}$

  $\begin{array}{c}=2\left(x+6\right)\\ =2x+12\end{array}$

The domain of $f\circ g$ is the set of all x in the domain of g such that $g\left(x\right)$ is in the domain of f .

Now, the domain of f is the set of real numbers and it includes all real values of g .

So, domain of $f\circ g$ is the entire domain of g which is, all real numbers x such that $x\ge -6$

b.

To determine

To find: $\left(g\circ f\right)\left(x\right)$ where $f\left(x\right)=2x^2$ and $g\left(x\right)=\sqrt{x+6}$

Answer to Problem 11CLT

  $\left(g\circ f\right)\left(x\right)=\sqrt{2x^2+6}$

Domain of $g\circ f$ : All real numbers

Explanation of Solution

Consider,

  $\begin{array}{c}\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(2x^2\right)\\ =\sqrt{2x^2+6}\end{array}$

The domain of $g\circ f$ is the set of all x in the domain of f such that $f\left(x\right)$ is in the domain of g .

Now, domain of g is all real numbers x such that $x\ge -6$

The domain of f is the set of real numbers. And $f\left(x\right)\ge 0$ for all x .

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