Q. 40 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 29 – 44, for the given functions f and g, find:

(a) $f\circ g$

 (b) $g\circ f$

(c) $f\circ f$

 (d)  $g\circ g$

State the domain of each composite function.    

$$\begin{gathered} f\left(x\right)=\sqrt{x-2} \\ g\left(x\right)=1-2x \end{gathered}$$

Step-by-Step Solution

Answer

a) $f\circ g=\sqrt{-1-2x}$and its domain is:$x\le -\frac{1}{2}$

b) $g\circ f=1-2\sqrt{x-2}$ , and its domain is: $x\ge 2$

c) $f\circ f=\sqrt{\sqrt{x-2}-2}$ and its domain is:$x\ge 6$.

d) $g\circ g=4x-1$ and its domain is:$\left(-\infty ,\infty \right)$

Step 1. Given information:

The functions are:

$$\begin{gathered} f\left(x\right)=\sqrt{x-2} \\ g\left(x\right)=1-2x \end{gathered}$$

The domain of $f\left(x\right)=x\ge 2$

The domain of $g\left(x\right)=\left(-\infty ,\infty \right)$

Part (a) Step 1. Find f ∘ g and its domain.

$$\begin{gathered} f\circ g=fg\left(x\right)) \\ =f(1-2x) \\ =\sqrt{1-2x-2} \\ =\sqrt{-1-2x} \end{gathered}$$

This is defined when $g\left(x\right)$ is defined and

 $$\begin{gathered} -1-2x\ge 0 \\ x\le -\frac{1}{2} \end{gathered}$$

So domain is:$x\le -\frac{1}{2}$

Part (b) Step 1. Find g ∘ f and its domain.

$$\begin{gathered} g\circ f=g\left(f\right(x\left)\right) \\ =g\left(\sqrt{x-2}\right) \\ =1-2\sqrt{x-2} \end{gathered}$$

This is defined only when $f\left(x\right)$ is defined and $$\begin{gathered} x-2\ge 0 \\ x\ge 2 \end{gathered}$$

So the domain is:$x\ge 2$

Part (c) Step 1. Find f ∘ f and its domain.

$$\begin{gathered} f\circ f=f\left(f\right(x\left)\right) \\ =f\left(\sqrt{x-2}\right) \\ =\sqrt{\sqrt{x-2}-2} \end{gathered}$$

This is defined when $f\left(x\right)$ is defined and 

 $$\begin{gathered} \sqrt{x-2}-2\ge 0 \\ \sqrt{x-2}\ge 2 \\ x-2\ge 4 \\ x\ge 4+2 \\ x\ge 6 \end{gathered}$$

So domain is:

$x\ge 6$

Part (d) Step 1. Find g ∘ g and its domain.

$$\begin{gathered} g\circ g=g\left(g\right(x\left)\right) \\ =g(1-2x) \\ =1-2(1-2x) \\ =1-2+4x \\ =4x-1 \end{gathered}$$

This is defined when $g\left(x\right)$ is defined.

The domain of this is $\left(-\infty ,\infty \right)$