Q 42. - Precalculus Enhanced with Graphing Utilities

Question

In Problems 29 – 44, for the given functions f and g, find$\left(a\right) f\circ g \left(b\right) g\circ f \left(c\right) f\circ f \left(d\right) g\circ g$

State the domain of each composite function.

$f\left(x\right)=x^2+4; g\left(x\right)=\sqrt{x-2}$

Step-by-Step Solution

Answer

a) $f\circ g=x+2$ and its domain is $\left\{x\overline{)x\ge 2}\right\}$.

b) $g\circ f=\sqrt{x^2+2}$ and its domain is all real numbers.

c) $f\circ f=x^4+8x^2+20$and its domain is all real numbers.

d) $g\circ g=\sqrt{\sqrt{x-2}-2}$ and its domain is $\left\{x\overline{)x\ge 6}\right\}$

Step 1. Given information

The given composite function is:

$$\begin{gathered} f\left(x\right)=x^2+4 \\ g\left(x\right)=\sqrt{x-2} \end{gathered}$$

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

The domain of $f$ is the set of all real numbers and g is $\left\{x\overline{)x\ge 1}\right\}$.

$(f\circ g)\left(x\right)=f\left(g\right(x\left)\right)$

Now substitute $g\left(x\right)=\sqrt{x-2}$ in the function $f\left(g\right(x\left)\right)$.

Then the function will become $f\left(\sqrt{x-2}\right)$.

Now replace x with $\sqrt{x-2}$ in$f\left(x\right)=x^2+4$,

$$\begin{gathered} f\left(\sqrt{x-2}\right)={(x-2)}^2+4 \\ =x-2+4 \\ =x+2 \end{gathered}$$

Therefore, the domain of $f\circ g$ is $\left\{x\overline{)x\ge 2}\right\}$ and $(f\circ g)\left(x\right)=x+2$.

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

$(g\circ f)\left(x\right)=g\left(f\right(x\left)\right)$

Now substitute $f\left(x\right)=x^2+4$ in the function $g\left(f\right(x\left)\right)$.

Then the function will become $g(x^2+4)$.

Now replace x with $(x^2+4)$ in $g\left(x\right)=\sqrt{x-2}$,

$$\begin{gathered} g(x^2+4)=\sqrt{x^2+4-2} \\ =\sqrt{x^2+2} \end{gathered}$$

Therefore, the domain of $g\circ f$ is all real numbers and $(g\circ f)\left(x\right)=\sqrt{x^2+2}$.

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

$(f\circ f)\left(x\right)=f\left(f\right(x\left)\right)$

Now substitute $f\left(x\right)=x^2+4$ in the function $f\left(f\right(x\left)\right)$.

Then the function will become $f(x^2+4)$.

Now replace x with $x^2+4$ in $f\left(x\right)=x^2+4$,

$$\begin{gathered} f(x^2+4)={(x^2+4)}^2+4 \\ =x^4+8x^2+20 \end{gathered}$$

Therefore the domain of $f\circ f$ is all real numbers and $(f\circ f)\left(x\right)=x^4+8x^2+20$. 

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

$(g\circ g)\left(x\right)=g\left(g\right(x\left)\right)$

Now substitute $g\left(x\right)=\sqrt{x-2}$ in the function $g\left(g\right(x\left)\right)$.

Then the function will become $g\left(\sqrt{x-2}\right)$..

Now replace x with $\sqrt{x-2}$ in $g\left(x\right)=\sqrt{x-2}$,

$g\left(\sqrt{x-2}\right)=\sqrt{\sqrt{x-2}-2}$

Solve the inequality to find x,

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

Therefore, the domain of $g\circ g$ is $\left\{x\overline{)x\ge 6}\right\}$ and $(g\circ g)\left(x\right)=\sqrt{\sqrt{x-2}-2}$.