Q. 36 - 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. 

$$\begin{gathered} f\left(x\right)=\frac{1}{x+3}, \\ g\left(x\right)=-\frac{2}{x} \end{gathered}$$

Step-by-Step Solution

Answer

a) $f\circ g=\frac{x}{3x-2}$, its domain is:$\left\{x|x\ne 0,x\ne \frac{2}{3}\right\}$.

b)$g\circ f=-2(x+3)$, its domain is:$\left\{x|x\ne -3\right\}$

c)$f\circ f=\frac{x+3}{3x+10}$,its domain is:$\left\{x|x\ne -3,x\ne -\frac{10}{3}\right\}$

d) $g\circ g=x$, its domain is:$\left\{x|x\ne 0\right\}$

Step 1. Given information:

The functions are:

$$\begin{gathered} f\left(x\right)=\frac{1}{x+3}, \\ g\left(x\right)=-\frac{2}{x} \end{gathered}$$

$f\left(x\right)$ is defined for $x\ne -3$.

$g\left(x\right)$ is defined for $x\ne 0$

Step 2. Find f ∘ g and its domain:

$$\begin{gathered} f\circ g=f\left(g\right(x\left)\right) \\ =\frac{1}{-{\displaystyle \frac{2}{x}}+3} \\ =\frac{1}{{\displaystyle \frac{-2+3x}{x}}} \\ =\frac{x}{3x-2} \end{gathered}$$

This is defined when,

$$\begin{gathered} 3x-2\ne 0 \\ x\ne \frac{2}{3} \end{gathered}$$ 

and 

for the $g\left(x\right)$ to be defined $x\ne 0$

So, Domain is:$\left\{x|x\ne \frac{2}{3},x\ne 0\right\}$

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

$$\begin{gathered} g\circ f=g\left(f\right(x\left)\right) \\ =-\frac{2}{{\displaystyle \frac{1}{x+3}}} \\ =-2(x+3) \end{gathered}$$

The domain of the function is the domain of $f\left(x\right)$

So, the domain is:$\left\{x|x\ne -3\right\}$.

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

$$\begin{gathered} f\circ f=f\left(f\right(x\left)\right) \\ =\frac{1}{{\displaystyle \frac{1}{x+3}}+3} \\ =\frac{1}{{\displaystyle \frac{1+3x+9}{x+3}}} \\ =\frac{x+3}{3x+10} \end{gathered}$$

This is defined when

 $$\begin{gathered} 3x+10\ne 0 \\ x\ne -\frac{10}{3} \end{gathered}$$

The domain is:

$\left\{x|x\ne -\frac{10}{3},x\ne -3\right\}$

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

$$\begin{gathered} g\circ g=g\left(g\right(x\left)\right) \\ =-\frac{2}{-{\displaystyle \frac{2}{x}}} \\ =x \end{gathered}$$

The domain is;$\left\{x|x\ne 0\right\}$