Q. 38 - 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.   

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

Step-by-Step Solution

Answer

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

b) $g\circ f=\frac{2(x+3)}{x}$, and its domain:$\left\{x|x\ne 0,x\ne -3\right\}$

c)$f\circ f=\frac{x}{4x+9}$  and its domain is:$\left\{x|x\ne -3,x\ne -\frac{9}{4}\right\}$.

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

Step 1. Given information:

The functions are:

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

The domain of $f\left(x\right)=\left\{x\right|x\ne -3\}$

The domain of $g\left(x\right)=\left\{x\right|x\ne 0\}$

Part (a) Step 1. To find f ∘ g and its domain:

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

This is defined when $g\left(x\right)$ is defined and $$\begin{gathered} 3x+2\ne 0 \\ x\ne -\frac{2}{3} \end{gathered}$$

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

Part (b) Step 1. To find g ∘ f and its domain:

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

This is defined when $f\left(x\right)$ is defined and $x\ne 0$.

So domain is:$\left\{x|x\ne -3,x\ne 0\right\}$

Part (c) Step 1. To find and its domain:

$$\begin{gathered} f\circ f=f\left(f\right(x\left)\right) \\ =\frac{{\displaystyle \frac{x}{x+3}}}{{\displaystyle \frac{x}{x+3}}+3} \\ =\frac{{\displaystyle \frac{x}{x+3}}}{{\displaystyle \frac{4x+9}{x+3}}} \\ =\frac{x}{4x+9} \end{gathered}$$

This is only defined when $f\left(x\right)$ is defined and $$\begin{gathered} 4x+9\ne 0 \\ x\ne -\frac{9}{4} \end{gathered}$$

So domain is:$\left\{x|x\ne -3,x\ne -\frac{4}{9}\right\}$

Part (d) Step 1. To find g ∘ g and its domain:

$$\begin{gathered} g\circ g=\frac{2}{{\displaystyle \frac{2}{x}}} \\ =x \end{gathered}$$

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

So, domain is:$\left\{x|x\ne 0\right\}$