Q. 39 - 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.   

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

Step-by-Step Solution

Answer

a) $f\circ g=\sqrt{2x+3}$and its domain is:$x\ge -\frac{3}{2}$

b) $g\circ f=2\sqrt{x}+3$ , and its domain:$x\ge 0$

c) $f\circ f=\sqrt[4]{x}$ and its domain is:$x\ge 0$.

d) $g\circ g=4x+9$ and its domain is:$\left(-\infty ,\infty \right)$

Step 1. Given information:

The functions are:

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

The domain of $f\left(x\right)=\left\{x\right|x\ge 0\}$

The domain of $g\left(x\right)$ is set of all real numbers.

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

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

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

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

So its domain is:$x\ge -\frac{3}{2}$

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

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

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

So its domain is:$x\ge 0$

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

$$\begin{gathered} f\circ f=f\left(\sqrt{x}\right) \\ =\sqrt{\sqrt{x}} \\ =\sqrt[4]{x} \end{gathered}$$

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

So domain is:$[0,\infty )$

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

$$\begin{gathered} g\circ g\circ =g\left(g\right(x\left)\right) \\ =g(2x+3) \\ =2(2x+3)+3 \\ =4x+6+3 \\ =4x+9 \end{gathered}$$

This is defined for the set of all real values of $x$.

So domain is:$(-\infty ,\infty )$