Q 51. - Precalculus Enhanced with Graphing Utilities

Question

In Problems 45–52, show that $(f\circ g)\left(x\right)=(g\circ f)\left(x\right)=x$

$f\left(x\right)=ax+b; g\left(x\right)=\frac{1}{a}(x-b) a\ne 0$

Step-by-Step Solution

Answer

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

Therefore, $(f\circ g)\left(x\right)=(g\circ f)\left(x\right)=x$.

Step 1. Given information.

The given composite function is:

$$\begin{gathered} f\left(x\right)=ax+b \\ g\left(x\right)=\frac{1}{a}(x-b) \end{gathered}$$

When we are given two functions f and g, the composite function which is denoted by $f\circ g$ is defined by $(f\circ g)\left(x\right)=f\left(g\right(x\left)\right)$.

Step 2. Find ( f ∘ g ) ( x ) .

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

Now substitute $g\left(x\right)=\frac{1}{a}(x-b)$ in the function $f\left(g\right(x\left)\right)$,

Then the function will become $f\left(\frac{1}{a}\right(x-b\left)\right)$,

$$\begin{gathered} f\left(\frac{1}{a}\right(x-b\left)\right)=a\left[\frac{1}{a}\right(x-b\left)\right]+b \\ =a\left[\frac{(x-b)}{a}\right]+b \\ =x-b+b \\ =x \end{gathered}$$

Therefore, $(f\circ g)\left(x\right)=x$.

Step 3. Find ( g ∘ f ) ( x ) .

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

Substitute $f\left(x\right)=ax+b$ in the function $g\left(f\right(x\left)\right)$,

$$\begin{gathered} (g\circ f)\left(x\right)=g\left(f\right(x\left)\right) \\ =g(ax+b) \\ =\frac{1}{a}(ax+b-b) \\ =x \end{gathered}$$

Therefore, $(g\circ f)\left(x\right)=x$.

It is shown that $(f\circ g)\left(x\right)=(g\circ f)\left(x\right)=x$.