Q. 63 - Precalculus Enhanced with Graphing Utilities

Question

If $f\left(x\right)=ax+b$ and $g\left(x\right)=cx+d$, then find: -

(a)$f\circ g$

(b)$g\circ f$

(c)the domain of $f\circ g$ and $g\circ f$

(d)the conditions for which $f\circ g=g\circ f$

Step-by-Step Solution

Answer

(a)$f\circ g\left(x\right)=acx+ad+b$

(b)$g\circ f\left(x\right)=acx+bc+d$

(c)Domain of $f\circ g$ and $g\circ f$ are same, that is set of real numbers.

(d)$f\circ g$and $g\circ f$ are same if $ad+b=bc+d.$

Step 1. Given Information

Given that $f\left(x\right)=ax+b$ and $g\left(x\right)=cx+d.$

Part (a) step 1. solution

we know that $f\circ g\left(x\right)=f\left(g\right(x\left)\right).$

Here, $f\left(x\right)=ax+b$, $g\left(x\right)=cx+d.$

then, 

$$\begin{gathered} f\circ g\left(x\right)=f\left(g\right(x\left)\right) \\ f\circ g\left(x\right)=a\left(g\right(x\left)\right)+b \\ f\circ g\left(x\right)=a\{cx+d\}+b \\ f\circ g\left(x\right)=acx+ad+b \end{gathered}$$

Part (b) Step 1. Solution

We know that $g\circ f\left(x\right)=g\left(f\right(x\left)\right).$

Here, $f\left(x\right)=ax+b$, $g\left(x\right)=cx+d$

$$\begin{gathered} g\circ f\left(x\right)=g\left(f\right(x\left)\right) \\ g\circ f\left(x\right)=c\left(f\right(x\left)\right)+d \\ g\circ f\left(x\right)=c\{ax+b\}+d \\ g\circ f\left(x\right)=acx+bc+d \end{gathered}$$

Part (c) Step 1. Solution

Since the domain of $f\left(x\right)$ and $g\left(x\right)$ are set of real number so domain of $f\circ g$ and $g\circ f$ are also set of real number.

Part (d) Step 1. Solution

We know that $f\circ g\left(x\right)=acx+ad+b$ and $g\circ f\left(x\right)=acx+bc+d$.

Now,

$$\begin{gathered} f\circ g\left(x\right)=g\circ f\left(x\right) \\ acx+ad+b=acx+bc+d \end{gathered}$$

Compare coefficients on both sides.

$ad+b=bc+d$.

So,$f\circ g=g\circ f$

if, $ad+b=bc+d.$