Question

If $f\left(x\right)=\frac{ax+b}{cx+d};g\left(x\right)=mx,$then find: -

(a)$f\circ g$

(b)$g\circ f$

(c)Domain of $f\circ g$and $g\circ f$.

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

Short answer

(a)$f\circ g\left(x\right)=\frac{amx+b}{cmx+d}$

(b)$g\circ f\left(x\right)=\frac{max+b}{cx+d}$

(c)Domain of $f\circ g$is set of real numbers except $x=-\frac{d}{c}$and domain of $g\circ f$is set of real numbers.

(d) $f\circ g\left(x\right)=g\circ f\left(x\right)$if,$m=1.$

Step-by-step solution

Given Information

Given that$f\left(x\right)=\frac{ax+b}{cx+d};g\left(x\right)=mx.$

Part (a) Step 1 Solution

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

Here, $f\left(x\right)=\frac{ax+b}{cx+d};g\left(x\right)=mx$

Now,

$$\begin{gathered} f\circ g\left(x\right)=f\left(g\right(x\left)\right) \\ f\circ g\left(x\right)=\frac{a\left(g\right(x\left)\right)+b}{c\left(g\right(x\left)\right)+d} \\ f\circ g\left(x\right)=\frac{a\left\{mx\right\}+b}{c\left\{mx\right\}+d} \\ f\circ g\left(x\right)=\frac{amx+b}{cmx+d}. \end{gathered}$$

Part (b) Step 1 Solution

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

Here, $f\left(x\right)=\frac{ax+b}{cx+d};g\left(x\right)=mx.$

Now,

$$\begin{gathered} g\circ f\left(x\right)=m\left(f\right(x\left)\right) \\ g\circ f\left(x\right)=m\left\{\frac{ax+b}{cx+d}\right\} \\ g\circ f\left(x\right)==\frac{max+b}{cx+d}. \end{gathered}$$

Part (C) Step 1 Solution

Domain of $f\left(x\right)=\frac{ax+b}{cx+d}$is set of real numbers except than $x=-\frac{d}{c}.$Domain of $g\left(x\right)=mx$is set of real numbers. So, domain of $f\circ g\left(x\right)$is set of real numbers except then $x=-\frac{d}{c}.$

Now, domain of $g\circ f\left(x\right)$is set of real numbers.

Part (d) Step 1 Solution

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

$$\begin{gathered} \frac{amx+b}{cmx+d}=\frac{max+b}{cx+d} \\ cmx+d=cx+d \end{gathered}$$

compare both sides

$$\begin{gathered} cm=c \\ m=1. \end{gathered}$$