Question

Let$f\left(x\right)=\frac{1}{1-x}$and$g\left(x\right)=\frac{1}{ax+b}$. Show that if$b\ne 0$, thenrole="math" localid="1649659462059" $g\left(x\right)=\frac{1}{b}f\left(-\frac{a}{b}x\right).$

Short answer

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

Hence proved

Step-by-step solution

Step 1. Given Information.

We have $f\left(x\right)=\frac{1}{1-x}$and $g\left(x\right)=\frac{1}{ax+b}$.

Show that$g\left(x\right)=\frac{1}{b}f\left(-\frac{a}{b}x\right)$.

Step 2. Explanation.

Substitute $x=-\frac{a}{b}x$in $f\left(x\right)$.

role="math" localid="1649660408537" $$\begin{gathered} f\left(-\frac{a}{b}x\right)=\frac{1}{1-\left(-\frac{a}{b}x\right)} \\ =\frac{1}{1+\frac{ax}{b}} \\ =\frac{1}{\frac{b+ax}{b}} \\ =\frac{b}{b+ax} \end{gathered}$$

Multiply $\frac{1}{b}$to $f\left(-\frac{a}{b}x\right)$.

role="math" localid="1649660532439" $$\begin{gathered} \frac{1}{b}f\left(-\frac{a}{b}x\right)=\frac{1}{b}\times \frac{b}{b+ax} \\ =\frac{1}{ax+b} \\ =g\left(x\right) \end{gathered}$$