Q. 7 - Calculus

Question

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

Step-by-Step Solution

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 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 $f\left(x\right)$.

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

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

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