Question

If f(x) = 1 + x + x² +.... for |x|< 1, then show that f⁻¹ (x) = $\frac{x - 1}{x}$

Answer 1

Functions and Geometric Progressions

We have a function here.

$\sf f(x) = 1+x+x^2+\dots$

Also, we are given | x | < 1. So the function f(x) forms an Infinite Geometric Progression (G.P.)

Here,

First Term = a = 1

Common Ratio = r = x

Sum of an Infinite G.P. is given by:

$\sf \displaystyle S_\infty = \frac{a}{1-r} \\\\\\ \implies S_\infty = \frac{1}{1-x} \\\\\\ \implies 1+x+x^2+\dots = \frac{1}{1-x} \\\\\\ \implies f(x) = \frac{1}{1-x}$

We need the inverse of f. We find it as follows:

$\bullet \sf \displaystyle\ \textsf{Let y=f(x)}\\\\\\ \implies y=f(x)=\frac{1}{1-x}\\\\\\ \bullet\ \textsf{We now need x in terms of y}\\\\\\ \implies y=\frac{1}{1-x}\\\\\\ \implies y(1-x)=1\\\\\\ \implies y-yx=1\\\\\\ \implies yx=y-1 \\\\\\ \implies x=\frac{y-1}{y} \\\\\\ \bullet\ \textsf{Interchange x and y to get inverse function \sf f^{-1}(x) } \\\\\\ \implies f^{-1}: y=\frac{x-1}{x} \\\\\\ \implies \Large \boxed{\sf f^{-1}(x)=\frac{x-1}{x}}$

$\mathcal{HENCE\ \ PROVED}$

Answer 2

$\huge\underline\textbf{Answer:-}$

Refer the given attachment:-