Question

If $f\left(x\right)=1+x+x^2+x^3+\dots \dots$ for $\left|x\right|<1$ then $f^{-1}\left(x\right)=$

• A. $\frac{x-1}{x+1}$
• B. $\frac{x+1}{x}$
• C. $\frac{x}{x-1}$
• D. $\frac{x-1}{x}$

Answer 1

The correct option is D $\frac{x-1}{x}$

$f\left(x\right)=1+x+x^2+\dots \dots$
$=\frac{1}{1-x}$
Let $y=\frac{1}{1-x}=f\left(x\right)$
$1-x=\frac{1}{y}$
$x=1-\frac{1}{y}=\frac{y-1}{y}$
But $x=f^{-1}\left(y\right)=\frac{y-1}{y}$