Discussion Forum

 Let   $x\neq0, |x|<\frac{1}{2}$  and 

$f(x)= 1+2x+4x^{2}+8x^{3}+....$  Then  $f^{-1}(x)=$ 

Answer:

Explanation:

 We have 

$f(x)= 1+2x+4x^{2}+8x^{3}+....$   and  |x| < $\frac{1}{2}$

Let f(x)=y

  $y= 1+2x+4x^{2}+8x^{3}+.....$

 = $1+(2x)+(2x)^{2}+(2x)^{3}+....$

 $y= \frac{1}{1-2x}$ [ $\because$  sum of infinite G.P,  $S_{\infty}= \frac{a}{1-r}]$

$\Rightarrow$ $y-2xy=1$  $\Rightarrow$  $2xy=y-1$

$\Rightarrow$    $x=\frac{y-1}{2y}  \Rightarrow f^{-1}(y)=\frac{y-1}{2y}$

 hence, $f^{-1}= \frac{x-1}{2x}$