Answer to Question #198205 in Calculus for desmond

Question

(b) By considering the derivative of the function f : [−1, 1] → R defined by f(x) = \frac{2x}{\:x^2+1} , show that f^{-1} exists and find \left(f^{-1}\right)^' ( \frac{4}{\:5} ).

Answer 1

The inverse function theorem helps us to validate that $f^{-1}$ exist.

And, we have that

$(f^{-1})'(x) = \dfrac{1}{f'(f^{-1}(x))}$

First, we find the derivative of f using quotient rule

$f'(x) = \dfrac{(x^2+1)(2)-(2x)(2x)}{(x^2+1)^2} = \dfrac{2(1-x^2)}{(x^2+1)^2}$

Next, we find the inverse of f.

Suppose y=f(x) for x in the defined domain and y in R. So, we have that

$y= \dfrac{2x}{x^2+1} \\ x^2y+y=2x\\x^2y-2x=-y\\ x^2-\frac{2x}{y}=-1\\(x-\frac{1}{y})^2=\dfrac{1-y^2}{y^2}\\ x-\frac{1}{y} = \pm \sqrt{\dfrac{1-y^2}{y^2}} \\ x= \dfrac{1 \pm \sqrt{1-y^2}}{y}$

By putting x everywhere, we see y, we know that the inverse function is

$f^{-1}(x)= \dfrac{1 \pm \sqrt{1-x^2}}{x}$

Taking positive square roots, we have that

$f^{-1}(x)= \dfrac{1 + \sqrt{1-x^2}}{x}$

So,

$(f^{-1})'(\frac{4}{5}) = \dfrac{1}{f'(f^{-1}(\frac{4}{5}))}$

Consider,

$f^{-1}(\frac{4}{5}) = \dfrac{1 + \sqrt{1-(\frac{4}{5})^2}}{\frac{4}{5}} = 2$

So, $(f^{-1})'(\frac{4}{5}) = \dfrac{1}{f'(f^{-1}(\frac{4}{5}))}$ becomes $(f^{-1})'(\frac{4}{5}) = \dfrac{1}{f'(2)}$

Also, $f'(2) = \dfrac{2(1-2^2)}{(2^2+1)^2} = -\dfrac{6}{25}$

Hence, $(f^{-1})'(\frac{4}{5}) = -\dfrac{25}{6}$