Question

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

Answer 1

The function $f(x)=\dfrac{2x}{x^2+1}$ is one­-to-one function on $[-1, 1].$

Hence the function $f(x)$ has inverse.

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

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

Inverse FunctionTheorem

Let $f(x)$ be a function that is both invertible and differentiable.

Let $y=f^{-1}(x)$  be the inverse of $f(x).$

For all $x$ satisfying $f'(f^{-1}(x))\not=0,$

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

$$f'(\dfrac{4}{5})=\dfrac{2(1-(\dfrac{4}{5})^2)}{((\dfrac{4}{5})^2+1)^2}=\dfrac{450}{1681}$$