Answer to Question #348645 in Real Analysis for Nikhil Singh

Question

Let f:[ 0,π/2] → [-1,1] be a function defined by f(x)= cos 2x . Verify that f satisfies the condition of the inverse function theorem. Hence, what can you conclude about the continuity of f^-1?

Answer 1

The function $f(x)=\cos2x$ is strictly decreasing on $(0, \pi/2).$ Then the function $f(x)=\cos2x$ is one-to-one on $[0, \pi/2].$

The function $f(x)=\cos2x$ is invertible on $[0, \pi/2].$

$$f'(x)=-2\sin(2x)$$

By  Inverse Function Theorem for all $x\in(0, \pi/2)$

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

$f^{-1}(x)$ is continuous on $[-1,1].$

$f(x)=\cos2x, x\in [0, \pi/2]$

$f^{-1}(x)=\dfrac{1}{2}\cos^{-1}x, x\in[-1,1]$

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

$$\dfrac{1}{f'(f^{-1}(x))}=\dfrac{1}{-2\sin(2(\dfrac{1}{2}\cos^{-1}x))}=-\dfrac{1}{2\sqrt{1-x^2}}$$