Answer to Question #283621 in Real Analysis for Akshay

Question

verify cauchy's mean value theorem for the function f(x) = x and g(x) = sinx in [0,π/2].

Answer 1

The functions $f(x)=x$ and $g(x)=\sin x$ are continuous on the interval $[0,\pi/2],$ differentiable on $(0,\pi/2),$ and $g'(x)=\cos x≠0$ for all $x\in(0,\pi/2).$

Then there is a point $x=c$ in this interval such that

$$\dfrac{f(\pi/2)-f(0)}{g(\pi/2)-g(0)}=\dfrac{f'(c)}{g'(c)}$$

$$f(\pi/2)=\pi/2, f(0)=0$$

$$g(\pi/2)=\sin(\pi/2)=1, g(0)=\sin(0)=0$$

$$f'(x)=1, f'(c)=1$$

$$g'(x)=\cos x, g'(c)=\cos c$$

$$\dfrac{\pi/2-0}{1-0}=\dfrac{1}{\cos c}$$

$$\cos c=2/\pi, c\in(0,\pi/2)$$

$$c=\cos^{-1}(2/\pi)$$

It is evident that this number $c=\cos^{-1}(2/\pi)$ lies in the interval $(0,\pi/2),$

 i.e. satisfies the Cauchy theorem.