If $f(x)=\max(1-\cos{x},2\sin{x})$ for all $x$ in $(0,\pi)$, Then $f'(x)$ is undefined at $x=x_0$ where $2\sin x_0+\cos x_0=1$

Question

Okay,I am new to Calculus. I tried to find Maxima of $1-\cos{x}$ and $2\sin{x}$ in the interval $(0,\pi)$ and it turns out that they have the same local Maxima ($=2$). I don't know how to proceed. Here is the problem

Answer 1

hint

To find $max (1-\cos (x),\sin (x)) $ note that

$$1-\cos (x)=2\sin^2 (\frac x2) $$

and $$\sin (x)=2\sin (\frac x2)\cos (\frac x2) $$

If $x\in (0,\frac \pi 2) $ then $\sin (\frac x2)<\cos (\frac x2) $ and $$f (x)=\sin (x ) $$ if $x\in (\frac \pi 2,\pi) $ then $$f (x)=1-\cos (x) $$

from here

$$f'(\frac \pi 2^+)=1$$ $$f'(\frac \pi 2^-)=0$$

Answer 2

There is an $x$ for which the derivative is undefined but I think it's not $x=\pi/2$.

Plot the two individual functions in the same axes. The point in which the two cross each other is the point where the derivative of $f$ is undefined.