Let $f$ be a differentiable function satisfying $$\sqrt[3]{f(x+y)}=\sqrt[3]{f(x)}+\sqrt[3]{f(y)}+1$$ $\forall \:\:x,y\in\mathbb{R}$ and $f'(0)=3$
If $h(x)=f(x)-x^3$ then find number of points where $y=h(|x|)$ is non derivable
If $x_0$ is a solution of the equation $f(x)=f^{-1}(x)$ then find the value of $$\cos^{-1}(\cos2x_0)+4\tan^{-1}\left(\tan\frac{x_0}{2}\right)$$
My initial thoughts were to differentiate the function partially with respect to $x$ and obtain a differential equation. Having said so, forming the differential equation is not helping me much and now I am stuck.
Any help is greatly appreciated.
Set $g(x) = \sqrt[3]{f(x)}$
$\Rightarrow g(x+y) = g(x) + g(y) + 1$
$\stackrel{h\neq 0}{\Rightarrow} \frac{g(x+h)-g(x)}{h} = \frac{g(h) + 1}h$
$g$ is differentiable except at the points where $f(x) = 0$:
$\Rightarrow g(0) = -1$ and $g'(x) = g'(0)$
$g'(0) = \frac{f'(0)}{3\sqrt[3]{(f(0))^2}} = \frac 1{\sqrt[3]{(f(0))^2}}$
Since $g(0) = -1\Rightarrow f(0) = -1$, we get
$g'(0)= 1 \Rightarrow g(x) = x-1 \Rightarrow \boxed{f(x) = (x-1)^3}$
It follows, $h(x) = -3x^2+3x-1$. So $h(|x|)$ is not differentiable at $x=0$ which is easy to check.
The solution $x_0$ of $f(x) = f^{-1}(x)$ must satisfy
$(x_0-1)^3 = x_0$ (Look at the graphs of $f$ and $f^{-1}$.)
Now, it's easily checked, that $x_0 \in (2,3)$.
$\Rightarrow 2x_0 \in (\pi,2\pi) \Rightarrow \cos^{-1}((\cos 2x_0)) = 2\pi - 2x_0$
$\Rightarrow \frac{x_0}2 \in (0,\frac{\pi}2) \Rightarrow \tan^{-1}(\tan \frac{x_0}2) = \frac{x_0}2$ Hence, $$\cos^{-1}(\cos2x_0)+4\tan^{-1}\left(\tan\frac{x_0}{2}\right)= 2\pi - 2x_0 + 4\frac{x_0}2 = \boxed{2\pi}$$