Find $f(10)$ for the following conditions Let $f(x)$ be a real valued function not identically zero satisfies the equation,
$f(x + y^n) = f(x) + (f(y))^n$  for all real $x$ & $y$ and $f'(0)\ge 0$ where $n>1$ is an odd natural number. Find $f(10)$
Putting $x=0,y=0$
$$f(0)=f(0)+f(0)$$
$$f(0)=0\tag{1}$$
Putting $x=0,y=1$
$$f(0+1)=f(0)+f(1)^n$$
$$f(1)^{n-1}=1 \text { where (n-1) is even }$$
$$f(1)=\pm1\tag{2}$$
$$f'(0)\ge 0$$
$$\lim_{h\to 0}\dfrac{f(h)-f(0)}{h}\ge0$$
$$\lim_{h\to 0}\dfrac{f(h)}{h}\ge0\tag{3}$$
I was not getting anything significant from it.
Putting $y=1$
Case $1:$ $f(1)=1$
$$f(x+1)=f(x)+f(1)^n$$
$$f(x+1)=f(x)+1\tag{4}$$
Case $2:$ $f(1)=-1$
$$f(x+1)=f(x)-1\tag{5}$$
So according to equation $(4)$
$$f(10)=10$$
According to equation $(5)$
$$f(10)=-10$$
But not able to determine which one to eliminate?

To complete your thoughts just note that by the mean value theorem $$f(10) = f(10) - f(0) = 10 \cdot f^\prime (t)$$ for some $t \in (0, 10)$.
