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?
 A: Note that by induction we have
$$f(ky^n)=kf(y)^n$$
for integer $k$. Letting $k\mapsto k^n$ gives us
$$f((ky)^n)=(kf(y))^n$$
This let's us take use of the derivative:
$$f'(0)^n=\lim_{h\to0}\frac{f(h)^n}{h^n}=\lim_{k\to\infty}(kf(1/k))^n=\lim_{k\to\infty}f(1)=f(1)$$
where we consider $y=1/k$. Since $f'(0)\ge0$, this gives us $f(1)\ge0$, and hence $f(10)=10$.
A: 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 $t \in (0, 10)$.
A: I got a very sophisticated method to solve this question, just thought of sharing it:-
Idea:- Try to find $f(x)$
$$f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$$
If $h\rightarrow 0$, $h^n\rightarrow 0$ as n is an natural no
Substituting $h$ with $h^n$
$$f'(x)=\lim_{h^n\to 0}\dfrac{f(x+h^n)-f(x)}{h^n}$$
$$f'(x)=\lim_{h^n\to 0}\dfrac{f(x)+f(h)^n-f(x)}{h^n}$$
$$f'(x)=\lim_{h^n\to 0}\dfrac{f(h)^n}{h^n}\tag{1}$$
$$f'(x)=\lim_{h^n\to 0}\left(\dfrac{f(h)}{h}\right)^{n}\tag{1}$$
We know from the original post, $f(0)=0$
$$f'(0)=\lim_{h\rightarrow 0}\dfrac{f(h)}{h}$$
If $h^n\rightarrow 0$, $h\rightarrow 0$ where n is natural no
$$f'(x)=f'(0)^n$$
Integrating both sides with respect to $x$
$$f(x)=f'(0)^nx+c$$
Putting $x=0$
$$f(0)=0+c$$
$$c=0$$
$$f(x)=f'(0)^nx$$
Putting $x=1$
$$f(1)=f'(0)^n$$
As $f'(0)\ge0$, $f(1)$ cannot be $-1$
$$1=f'(0)^n$$
As $n$ is odd natural no
$$f'(0)=1$$
So finally $f(x)=x$
$$f(10)=10$$
