Question

Let $f\left(x\right)$ be a real value function not identically zero satisifes the equation,
$f(x+y^n)=f\left(x\right)+{f\left(y\right)}^n$ for all real $x,y$ and $f^{\prime }\left(0\right)\ge 0$ where $n(>1)$ is an odd natural number. $f\left(10\right)=10k$.Find $k$ value

Answer 1

$f(x+y^n)=f\left(x\right)+{\left(f\right(y\left)\right)}^n$
$f(0+0)=f\left(0\right)+{\left(f\right(0\left)\right)}^n$ $\Rightarrow f\left(0\right)=0$
also $f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}$
Let $I=f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0+{\left(h^{1/n}\right)}^n)-f\left(0\right)}{{\left(h^{1/n}\right)}^n}=\underset{h\to 0}{lim}\frac{f\left({\left(h^{1/n}\right)}^n\right)}{{\left(h^{1/n}\right)}^n}=\underset{h\to 0}{lim}{\left(\frac{f\left(h^{1/n}\right)}{h^{1/n}}\right)}^n=1^n$
$\Rightarrow I=I^n$ or $I=0,1,-1$
since $f^{\prime }\left(0\right)\ge 0$ and $f\left(x\right)$ is not identically zero
So $I=1$ $\therefore f^{\prime }\left(0\right)=1........\left(i\right)$
Thus $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}=\underset{h\to 0}{lim}\frac{f(x+{\left(h^{1/n}\right)}^n)-f\left(x\right)}{{\left(h^{1/n}\right)}^n}=\underset{h\to 0}{lim}\frac{f(x+{\left(h^{1/n}\right)}^n)-f\left(x\right)}{{\left(h^{1/n}\right)}^n}=\underset{h\to 0}{lim}\frac{f\left(x\right)+f(x+{\left(h^{1/n}\right)}^n)-f\left(x\right)}{{\left(h^{1/n}\right)}^n}$
$=\underset{h\to 0}{lim}{\left(\frac{f\left(h^{1/n}\right)}{h^{1/n}}\right)}^n={\left(f^{\prime }\right(0\left)\right)}^n$
$\Rightarrow f^{\prime }\left(x\right)=1$ ------(using $\left(i\right)$)
Integrating both sides
$f\left(x\right)=x+c$
$\Rightarrow f\left(x\right)=x$ [$f\left(0\right)=0$]
$\Rightarrow f\left(10\right)=10$
$\Rightarrow k=1$