calculus - Find $f(x)$ for a function $f: R to R$, which satisfies condition $f(x+y^{3}) = f(x) + [f(y)]^{3}$ for all $x,y in R$ and $f'(0)≥0$.

Find $f(x)$ for a function $f: R \to R$, which satisfies condition $f(x+y^{3}) = f(x) + [f(y)]^{3}$ for all $x,y \in R$ and $f'(0)≥0$

My attempt:

Replacing $x$ and $y$ by $0$, $f(0)=0$

Replacing only x by $0$, $ f(y^{3}) = [f(y)]^{3}$

So $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} $

$= \lim_{h \to 0} \frac{f(x) + [f(h^{1/3})]^{3} - f(x)}{h}$

$= \lim_{h \to 0} \frac{f(h)}{h}$

= $f'(0)$

Then I'm stuck.

How to proceed$?$

Answer

What you've shown so far is that $f'(x)$ is a constant, since $f'(x) = f'(0)$. And we know that $f(0) = 0$. This means that your solution is going to be something in the form $f(x) = ax$ for a non-negative constant $a$ (since you've specified that $f'(x) \geq 0$).

So what constants work? Well, we know that $f(x^3) = ax^3 = f(x)^3 = a^3x^3$. Cancel out the $x^3$ (since this must hold for any $x$, we can just pick whatever $x \neq 0$ we like) and you're left with $a = a^3$.

This equation has only two non-negative solutions: $a=0$ and $a=1$. These correspond to the functions $f(x) = 0$ and $f(x) = x$, respectively.

(There is of course a third option if you remove the $f'(x) \geq 0$ constraint, namely, $f(x) = -x$. Whoever wrote the question wanted to exclude this one specifically, for whatever unknown reason.)