If $f(x+y^3)=f(x)+[f(y)]^3$ and $f'(0)\ge0$, what can $f(10)$ be?

Question

A real valued function satisfies the condition: $f(x+y^3)=f(x)+[f(y)]^3$ for all real $x$, $y$.

If $f'(0)\ge0$ how to find $f(10)$ ?

Answer 1

Using the condition $f(x+y^3)=f(x)+[f(y)]^3$, we note that for any $x$, we have $$ f'(x) = \lim_{h \to 0} \frac{f(x+h^3) - f(x)}{h^3} = \lim_{h \to 0} \frac{(f(x) +[f(h)]^3) - f(x)}{h^3} =\lim_{h \to 0} \frac{[f(h)]^3}{h^3} $$ We may deduce that $f(0) = 0$ and that $f'(x)$ does not depend on $x$, so that $f$ has constant slope. In particular, for any $x$, we have $f'(x) = f'(0)$.

Since $f$ has constant slope and $f(0) = 0$, we must have $f(x) = [f'(0)]x$. Moreover, we also have $$ f'(0) = \lim_{h \to 0} \frac{[f(h)]^3}{h^3} = \left[\lim_{h \to 0} \frac{[f(h)]}{h}\right]^3 = [f'(0)]^3 $$ From which we may deduce that we either have $f'(0) = 0$ or $f'(0) = 1$.

Answer 2

Here is my attempt:

Let $(x,y)=(0,t)$:$$ f(0+t^3) = f(t^3) = f(0) + [f(t)]^3 $$We can use this to find $f(0)$:$$ f(0^3)=f(0)+[f(0)]^3 \implies f(0) - f(0) = [f(0)]^3 \implies f(0) = 0 $$This means our function simplifies to:$$ f(t^3) = [f(t)]^3 $$Next let $(x,y)=(t,c)$ where $c$ is some arbitrary constant with respect to $t$:$$ f(t+c^3) = f(t) + [f(c)]^3 $$Differentiate both sides with respect to $t$:$$ f'(t+c^3) = f'(t) \implies f'(c^3) = f'(0) \geq 0 $$Since $c$ is any arbitrary constant, this means that $f'$ is always non-negative and so $f$ is always increasing or constant. So $0<1 \implies f(0)\leq f(1) \implies 0 \leq f(1)$.

Next let $(x,y)=(0,1)$:$$ f(1^3)=f(1) = [f(1)]^3 \implies f(1) = 1 \lor f(1) = 0 $$The fact that $0 \leq f(1)$ rules out $f(1)=-1$.

Next let $(x,y)=(1,1)$:$$ f(1+1^3) = f(1) + [f(1)]^3 \implies f(2) = 1 + 1^3 = 2 \lor f(2) = 0 $$Next let $(x,y)=(2,2)$:$$ f(2+2^3) = f(2) + [f(2)]^3 \implies f(10) = 2 + 2^3 = 10 \lor f(10) = 0 $$

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It would appear that $f(x)=x$ satisfies both conditions of the provided function.$$ f'(0) = 1 \geq 0 \quad \land\quad f(x+y^3) = x+y^3 = f(x) + [f(y)]^3 $$

It would appear that $f(x)=0$ also satisfies both conditions.$$ f'(0) = 0 \geq 0 \quad \land\quad f(x+y^3) = 0 = 0 + [0]^3 $$