Let $f$ be a function which satisfies:
(I) $f$ is continuous $\forall x \geq 0$
(II) $f'(x)$ exists for $x>0$
(III)$f(0)=0$
(IV) $f'$ is a strictly increasing function
If $$g(x) =\frac {f(x)}{x}, x>0$$
Prove that $g(x)$ is a strictly increasing function.
I don't know if it's a good path to prove this, but my thought was making $g'(x)=\dfrac{f'(x)\cdot x – f(x)}{x^2}$ and showing that $g'(x) > 0, \forall x>0$, but I can't see or show that this is true.
Can anybody help me? Thanks.
Best Answer
By the Mean Value Theorem, for every $x$ there exists $\xi\in(0,x)$ such that $\frac{f(x)}{x}=f'(\xi)$. Since $f'$ is strictly increasing, $\frac{f(x)}{x}<f'(x)$, which implies that $g'>0$.
(If $f'$ is just increasing, not strictly, then $g'\geq 0$, so $g$ is just increasing, not strictly)