Show that $x \mapsto \frac{f(x)}{x}$ is strictly increasing on (0,1) given that f '(x) is strictly increasing on (0,1) and that f(0)=0 Let f : [0, 1] → R be a continuous function with f (0) = 0 and suppose that it is differentiable for all x ∈ (0, 1).
Further assume that $x \mapsto f'(x)$ is a strictly increasing function on (0, 1). Show that $x \mapsto \frac{f(x)}{x}$ is also a strictly increasing function on (0, 1).
 A: Suppose $\frac{f(x)}{x}=\frac{f(y)}{y}$. By the Mean Value Theorem we have that there exists $z\in(x,y)$ such that $0=\Bigl(\frac{f(z)}{z}\Bigr)'=-\frac{f(z)}{z^2}+\frac{f'(z)}{z}$, therefore $f(z)=zf'(z)$.
However, $f(z)=\int_0^z f'(u) du < \int_0^z f'(z) du = zf'(z)$, where the strict inequality follows from the strict monotonicity of $f'$.
Therefore $\frac{f(x)}{x}$ is monotone.
A: If you take the derivative of $\frac{f\left(x\right)}{x}$ you have $\frac{f'\left(x\right)x-f\left(x\right)}{x^{2}}$. It is enough to show the numerator is positive for all $x \in \left( 0 , 1 \right)$. Since $f'\left(x\right)$ is strictly increasing you have
$f\left(x\right)= \int_{0}^{x} f'\left(y\right) dy \leq \int_{0}^{x} f'\left(x\right) dy = xf'\left(x\right)$. But we have more, i.e. the inequality is strict, just split $\int_{0}^{x} f'\left(y\right) dy=\int_{0}^{x/2} f'\left(y\right) dy +\int_{x/2}^{x} f'\left(y\right) dy \leq x\left(f'\left(x/2\right)+f'\left(x\right)\right)<xf'\left(x\right)$, that's what we need to conclude, since it gives us $\frac{f'\left(x\right)x-f\left(x\right)}{x^{2}}>0$.
