Assume $f(x)$ is continuous on $[0,1]$ and differentiable on $(0,1)$.

Also we know $f(0)=0$ and $f'(x)$ is increasing.

How to prove $\frac{f(x)}{x}$ is also increasing on $(0,1)$?


1 Answer 1


Consider $[0,x], x \le 1$. Then applying the mean value theorem, $f(x) - f(0) = xf'(c)$ for $c \in (0,x)$. But since $f'$ is increasing, $f(0) = 0, $ and $c \lt x$, we see $f(x) \lt xf'(x)$.

Then by the quotient rule, letting $g(x) = \frac{f(x)}{x}, g'(x) = \frac{xf'(x) - f(x)}{x^2} \gt 0$. Thus $g$ is increasing.


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