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Suppose $f$ is a differentiable function over $\mathbb{R}$ satisfying $f(0) = 0$. Show that if $f'$ is strictly increasing, then $\frac{f(x)}{x}$ is increasing over $(0,\infty)$.

Attempt:

Since $f'$ is increasing we know that $f'' > 0$ for all $x$. Now let $g(x) = \frac{f(x)}{x}$ and thus $f(x) = xg(x)$. Then we wish to show that $g'(x) > 0$ on $(0,\infty)$ and so since $f'(x) = xg'(x)+g(x)$ and $f''(x) = xg''(x)+2g'(x) > 0$, we have $g''(x) > \frac{-2g'(x)}{x}$. I am not sure how to use this though to prove the statement.

user19405892
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2 Answers2

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By MVT, for some $c \in (0,x)$ such that $$\frac{f(x)}{x}=\frac{f(x)-f(0)}{x-0}=f'(c)$$

Since $x>c$, this gives us that $$f'(x)>f'(c)=\frac{f(x)}{x}$$

Thus, $f'(x)x-f(x)>0$ if $x>0$.

But the derivative of $\frac{f(x)}{x}$ is $$\frac{f'(x)x-f(x)}{x^2}>0$$Our proof is done.

S.C.B.
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This proof is valid only if $f$ is twice differentiable. By Taylor, for some $c\in(0,x)$ and $x>0$,($c=c(x)$) : $$f(x)=f'(0)x+f''(c)x^2$$ then $$\frac{f(x)}{x}=f'(0)+f''(c)x$$ differentiating on both sides $$g'(x)=f''(c)>0$$ for every $c(x)$.

mrprottolo
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