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Suppose that $f$ is continuous on $[0, 1]$, differentiable on $(0, 1)$ and $f(0) = 0$. Prove that if $f'$ is decreasing on $(0, 1)$ then the function $g : (0, 1) → R$ given by $g(x) = f(x)/x$ is decreasing on $(0, 1)$.

Any ideas? The first sentence suggests MVT, but $f'(c) = f(1)$, for some $c$ in $(0,1)$ does not help me. I considered $g'(x) = (f'(x)x-f(x))/x^2$ (since g is differentiable), and saw that I could try to prove that $f'(x)x-f(x)<=0$, but the problem is that I do not see how to relate $f'$ and $f$, given the information I have. On the other hand, I really cannot see how to use the fact that $f'$ is decreasing. Thanks!

1 Answers1

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hint

For an $ x$ in $ (0,1)$, by MVT, there exists $ c $ in $(0,x) $ such that

$$f(x)=f(0)+xf'(c) =xf'(c)$$

or

$$\frac{f(x)}{x}=f'(c)$$

hamam_Abdallah
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  • I see now, thank you. I had recently seen that trick (to consider the interval (0,x)), but never passed my mind in this for some reason. Thanks! –  Feb 02 '20 at 22:05