Comment on the monotonicity of $g(x)={f(x)\over x}$

Asked Apr 14 '19 at 12:28

Active Apr 14 '19 at 13:14

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If $f(x)$ is a real valued function defined on $[0,\infty)$ such that $f(0)=0$ and $f''(x)>0$ for all $x$.

Consider a new function $g(x)={f(x)\over x}$. I want to comment on the monotonicity of function $g$. I mean I want to find whether $g$ is increasing or decreasing and where.

Here are the things I observed. Don't know if it can help or not.

We can find that $g(0)=f'(0)$

Since second derivative of $f$ is positive, it means $f'$ is an increasing function.

$g'(x)=\dfrac{xf'(x)-f(x)}{x^2}$

But I am not able to proceed.

Thanks for hint.

edited Apr 14 '19 at 13:14

egreg

244,946

asked Apr 14 '19 at 12:28

Shweta Aggrawal

5,659

For the function $f(x)=-\log(1+x)$ $g$ is decreasing everywhere and for $f(x)=x^{2}$ is is increasing everywhere. – Kavi Rama Murthy Apr 14 '19 at 12:39

Comment on the monotonicity of $g(x)={f(x)\over x}$

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