Question

If $f\left(0\right)=0$ and $f^{\prime \prime }\left(x\right)>0$ for all $x>0$, then $\frac{f\left(x\right)}{x}$

• A. decreases on $(0,\infty )$
• B. increases on $(0,\infty )$
• C. decreases on $(1,\infty )$
• D. neither increases nor decreases on $(0,\infty )$

Answer 1

The correct option is B increases on $(0,\infty )$

Given, $f\left(0\right)=0$

$f^{\prime \prime }\left(x\right)>0$, it means $f^{\prime }\left(x\right)$ is also increasing

Let $g\left(x\right)=\frac{f\left(x\right)}{x}$

$g^{\prime }\left(x\right)=\frac{x{f}^{\prime }\left(x\right)-f\left(x\right)}{x^2}$

$f^{\prime }\left(x\right)$ is increasing and $xϵ(0,\infty )$ thus

$f^{\prime }\left(x\right)=$positive

$f\left(x\right)=$positive

$x^2=$positive

Therefore, $g\left(x\right)=$positive $>0$

Thus, $\frac{f\left(x\right)}{x}$ increases for $xϵ(0,\infty )$