Question

If $f$ is a differentiable function for all real $x$ and $f^{{}^{\prime }}\left(x\right)\le 5\forall x\in R$ . If $f\left(2\right)=0$ and $f\left(5\right)=15$ value of $f\left(3\right)$ is

Answer 1

Answer: 5

$\frac{f\left(3\right)-f\left(2\right)}{3-2}=f^{{}^{\prime }}\left(4\right)$
$f\left(3\right)\le \mathrm{5...}\left(1\right)$
$\frac{f\left(5\right)-f\left(3\right)}{5-3}=f^{{}^{\prime }}\left(2\right)$
$f\left(3\right)\ge \mathrm{5...}\left(2\right)$
Hence $f\left(3\right)=5$