Question

Let $f$ be a twice differentiable function defined in $[-3,3]$ such that $f\left(0\right)=-4,f^{\prime }\left(3\right)=0,$ $f^{\prime }(-3)=12$ and $f^{\prime \prime }\left(x\right)\ge -2\forall x\in [-3,3].$ If $g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt,$ then the maximum value of $g\left(x\right)$ is

Answer 1

Applying LMVT to $y=f^{\prime }\left(x\right)$ in $[-3,3]$, we get
$\frac{f^{\prime }\left(3\right)-f^{\prime }(-3)}{3-(-3)}=f^{\prime \prime }\left(c\right)$ for some $c\in (-3,3)$
$\Rightarrow f^{\prime \prime }\left(c\right)=-2$ for some $c\in (-3,3)$
But $f^{\prime \prime }\left(x\right)\ge -2\forall x\in [-3,3]$
$\therefore f^{\prime \prime }\left(x\right)=-2\forall x\in [-3,3]$

Now, $f^{\prime }\left(x\right)=-2x+C$
$f^{\prime }\left(3\right)=C-6=0\Rightarrow C=6$
$\therefore f^{\prime }\left(x\right)=6-2x$
$f\left(x\right)=6x-x^2+C_1$
$f\left(0\right)=C_1=-4$
$\therefore f\left(x\right)=6x-x^2-4$

Now, $g\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$
$\Rightarrow g\left(x\right)=\underset{0}{\overset{x}{\int }}(-t^2+6t-4)dt$
$\therefore g\left(x\right)=\frac{-x^3}{3}+3x^2-4x$ in $[-3,3]$
$g^{\prime }\left(x\right)=-x^2+6x-4$

$g(-3)=48$ which is the maximum value in $[-3,3]$.