The absolute maximum value of the function $f(x) = 4x-\frac{1}{2}x^2$ in the interval $\left[-2,\frac{9}{2}\right]$ is

8

The correct answer is Option (3) → 8

$f(x)=4x-\frac{1}{2}x^2$

$\frac{df}{dx}=4-x$

Critical point from $4-x=0$

$x=4$

Given interval is $\left[-2,\frac{9}{2}\right]$ and $4$ lies in it

Evaluate $f(x)$ at endpoints and critical point

$f(-2)=4(-2)-\frac{1}{2}(4)=-8-2=-10$

$f(4)=16-\frac{1}{2}(16)=16-8=8$

$f\left(\frac{9}{2}\right)=18-\frac{1}{2}\cdot\frac{81}{4}=18-\frac{81}{8}=\frac{63}{8}$

Compare values

$-10,\;8,\;\frac{63}{8}$

Maximum value is $8$

The absolute maximum value is $8$.