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
Answer & explanation
Correct answer: option 3
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 $8$.