Question

The absolute minimum and maximum values of $f\left(x\right)=4x-\frac{x^2}{2},xϵ[-2,\frac{9}{2}]$ are respectively

• A. $-10,8$
• B. $\frac{63}{8},-10$
• C. $25,16$
• D. $-10\frac{63}{8}$

Answer 1

The correct option is A $-10,8$

Differentiate the given function $f\left(x\right)=4x-\frac{x^2}{2}$.

$f^{\prime }\left(x\right)=4-x$

Put $f^{\prime }\left(x\right)=0$,

$4-x=0$

$x=4$

Put the value of $x$ and the end points of the given interval in the given function.

$f\left(4\right)=4\left(4\right)-\frac{4^2}{2}$

$=8$

$f\left(-2\right)=4\left(-2\right)-\frac{{\left(-2\right)}^2}{2}$

$=-10$

$f\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{{\left(\frac{9}{2}\right)}^2}{2}$

$=\frac{63}{8}$

$=7.8$

The absolute maximum value of the given function is 8 and the absolute minimum value is $-10$.