Question

The absolute maxima/minima value of following functions$f\left(x\right)=4x-\frac{x^2}{2};x\in [-2,\frac{9}{2}]$ be max,min. Find $max-min$?

Answer 1

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

$\frac{dy}{dx}=4-x=0$ for min. or max.

$\Rightarrow x=4$ also $f^{\prime \prime }\left(x\right)=-1<0$

Thus $f\left(x\right)$ will attain it's maxima at $x=4$

$\Rightarrow y_{max}=16-8=8$

Now $y(-2)=-10,y\left(\frac{9}{2}\right)=18-\frac{81}{4}=-\frac{9}{4}$

Thus $y_{min}=-10$ in the given interval.

$max-min=8-(-10)=18$