Question

Find the absolue maximum value and the absolute minimum value of the following function in the given intervals:

$f\left(x\right)=4x-\frac{1}{2}{x}^2,xϵ[-2,\frac{9}{2}]$

Answer 1

Given function is $f\left(x\right)=4x-\frac{1}{2}{x}^2,\Rightarrow f^{\prime }\left(x\right)=4-\frac{1}{2}\left(2x\right)=4-x$
For maxima or minima put $f^{\prime }\left(x\right)=0,4-x=0\Rightarrow x=4ϵ[-2,\frac{9}{2}]$
Now, we evaluate the value of f at critical point x=4 and at the end points of the interval $[-2,\frac{9}{2}]$
$\begin{array}{cc}Atx=4& f\left(4\right)=4\left(4\right)-\frac{1}{2}(4{)}^2=16-8=8\\ Atx=-2& f(-2)=4(-2)-\frac{1}{2}(-2{)}^2=-8-2=-10\\ Atx=\frac{9}{2},& f\left(\frac{9}{2}\right)=4\left(\frac{9}{2}\right)-\frac{1}{2}{\left(\frac{9}{2}\right)}^2=18-\frac{81}{8}=\frac{63}{8}=7.875\end{array}$
Thus, absolute maximum value is 8 at x=4 and absolute minimum value is -10 at x=-2.