(i) $f(x)=4 x-\frac{x^{2}}{2}$ in $[-2,4,5]$
(ii) $f(x)=(x-1)^{2}+3$ in $[-3,1]$
(iii) $f(x)=3 x 4-8 x^{3}+12 x^{2}-48 x+25$ in $[0,3]$
(iv) $f(x)=(x-2) \sqrt{x-1}$ in $[1,9]$
(i)
Given: $f(x)=4 x-\frac{x^{2}}{2}$
$\Rightarrow f^{\prime}(x)=4-x$
For a local maximum or a local minimum, we must have
$f^{\prime}(x)=0$
$\Rightarrow 4-x=0$
$\Rightarrow x=4$
Thus, the critical points of $f$ are $-2,4$ and $4.5$.
Now,
$f(-2)=4(-2)-\frac{(-2)^{2}}{2}=-8-2=-10$
$f(4)=4(4)-\frac{(4)^{2}}{2}=16-8=8$
$f(4.5)=4(4.5)-\frac{(4.5)^{2}}{2}=18-10.125=7.875$
Hence, the absolute maximum value when $x=4$ is 8 and the absolute minimum value when $x=-2$ is $-10$.
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