If the solve the problem


(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 $[1,9]$



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


$\Rightarrow 4-x=0$

$\Rightarrow x=4$

Thus, the critical points of $f$ are $-2,4$ and $4.5$.





Hence, the absolute maximum value when $x=4$ is 8 and the absolute minimum value when $x=-2$ is $-10$.

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