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The sum of the absolute minimum and the absolute maximum values of the function $f(x)=\left|3 x-x^{2}+2\right|-x$ in the interval $[-1,2]$ is :
Option: 1
$\frac{\sqrt{17}+3}{2}$

Option: 2
$\frac{\sqrt{17}+5}{2}$

Option: 3
$5$

Option: 4
$\frac{9-\sqrt{17}}{2}$

Answers (1)

best_answer

$\mathrm{f(x)= \begin{cases}x^{2}-4 x-2, & \forall x \in\left(-1, \frac{3-\sqrt{17}}{2}\right) \\ -x^{2}+2 x+2, & \forall x \in\left(\frac{3-\sqrt{17}}{2}, 2\right)\end{cases}}$

$\mathrm{f^{\prime}(x) \text { when } x \in\left(-1, \frac{3-\sqrt{17}}{2}\right)} \\$

$\mathrm{f^{\prime}(x)=2 x-4=0 \Rightarrow x=2} \\$

$\mathrm{f^{\prime}(x)=2(x-2) } \\$

$\mathrm{\Rightarrow f^{\prime}(x) \text { is always } \downarrow } \\$

$\mathrm{f(2)=2} \\$

$\mathrm{f(-1)=3}$

$\dot{f}\left(\frac{3-\sqrt{17}}{2}\right)=\frac{\sqrt{17}-3}{2} \\$

$\mathrm{f^{\prime}(x) \text { when } x \in\left(\frac{3-\sqrt{17}}{2}, 2\right) }\\$

$\mathrm{f^{\prime}(x)=-2 x+2} \\$

$\mathrm{f^{\prime}(x)=-2(x-1)} \\$

$\mathrm{f^{\prime}(x)=0 \qquad\text { when } x=1}$

$\mathrm{f(1)=3}$

Absolute minimum value $\mathrm{=\frac{\sqrt{17}-3}{2}}$

Absolute maximum value $\mathrm{=3}$

Sum $\mathrm{=\frac{\sqrt{17}-3}{2}+3=\frac{\sqrt{17}+3}{2}}$

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