Question

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 :

• A. $\frac{\sqrt{17}+3}{2}$
• B. $\frac{\sqrt{17}+5}{2}$
• C. $5$
• D. $\frac{9-\sqrt{17}}{2}$

Answer 1

Answer: (A)

$f(x)= \begin{cases}x^{2}-4 x-2, & \forall x \in\left(-1, \frac{3-\sqrt{17}}{2}\right) \\ -x^{2}+2 x+2, & \forall x \in\left(\frac{3-\sqrt{17}}{2}, 2\right)\end{cases}$
$f^{\prime}(x)$ when $x \in\left(-1, \frac{3-\sqrt{17}}{2}\right)$
$f^{\prime}(x)=2 x-4=0 \rightarrow x=2$
$f^{\prime}(x)=2(x-2) \Rightarrow f^{\prime}(x)$ is always $\downarrow$
$f(2)=2$
$f(-1)=3$
$f\left(\frac{3-\sqrt{17}}{2}\right)=\frac{\sqrt{17}-3}{2}$
$f^{\prime}(x)$ when $x \in\left(\frac{3-\sqrt{17}}{2}, 2\right)$
$f^{\prime}( x )=-2 x +2$
$f^{\prime}(x)=-2(x-1)$
$f^{\prime}(x)=0$ when $x=1$
$f(1)=3$
absolute minimum value $=\frac{\sqrt{17}-3}{2}$
absolute maximum value $=3$
Sum $=\frac{\sqrt{17}-3}{2}+3=\frac{\sqrt{17}+3}{2}$