Question

The sum of the absolute minimum and the absolute maximum values of the function $f(x)=\left|3x-x^2+2\right|-x$ 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)

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