Question

The sum of the absolute minimum and the absolute maximum values of the function ƒ(x) = |3x - x2 + 2| - x in the interval [-1, 2] is

• A. $\frac{\sqrt17+3}{2}$
• B. $\frac{\sqrt{17}+5}{2}$
• C. 5
• D. $\frac{9-\sqrt{17}}{2}$

Answer 1

Answer: (A)

$\frac{\sqrt17+3}{2}$

Answer 2

The correct answer is (A) : $\frac{\sqrt17+3}{2}$
ƒ(x) = |x2– 3x – 2| – x
$∀x∈[−1,2]$
$f(x) = \begin{cases} x^2 - 4x - 2 & \text{if } -1 \leq x < \frac{3 - \sqrt{17}}{2} \\\ -x^2 + 2x + 2 & \text{if } \frac{3 - \sqrt{17}}{2} \leq x \leq 2 \end{cases}$

Fig.

$ƒ(x)_{max} = 3$
$ƒ(x)_{min}=ƒ(\frac{3−\sqrt{17}}{2})$
$=\frac{\sqrt{17}-3}{2}$