The sum of the absolute minimum and the absolute maximum values of the function $f (x) = ∣ ∣ 3 x - x^{2} + 2 ∣ ∣ - x$ in the interval $[- 1, 2]$ is :

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Solution

$f (x) = ∣ ∣ x^{2} - 3 x - 2 ∣ ∣ - x \forall x \in [- 1, 2]$

$\Rightarrow f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x^{2} - 4 x - 2 if - 1 \leq x < \frac{3 - \sqrt{17}}{2} - x^{2} + 2 x + 2 if \frac{3 - \sqrt{17}}{2} \leq x \leq 2 \end{matrix}$

$f (x)_{max} = 3$

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

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