The sum of the absolute minimum and the absol | vedclass

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(A) First,we analyze the expression inside the absolute value: $g(x) = -x^2 + 3x + 2$. The roots of $g(x) = 0$ are $x = \frac{-3 \pm \sqrt{9 - 4(-1)(2

Vedclass · Accepted Answer

$\frac{\sqrt{17} + 3}{2}$

Vedclass · Answer

(A) First,we analyze the expression inside the absolute value: $g(x) = -x^2 + 3x + 2$. The roots of $g(x) = 0$ are $x = \frac{-3 \pm \sqrt{9 - 4(-1)(2)}}{2(-1)} = \frac{-3 \pm \sqrt{17}}{-2} = \frac{3 \mp \sqrt{17}}{2}$.Since $g(x) \ge 0$ for $x \in [\frac{3-\sqrt{17}}{2}, \frac{3+\sqrt{17}}{2}]$,and the interval is $[-1, 2]$,we note that $\frac{3-\sqrt{17}}{2} \approx -0.56$ and $\frac{3+\sqrt{17}}{2} \approx 3.56$.Thus,$f(x) = -x^2 + 2x + 2$ for $x \in [\frac{3-\sqrt{17}}{2}, 2]$ and $f(x) = x^2 - 4x - 2$ for $x \in [-1, \frac{3-\sqrt{17}}{2}]$.Evaluating at critical points and endpoints:$f(-1) = |3(-1) - (-1)^2 + 2| - (-1) = |-3 - 1 + 2| + 1 = |-2| + 1 = 3$.$f(2) = |3(2) - (2)^2 + 2| - 2 = |6 - 4 + 2| - 2 = 4 - 2 = 2$.At $x = \frac{3-\sqrt{17}}{2}$,$f(x) = 0 - \frac{3-\sqrt{17}}{2} = \frac{\sqrt{17}-3}{2}$.For $x \in (\frac{3-\sqrt{17}}{2}, 2)$,$f'(x) = -2x + 2 = 0 \Rightarrow x = 1$. $f(1) = |3 - 1 + 2| - 1 = 4 - 1 = 3$.Comparing values: $f(-1)=3, f(2)=2, f(1)=3, f(\frac{3-\sqrt{17}}{2}) = \frac{\sqrt{17}-3}{2}$.Absolute maximum $= 3$,absolute minimum $= \frac{\sqrt{17}-3}{2}$.Sum $= 3 + \frac{\sqrt{17}-3}{2} = \frac{6 + \sqrt{17} - 3}{2} = \frac{\sqrt{17}+3}{2}$.

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

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