The sum of the absolute maximum and minimum v | vedclass

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(A) Given $f(x) = |x^2 - 5x + 6| - 3x + 2$. The roots of $x^2 - 5x + 6 = 0$ are $x = 2$ and $x = 3$.For $x \in [-1, 2]$,$x^2 - 5x + 6 \ge 0$,so $f(x)

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$10$

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(A) Given $f(x) = |x^2 - 5x + 6| - 3x + 2$. The roots of $x^2 - 5x + 6 = 0$ are $x = 2$ and $x = 3$.For $x \in [-1, 2]$,$x^2 - 5x + 6 \ge 0$,so $f(x) = x^2 - 5x + 6 - 3x + 2 = x^2 - 8x + 8$.For $x \in [2, 3]$,$x^2 - 5x + 6 \le 0$,so $f(x) = -(x^2 - 5x + 6) - 3x + 2 = -x^2 + 2x - 4$.Now,check critical points and endpoints:$1$. For $x \in [-1, 2]$,$f'(x) = 2x - 8$. Setting $f'(x) = 0$ gives $x = 4$,which is outside the interval. Thus,check endpoints: $f(-1) = (-1)^2 - 8(-1) + 8 = 1 + 8 + 8 = 17$ and $f(2) = 2^2 - 8(2) + 8 = 4 - 16 + 8 = -4$.$2$. For $x \in [2, 3]$,$f'(x) = -2x + 2$. Setting $f'(x) = 0$ gives $x = 1$,which is outside the interval. Thus,check endpoints: $f(2) = -4$ and $f(3) = -(3)^2 + 2(3) - 4 = -9 + 6 - 4 = -7$.The absolute maximum value is $17$ and the absolute minimum value is $-7$.The sum of the absolute maximum and minimum values is $17 + (-7) = 10$.

The sum of the absolute maximum and minimum values of the function $f(x) = |x^2 - 5x + 6| - 3x + 2$ in the interval $[-1, 3]$ is equal to:

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