Question:

Step 1: Define the given function and its interval. The given function is $f(x)=|x^2-5x+6|-3x+2$, and we need to find its absolute maximum and minimum values in the interval $[-1,3]$. Step 2: Rewrite the function by analyzing the absolute value term. First, factor the quadratic expression inside the absolute value: $$x^2-5x+6 = (x-2)(x-3)$$ The roots are $x=2$ and $x=3$. This means: \begin{itemize} \item $x^2-5x+6 \ge 0$ for $x \in (-\infty, 2] \cup [3, \infty)$ \item $x^2-5x+6 < 0$ for $x \in (2, 3)$ \end{itemize} Considering the given interval $[-1,3]$, we can define $f(x)$ piecewise: \begin{enumerate} \item For $x \in [-1, 2]$: Here, $x^2-5x+6 \ge 0$. $$f(x) = (x^2-5x+6) - 3x + 2 = x^2 - 8x + 8$$ \item For $x \in (2, 3]$: Here, $x^2-5x+6 < 0$. $$f(x) = -(x^2-5x+6) - 3x + 2 = -x^2 + 5x - 6 - 3x + 2 = -x^2 + 2x - 4$$ \end{enumerate} Step 3: Find the derivative of the function for each piece. For the first piece, $f(x) = x^2 - 8x + 8$ for $x \in [-1, 2]$: $$f'(x) = 2x - 8$$ For the second piece, $f(x) = -x^2 + 2x - 4$ for $x \in (2, 3]$: $$f'(x) = -2x + 2$$ Step 4: Identify critical points within the interval $[-1,3]$. Critical points occur where $f'(x)=0$ or $f'(x)$ is undefined. \begin{enumerate} \item From $f'(x) = 2x - 8$: Setting $2x-8=0$ gives $x=4$. This point is not in the interval $[-1, 2]$, so it's not a critical point for this piece. \item From $f'(x) = -2x + 2$: Setting $-2x+2=0$ gives $x=1$. This point is not in the interval $(2, 3]$, so it's not a critical point for this piece. \item We also need to check points where the derivative might be undefined. This occurs at $x=2$, where the definition of $f(x)$ changes. Let's check the left and right hand derivatives at $x=2$: $f'(2^-) = 2(2) - 8 = -4$ $f'(2^+) = -2(2) + 2 = -2$ Since $f'(2^-) \neq f'(2^+)$, the derivative $f'(x)$ does not exist at $x=2$. Thus, $x=2$ is a critical point. \end{enumerate} The critical points within the interval $[-1,3]$ are $x=2$. The endpoints of the interval are $x=-1$ and $x=3$. Step 5: Evaluate the function at the critical points and the interval endpoints. \begin{itemize} \item At $x=-1$ (endpoint, falls in $x \in [-1,2]$): $f(-1) = (-1)^2 - 8(-1) + 8 = 1 + 8 + 8 = 17$ \item At $x=2$ (critical point): Using $f(x) = x^2 - 8x + 8$: $f(2) = (2)^2 - 8(2) + 8 = 4 - 16 + 8 = -4$ (Alternatively, using $f(x) = -x^2 + 2x - 4$: $f(2) = -(2)^2 + 2(2) - 4 = -4 + 4 - 4 = -4$) \item At $x=3$ (endpoint, falls in $x \in (2,3]$): $f(3) = -(3)^2 + 2(3) - 4 = -9 + 6 - 4 = -7$ \end{itemize} The values of the function at these points are $17, -4, -7$. Step 6: Determine the absolute maximum and minimum values. Comparing the values obtained in Step 5: Absolute maximum value $= \max\{17, -4, -7\} = 17$ Absolute minimum value $= \min\{17, -4, -7\} = -7$ Step 7: Calculate the sum of the absolute maximum and minimum values. Sum of absolute maximum and minimum values $= 17 + (-7) = 10$. The final answer is $\boxed{10}$.