Key Concepts and Formulas

• Extreme Value Theorem: A continuous function on a closed interval $[a, b]$ attains its absolute maximum and minimum values on that interval. These values occur either at critical points in $(a, b)$ or at the endpoints $a$ and $b$.
• Critical Points: Points where the derivative of a function is either zero or undefined. For functions involving absolute values, points where the expression inside the absolute value is zero are also critical points as they represent points of non-differentiability.
• Absolute Value Function: $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.

Step-by-Step Solution

Step 1: Analyze the Absolute Value Function and Define $f(x)$ piecewise

The expression inside the absolute value is $x^2 - 5x + 6$. We factor this quadratic to find its roots: $x^2 - 5x + 6 = (x-2)(x-3)$ The roots are $x=2$ and $x=3$. This means the expression changes sign at $x=2$ and $x=3$. We can now write $f(x)$ as a piecewise function:

• For $x \in [-1, 2)$: $x^2 - 5x + 6 > 0$, so $|x^2 - 5x + 6| = x^2 - 5x + 6$. Thus, $f(x) = x^2 - 5x + 6 - 3x + 2 = x^2 - 8x + 8$.
• For $x \in [2, 3]$: $x^2 - 5x + 6 \le 0$, so $|x^2 - 5x + 6| = -(x^2 - 5x + 6) = -x^2 + 5x - 6$. Thus, $f(x) = -x^2 + 5x - 6 - 3x + 2 = -x^2 + 2x - 4$.

Therefore, $f(x) = \begin{cases} x^2 - 8x + 8 & \text{if } -1 \le x < 2 \\ -x^2 + 2x - 4 & \text{if } 2 \le x \le 3 \end{cases}$

Step 2: Find Critical Points in the Interval $(-1, 3)$

We need to find the critical points for each piece of the function.

• For $x^2 - 8x + 8$, the derivative is $f'(x) = 2x - 8$. Setting $f'(x) = 0$, we get $2x - 8 = 0$, so $x = 4$. Since $4$ is not in the interval $[-1, 2)$, we discard this critical point.
• For $-x^2 + 2x - 4$, the derivative is $f'(x) = -2x + 2$. Setting $f'(x) = 0$, we get $-2x + 2 = 0$, so $x = 1$. Since $1$ is not in the interval $[2, 3]$, we discard this critical point. However, we need to consider $x=2$ as a point of potential non-differentiability because it is where the absolute value changes sign, and it is in our interval.

Step 3: Evaluate $f(x)$ at Endpoints and Critical Points

We evaluate $f(x)$ at the endpoints $x = -1$ and $x = 3$, and at the point $x=2$ where the definition of the function changes.

• $f(-1) = (-1)^2 - 8(-1) + 8 = 1 + 8 + 8 = 17$.
• $f(2) = -2^2 + 2(2) - 4 = -4 + 4 - 4 = -4$.
• $f(3) = -3^2 + 2(3) - 4 = -9 + 6 - 4 = -7$.

Step 4: Determine Absolute Maximum and Minimum Values

Comparing the values $17, -4, -7$, we see that the absolute maximum value is $17$ and the absolute minimum value is $-7$.

Step 5: Calculate the Sum of Absolute Maximum and Minimum Values

The sum of the absolute maximum and minimum values is $17 + (-7) = 10$.

Common Mistakes & Tips

• Remember to consider the points where the absolute value expression changes sign as potential points of non-differentiability.
• Always check that the critical points you find are within the given interval.
• Be careful with the piecewise definition of the function, and make sure you use the correct expression for $f(x)$ when evaluating at different points.

Summary

We found the absolute maximum and minimum values of the function $f(x) = |x^2 - 5x + 6| - 3x + 2$ on the interval $[-1, 3]$ by first expressing $f(x)$ as a piecewise function, then finding critical points and evaluating the function at the endpoints and critical points. The absolute maximum value was 17 and the absolute minimum value was -7. Their sum is 10.

Final Answer

The final answer is $\boxed{10}$, which corresponds to option (C).