Key Concepts and Formulas

• Absolute Extrema: The absolute maximum and minimum values of a continuous function on a closed interval occur either at critical points within the interval or at the endpoints of the interval.
• Critical Points: Critical points are the points where the derivative of the function is either zero or undefined.
• Absolute Value Function: The absolute value function $|g(x)|$ can be expressed as a piecewise function: $g(x)$ if $g(x) \ge 0$, and $-g(x)$ if $g(x) < 0$.

Step-by-Step Solution

Step 1: Analyze the function inside the absolute value

We want to determine when $3x - x^2 + 2$ is positive or negative within the interval $[-1, 2]$. Let $g(x) = 3x - x^2 + 2$. We find the roots of $g(x) = 0$: $x = \frac{-3 \pm \sqrt{3^2 - 4(-1)(2)}}{2(-1)} = \frac{-3 \pm \sqrt{9 + 8}}{-2} = \frac{-3 \pm \sqrt{17}}{-2} = \frac{3 \pm \sqrt{17}}{2}$ The roots are $x_1 = \frac{3 - \sqrt{17}}{2} \approx -0.56$ and $x_2 = \frac{3 + \sqrt{17}}{2} \approx 3.56$. Since the parabola $g(x) = -x^2 + 3x + 2$ opens downward, $g(x)$ is positive between the roots. In the interval $[-1, 2]$, $x_1 = \frac{3 - \sqrt{17}}{2}$ lies within the interval, while $x_2 = \frac{3 + \sqrt{17}}{2}$ is outside the interval. Therefore, $g(x) \ge 0$ for $x \in [\frac{3 - \sqrt{17}}{2}, 2]$ and $g(x) < 0$ for $x \in [-1, \frac{3 - \sqrt{17}}{2})$.

Step 2: Express the function as a piecewise function

Based on the analysis in Step 1, we can write $f(x)$ as a piecewise function: $f(x) = \begin{cases} -(3x - x^2 + 2) - x = x^2 - 4x - 2 & \text{if } -1 \le x < \frac{3 - \sqrt{17}}{2} \\ (3x - x^2 + 2) - x = -x^2 + 2x + 2 & \text{if } \frac{3 - \sqrt{17}}{2} \le x \le 2 \end{cases}$

Step 3: Find critical points

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

For $x^2 - 4x - 2$, the derivative is $2x - 4$. Setting $2x - 4 = 0$, we get $x = 2$. However, this critical point is not within the interval $[-1, \frac{3 - \sqrt{17}}{2})$.

For $-x^2 + 2x + 2$, the derivative is $-2x + 2$. Setting $-2x + 2 = 0$, we get $x = 1$. This critical point is within the interval $[\frac{3 - \sqrt{17}}{2}, 2]$.

Step 4: Evaluate the function at the endpoints, critical points, and the point where the piecewise function changes

We need to evaluate $f(x)$ at $x = -1$, $x = 1$, $x = 2$, and $x = \frac{3 - \sqrt{17}}{2}$.

• $f(-1) = |-3 - 1 + 2| - (-1) = |-2| + 1 = 2 + 1 = 3$
• $f(1) = |3 - 1 + 2| - 1 = |4| - 1 = 4 - 1 = 3$
• $f(2) = |6 - 4 + 2| - 2 = |4| - 2 = 4 - 2 = 2$
• $f(\frac{3 - \sqrt{17}}{2}) = 0 - (\frac{3 - \sqrt{17}}{2}) = \frac{\sqrt{17} - 3}{2} \approx 0.56$

Step 5: Determine the absolute maximum and minimum values

The values we found are $f(-1) = 3$, $f(1) = 3$, $f(2) = 2$, and $f(\frac{3 - \sqrt{17}}{2}) = \frac{\sqrt{17} - 3}{2}$. The absolute maximum is 3, and the absolute minimum is $\frac{\sqrt{17} - 3}{2}$.

Step 6: Calculate the sum of the absolute maximum and minimum values

The sum is $3 + \frac{\sqrt{17} - 3}{2} = \frac{6 + \sqrt{17} - 3}{2} = \frac{3 + \sqrt{17}}{2}$.

Common Mistakes & Tips

• Remember to consider the point where the absolute value function changes its behavior. This point may not be a critical point in the traditional sense (where the derivative is zero), but it can still be a location of an absolute extremum.
• Be careful with the intervals when dealing with piecewise functions. Make sure to use the correct expression for $f(x)$ when evaluating at different points.
• When finding critical points, solve $f'(x) = 0$ separately for each piece of the function.

Summary

We analyzed the function $f(x) = |3x - x^2 + 2| - x$ on the interval $[-1, 2]$ by breaking it into a piecewise function based on the sign of $3x - x^2 + 2$. We then found the critical points of each piece and evaluated the function at the critical points, endpoints, and the point where the piecewise function changes. Finally, we determined the absolute maximum and minimum values and calculated their sum, which is $\frac{\sqrt{17} + 3}{2}$.

Final Answer

The final answer is $\boxed{\frac{\sqrt{17} + 3}{2}}$, which corresponds to option (A).