Key Concepts and Formulas

• Critical Points: Points where the derivative of a function is zero or undefined.
• Absolute Extrema: The maximum and minimum values of a function on a given interval. These occur at critical points within the interval or at the endpoints of the interval.
• Derivatives: The derivative of $\log_e |x|$ is $\frac{1}{x}$.

Step-by-Step Solution

Step 1: Differentiating the Function

We are given the function $f(x) = x^3 + ax^2 + b \log_e |x| + 1$ for $x \neq 0$. To find the critical points, we need to find the derivative $f'(x)$.

$f'(x) = \frac{d}{dx}(x^3 + ax^2 + b \log_e |x| + 1)$ $f'(x) = 3x^2 + 2ax + \frac{b}{x}$

Step 2: Using the Critical Points to Find $a$ and $b$

We are given that $x=-1$ and $x=2$ are critical points, meaning $f'(-1) = 0$ and $f'(2) = 0$. We will use these conditions to solve for $a$ and $b$.

For $x = -1$: $f'(-1) = 3(-1)^2 + 2a(-1) + \frac{b}{-1} = 0$ $3 - 2a - b = 0 \quad (*)$

For $x = 2$: $f'(2) = 3(2)^2 + 2a(2) + \frac{b}{2} = 0$ $12 + 4a + \frac{b}{2} = 0 \quad (**)$

Now we have a system of two linear equations with two unknowns. Multiplying equation $(**)$ by 2, we get: $24 + 8a + b = 0 \quad (***)$

Adding equations $(*)$ and $(***)$, we eliminate $b$: $(3 - 2a - b) + (24 + 8a + b) = 0$ $27 + 6a = 0$ $6a = -27$ $a = -\frac{27}{6} = -\frac{9}{2}$

Substitute $a = -\frac{9}{2}$ into equation $(*)$: $3 - 2\left(-\frac{9}{2}\right) - b = 0$ $3 + 9 - b = 0$ $12 - b = 0$ $b = 12$

Thus, we have $a = -\frac{9}{2}$ and $b = 12$.

Step 3: Finding the Function $f(x)$

Now we have the function: $f(x) = x^3 - \frac{9}{2}x^2 + 12 \log_e |x| + 1$

Step 4: Evaluating $f(x)$ at the endpoints and critical points within the interval

We are given the interval $[-2, -\frac{1}{2}]$. The critical points are $x=-1$ and $x=2$. Since $2$ is not in the interval $[-2, -\frac{1}{2}]$, we only need to consider $x=-1$. We also need to evaluate $f(x)$ at the endpoints $x=-2$ and $x=-\frac{1}{2}$.

$f(-2) = (-2)^3 - \frac{9}{2}(-2)^2 + 12 \log_e |-2| + 1 = -8 - \frac{9}{2}(4) + 12 \log_e 2 + 1 = -8 - 18 + 12(0.7) + 1 = -26 + 8.4 + 1 = -16.6$ $f(-1) = (-1)^3 - \frac{9}{2}(-1)^2 + 12 \log_e |-1| + 1 = -1 - \frac{9}{2} + 12(0) + 1 = -1 - 4.5 + 0 + 1 = -4.5$ $f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^3 - \frac{9}{2}\left(-\frac{1}{2}\right)^2 + 12 \log_e \left|-\frac{1}{2}\right| + 1 = -\frac{1}{8} - \frac{9}{2}\left(\frac{1}{4}\right) + 12 \log_e \left(\frac{1}{2}\right) + 1 = -\frac{1}{8} - \frac{9}{8} + 12(-\log_e 2) + 1 = -\frac{10}{8} - 12(0.7) + 1 = -\frac{5}{4} - 8.4 + 1 = -1.25 - 8.4 + 1 = -8.65$

Step 5: Determining the Absolute Maximum (M) and Absolute Minimum (m)

Comparing the values of $f(x)$ at the endpoints and critical point: $f(-2) = -16.6$ $f(-1) = -4.5$ $f(-\frac{1}{2}) = -8.65$

The absolute maximum is $M = -4.5$ and the absolute minimum is $m = -16.6$.

Step 6: Calculating $|M + m|$

$|M + m| = |-4.5 + (-16.6)| = |-4.5 - 16.6| = |-21.1| = 21.1$

Common Mistakes & Tips

• Sign Errors: Be very careful with negative signs, especially when substituting values into equations.
• Logarithm Properties: Remember that $\log_e(1/x) = -\log_e(x)$.
• Endpoint Evaluation: Don't forget to evaluate the function at the endpoints of the interval.

Summary

We found the derivative of the given function and used the critical points to solve for the unknown coefficients $a$ and $b$. Then, we determined the function $f(x)$ and evaluated it at the endpoints and relevant critical point within the interval $[-2, -\frac{1}{2}]$. Finally, we identified the absolute maximum and minimum values and calculated the absolute value of their sum.

The final answer is \boxed{21.1}, which corresponds to option (A).