Key Concepts and Formulas

• Extreme Value Theorem: A continuous function on a closed interval $[a, b]$ attains an absolute maximum and an absolute minimum value on that interval.
• Critical Points: Points where the derivative of a function is either zero or undefined.
• Finding Absolute Extrema: Evaluate the function at all critical points within the interval and at the endpoints of the interval. The largest value is the absolute maximum, and the smallest is the absolute minimum.

Step-by-Step Solution

Step 1: Find the first derivative of the function, $f'(x)$. We differentiate the given function $f(x)$ with respect to $x$ to find the critical points where the function's slope is zero. This helps us identify potential locations of maxima and minima.

Given function: $f(x) = 2x^3 - 9x^2 + 12x + 5$ Differentiating term by term: $f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(9x^2) + \frac{d}{dx}(12x) + \frac{d}{dx}(5)$ Using the power rule, $\frac{d}{dx}(x^n) = nx^{n-1}$: $f'(x) = 6x^2 - 18x + 12$

Step 2: Find the critical points by setting $f'(x) = 0$. We set the derivative equal to zero to find the x-values where the function has a horizontal tangent, which are potential locations for local maxima or minima.

Set $f'(x) = 0$: $6x^2 - 18x + 12 = 0$ Divide the entire equation by 6 to simplify: $x^2 - 3x + 2 = 0$ Factor the quadratic equation: $(x - 1)(x - 2) = 0$ Solve for $x$: $x = 1, \quad x = 2$ Both critical points, $x = 1$ and $x = 2$, lie within the given interval $[0, 3]$, so we consider them.

Step 3: Evaluate the original function $f(x)$ at the critical points and at the endpoints of the interval. We substitute the critical points and the endpoints of the interval into the original function to determine the function's values at these points.

1. At $x = 0$: $f(0) = 2(0)^3 - 9(0)^2 + 12(0) + 5 = 5$

2. At $x = 1$: $f(1) = 2(1)^3 - 9(1)^2 + 12(1) + 5 = 2 - 9 + 12 + 5 = 10$

3. At $x = 2$: $f(2) = 2(2)^3 - 9(2)^2 + 12(2) + 5 = 16 - 36 + 24 + 5 = 9$

4. At $x = 3$: $f(3) = 2(3)^3 - 9(3)^2 + 12(3) + 5 = 54 - 81 + 36 + 5 = 14$

Step 4: Identify the absolute maximum (M) and absolute minimum (m). Compare the function values calculated in the previous step to find the largest (absolute maximum) and smallest (absolute minimum) values.

The function values are: $f(0) = 5, \quad f(1) = 10, \quad f(2) = 9, \quad f(3) = 14$ The absolute maximum is $M = 14$, and the absolute minimum is $m = 5$.

Step 5: Calculate $M - m$. Subtract the absolute minimum from the absolute maximum to find the difference.

$M - m = 14 - 5 = 9$

Common Mistakes & Tips:

• Forgetting Endpoints: Remember to check the function values at the endpoints of the interval, as the absolute extrema can occur there.
• Invalid Critical Points: Make sure the critical points you find lie within the given interval. Critical points outside the interval are not relevant.
• Arithmetic Errors: Double-check your calculations to avoid arithmetic mistakes.

Summary

To find the absolute maximum and minimum values of the given function on the closed interval $[0, 3]$, we found the derivative, identified critical points within the interval, evaluated the function at the critical points and endpoints, and then compared the function values. The absolute maximum was found to be 14, and the absolute minimum was 5. Therefore, the difference $M - m$ is 9.

The final answer is \boxed{9}, which corresponds to option (B).