Key Concepts and Formulas

• Extreme Value Theorem: A continuous function on a closed interval $[a, b]$ attains both an absolute maximum and an absolute minimum on that interval.
• Critical Points: Points where the derivative of a function is zero or undefined.
• Product Rule: $\frac{d}{dx}(uv) = u'v + uv'$

Step-by-Step Solution

Step 1: Find the derivative of $f(x)$

We are given the function $f(x) = (x+3)^2(x-2)^3$ and we want to find its derivative, $f'(x)$, to locate critical points. We will use the product rule, where $u = (x+3)^2$ and $v = (x-2)^3$.

$u' = \frac{d}{dx}(x+3)^2 = 2(x+3)$ $v' = \frac{d}{dx}(x-2)^3 = 3(x-2)^2$

Applying the product rule: $f'(x) = u'v + uv' = 2(x+3)(x-2)^3 + (x+3)^2 \cdot 3(x-2)^2$

Now, we factor out the common terms $(x+3)$ and $(x-2)^2$: $f'(x) = (x+3)(x-2)^2[2(x-2) + 3(x+3)]$ $f'(x) = (x+3)(x-2)^2[2x - 4 + 3x + 9]$ $f'(x) = (x+3)(x-2)^2[5x + 5]$ $f'(x) = 5(x+3)(x-2)^2(x+1)$

The derivative is $f'(x) = 5(x+3)(x-2)^2(x+1)$.

Step 2: Find the critical points

Critical points occur where $f'(x) = 0$ or $f'(x)$ is undefined. Since $f'(x)$ is a polynomial, it is defined for all real numbers. Thus, we only need to solve $f'(x) = 0$:

$5(x+3)(x-2)^2(x+1) = 0$

This gives us the following critical points:

• $x+3 = 0 \implies x = -3$
• $(x-2)^2 = 0 \implies x = 2$
• $x+1 = 0 \implies x = -1$

The critical points are $x = -3, -1, 2$. All these points lie within the interval $[-4, 4]$.

Step 3: Evaluate $f(x)$ at the critical points and endpoints

We need to evaluate $f(x) = (x+3)^2(x-2)^3$ at the critical points $x = -3, -1, 2$ and the endpoints $x = -4, 4$.

1. $f(-4) = (-4+3)^2(-4-2)^3 = (-1)^2(-6)^3 = 1 \cdot (-216) = -216$
2. $f(-3) = (-3+3)^2(-3-2)^3 = (0)^2(-5)^3 = 0$
3. $f(-1) = (-1+3)^2(-1-2)^3 = (2)^2(-3)^3 = 4 \cdot (-27) = -108$
4. $f(2) = (2+3)^2(2-2)^3 = (5)^2(0)^3 = 25 \cdot 0 = 0$
5. $f(4) = (4+3)^2(4-2)^3 = (7)^2(2)^3 = 49 \cdot 8 = 392$

So we have the following values: $f(-4) = -216$, $f(-3) = 0$, $f(-1) = -108$, $f(2) = 0$, and $f(4) = 392$.

Step 4: Identify the maximum ($M$) and minimum ($m$) values

From the values calculated in Step 3, we can see that the maximum value is $M = 392$ and the minimum value is $m = -216$.

Therefore, $M - m = 392 - (-216) = 392 + 216 = 608$.

Common Mistakes & Tips

• Forgetting Endpoints: Always remember to evaluate the function at the endpoints of the interval.
• Sign Errors: Be careful with negative signs when evaluating the function, especially when raising negative numbers to odd powers.
• Factoring: Factoring the derivative simplifies finding the critical points and reduces the risk of algebraic errors.

Summary

To find the maximum and minimum values of the function $f(x) = (x+3)^2(x-2)^3$ on the interval $[-4, 4]$, we first found the derivative $f'(x)$, then determined the critical points by setting $f'(x) = 0$. We evaluated $f(x)$ at the critical points and the endpoints of the interval. Finally, we identified the maximum and minimum values and calculated their difference, $M - m = 608$.

Final Answer

The final answer is \boxed{608}, which corresponds to option (C).