Question

Let \( f(x) = (x + 3)^2 (x - 2)^3 \), \( x \in [-4, 4] \). If \( M \) and \( m \) are the maximum and minimum values of \( f \), respectively in \([-4, 4]\), then the value of \( M - m \) is:

• A. 600
• B. 392
• C. 608
• D. 108

Answer 1

The Correct Option is C

To determine the maximum and minimum values of the function \( f(x) = (x + 3)^2 (x - 2)^3 \) within the interval \([-4, 4]\), we calculate \( M - m \), where \( M \) represents the maximum value and \( m \) represents the minimum value.

The initial phase involves identifying the function's critical points by computing its derivative with respect to \( x \) and equating it to zero.

Applying the product rule for differentiation, with \( u(x) = (x + 3)^2 \) and \( v(x) = (x - 2)^3 \), we get:

• \( u'(x) = 2(x + 3) \)
• \( v'(x) = 3(x - 2)^2 \)

The product rule yields:

\(f'(x) = u'(x) v(x) + u(x) v'(x) = 2(x + 3)(x - 2)^3 + 3(x + 3)^2(x - 2)^2\)

Setting \( f'(x) = 0 \) requires solving:

\(2(x + 3)(x - 2)^3 + 3(x + 3)^2(x - 2)^2 = 0\)

Factoring out the common term \((x + 3)(x - 2)^2\):

\((x + 3)(x - 2)^2 [2(x - 2) + 3(x + 3)] = 0\)

This simplifies to:

\((x + 3)(x - 2)^2 (5x + 9) = 0\)

The solutions for each factor are:

• \( x + 3 = 0 \) implies \( x = -3 \)
• \( (x - 2)^2 = 0 \) implies \( x = 2 \)
• \( 5x + 9 = 0 \) implies \( x = -\frac{9}{5} \)

The function \( f(x) \) is then evaluated at these critical points and the interval's endpoints \( [-4, 4] \):

• \( f(-4) = ((-4) + 3)^2((-4) - 2)^3 = 1 \times (-6)^3 = -216 \)
• \( f(-3) = ((-3) + 3)^2((-3) - 2)^3 = 0 \)
• \( f(2) = (2 + 3)^2(2 - 2)^3 = 0 \)
• \( f(-\frac{9}{5}) = ((-\frac{9}{5}) + 3)^2((-\frac{9}{5}) - 2)^3 = (\frac{6}{5})^2(-\frac{19}{5})^3 = \frac{36}{25} \times -\frac{6859}{125} = -\frac{246924}{3125} \approx -79.006 \)
• \( f(4) = (4 + 3)^2(4 - 2)^3 = 49 \times 8 = 392 \)

Based on these evaluations, \( M = 392 \) and \( m = -216 \). Consequently, \( M - m = 392 - (-216) = 608 \).

The final result is \(608\).