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:

Updated On: Nov 4, 2025
  • 600
  • 392
  • 608
  • 108
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The Correct Option is C

Approach Solution - 1

Let's solve the problem to find the maximum and minimum values of the function \( f(x) = (x + 3)^2 (x - 2)^3 \) in the interval \([-4, 4]\). We need to find \( M - m \), where \( M \) is the maximum value, and \( m \) is the minimum value.

The first step is to find the critical points of the function by differentiating it with respect to \( x \) and setting the derivative to zero.

Using the product rule of differentiation, if \( u(x) = (x + 3)^2 \) and \( v(x) = (x - 2)^3 \), then:

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

Applying the product rule:

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

Setting \( f'(x) = 0 \), we solve:

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

Factor out \((x + 3)(x - 2)^2\):

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

This simplifies to:

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

Solve each factor:

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

Evaluate \( f(x) \) at the critical points and endpoints of the interval \( [-4, 4] \):

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

From the above calculations, we find that \( M = 392 \) and \( m = -216 \). Therefore, the value of \( M - m = 392 - (-216) = 608 \).

Thus, the correct answer is \(608\).

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Approach Solution -2

To find the maximum and minimum values of \( f(x) \):

Take the derivative \( f'(x) \) and find the critical points.

Evaluate \( f(x) \) at critical points and endpoints \( x = -4, -3, -2, -1, 1, 2, 3, 4 \).

The maximum value \( M = 392 \) and the minimum value \( m = -216 \).

The value of \( M - m \) is:

\[ M - m = 392 - (-216) = 608. \]

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