Question

let ( f ) be the function defined by ( f(x) = x^3 - \frac{3}{2}x^2 - 6x ). what is the absolute maximum value of ( f ) on the interval ( -2, 3 )?

a ( -10 )

b ( -\frac{9}{2} )

c ( \frac{7}{2} )

d ( 3 )

Explanation:

Step1: Find the derivative of \( f(x) \)

To find critical points, we first find the derivative of \( f(x) = x^3 - \frac{3}{2}x^2 - 6x \). Using the power rule, the derivative \( f'(x) \) is:
\( f'(x) = 3x^2 - 3x - 6 \)

Step2: Find critical points

Set \( f'(x) = 0 \) to find critical points:
\( 3x^2 - 3x - 6 = 0 \)
Divide both sides by 3:
\( x^2 - x - 2 = 0 \)
Factor the quadratic:
\( (x - 2)(x + 1) = 0 \)
So the critical points are \( x = 2 \) and \( x = -1 \), both of which lie in the interval \([-2, 3]\).

Step3: Evaluate \( f(x) \) at critical points and endpoints

We evaluate \( f(x) \) at \( x = -2 \), \( x = -1 \), \( x = 2 \), and \( x = 3 \).

- For \( x = -2 \):
\( f(-2) = (-2)^3 - \frac{3}{2}(-2)^2 - 6(-2) = -8 - \frac{3}{2}(4) + 12 = -8 - 6 + 12 = -2 \) Wait, no, let's recalculate:
\( (-2)^3 = -8 \), \( \frac{3}{2}(-2)^2 = \frac{3}{2}(4) = 6 \), \( -6(-2) = 12 \). So \( f(-2) = -8 - 6 + 12 = -2 \)? Wait, no, original function: \( x^3 - \frac{3}{2}x^2 - 6x \). So \( (-2)^3 = -8 \), \( -\frac{3}{2}(-2)^2 = -\frac{3}{2}(4) = -6 \), \( -6(-2) = 12 \). So \( -8 -6 +12 = -2 \). Wait, maybe I made a mistake earlier. Wait, let's check again.

Wait, no, let's do each point carefully:

1. \( x = -2 \):
\( f(-2) = (-2)^3 - \frac{3}{2}(-2)^2 - 6(-2) = -8 - \frac{3}{2}(4) + 12 = -8 - 6 + 12 = -2 \)

2. \( x = -1 \):
\( f(-1) = (-1)^3 - \frac{3}{2}(-1)^2 - 6(-1) = -1 - \frac{3}{2}(1) + 6 = -1 - \frac{3}{2} + 6 = \frac{-2 - 3 + 12}{2} = \frac{7}{2} = 3.5 \)

3. \( x = 2 \):
\( f(2) = (2)^3 - \frac{3}{2}(2)^2 - 6(2) = 8 - \frac{3}{2}(4) - 12 = 8 - 6 - 12 = -10 \)

4. \( x = 3 \):
\( f(3) = (3)^3 - \frac{3}{2}(3)^2 - 6(3) = 27 - \frac{3}{2}(9) - 18 = 27 - \frac{27}{2} - 18 = \frac{54 - 27 - 36}{2} = \frac{-9}{2} = -4.5 \)

Now, we compare the values: \( f(-2) = -2 \), \( f(-1) = \frac{7}{2} \), \( f(2) = -10 \), \( f(3) = -\frac{9}{2} \). The largest among these is \( \frac{7}{2} \)? Wait, no, wait \( f(-2) \) was calculated as -2, but let's recheck \( f(-2) \):

Wait, \( f(x) = x^3 - \frac{3}{2}x^2 - 6x \). So \( x = -2 \):

\( (-2)^3 = -8 \)

\( -\frac{3}{2}(-2)^2 = -\frac{3}{2}(4) = -6 \)

\( -6(-2) = 12 \)

So \( -8 -6 +12 = -2 \). Correct.

\( x = -1 \):

\( (-1)^3 = -1 \)

\( -\frac{3}{2}(-1)^2 = -\frac{3}{2}(1) = -\frac{3}{2} \)

\( -6(-1) = 6 \)

So \( -1 - \frac{3}{2} + 6 = (-1 + 6) - \frac{3}{2} = 5 - \frac{3}{2} = \frac{10 - 3}{2} = \frac{7}{2} \). Correct.

\( x = 2 \):

\( 8 - \frac{3}{2}(4) - 12 = 8 - 6 - 12 = -10 \). Correct.

\( x = 3 \):

\( 27 - \frac{27}{2} - 18 = (27 - 18) - \frac{27}{2} = 9 - \frac{27}{2} = \frac{18 - 27}{2} = -\frac{9}{2} \). Correct.

Now, the values are: -2, \( \frac{7}{2} \) (3.5), -10, -4.5. Wait, but -2 is greater than -4.5 and -10, but \( \frac{7}{2} \) is 3.5, which is greater than -2. Wait, but the options include \( \frac{7}{2} \) (option C) and 3 (option D). Wait, did I make a mistake in calculating \( f(-2) \)?

Wait, wait, let's recalculate \( f(-2) \) again:

\( f(-2) = (-2)^3 - \frac{3}{2}(-2)^2 - 6(-2) \)

\( = -8 - \frac{3}{2}(4) + 12 \)

\( = -8 - 6 + 12 \)

\( = (-8 -6) + 12 \)

\( = -14 + 12 = -2 \). Correct.

Wait, but \( f(-1) = \frac{7}{2} = 3.5 \), \( f(-2) = -2 \), \( f(3) = -4.5 \), \( f(2) = -10 \). So the maximum value among these is \( \frac{7}{2} \), which is option C. But wait, the option D is 3. Did I make a mistake?

Wait, let's check \( f(-1) \) again:

\( f(-1) = (-1)^3 - \frac{3}{2}(-1)^2 - 6(-1) \)

\( = -1 - \frac{3}{2}(1) + 6 \)

\( = -1 - 1.5 + 6 \)

\( = 3.5 \), which is \( \frac{7}{2} \). Correct.

Wait, maybe the original function was written incorrectly? Let me check the problem again. The function is \( f(x) = x^3 - \frac{3}{2}x^2 - 6x \). Interval [-2, 3].

Wait, maybe I made a mistake in the critical points. Let's recheck the derivative:

\( f(x) = x^3 - \frac{3}{2}x^2 - 6x \)

\( f'(x) = 3x^2 - 3x - 6 \). Correct.

Set to zero: \( 3x^2 - 3x - 6 = 0 \) => \( x^2 - x - 2 = 0 \) => (x - 2)(x + 1) = 0. So x = 2 and x = -1. Correct.

So evaluating at x = -2, -1, 2, 3:

- x = -2: -8 - 6 + 12 = -2
- x = -1: -1 - 1.5 + 6 = 3.5 = 7/2
- x = 2: 8 - 6 - 12 = -10
- x = 3: 27 - 13.5 - 18 = -4.5 = -9/2

So the maximum value is 7/2, which is option C.

Answer:

C. \( \frac{7}{2} \)