Find the absolute maximum and absolute minimum of values of
f (x) = 2x3 − 3x2 − 12x + 1
on [−2, 3]

To find the absolute maximum and minimum values of the function \( f(x) = 2x^3 - 3x^2 - 12x + 1 \) on the interval \( [-2, 3] \), we follow these steps:

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### Step 1: Find the critical points of \( f(x) \)  
The critical points occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined.

1. **Compute the derivative:**
   \[
   f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 1) = 6x^2 - 6x - 12
   \]

2. **Set the derivative equal to 0:**
   \[
   6x^2 - 6x - 12 = 0
   \]
   Divide the equation by 6:
   \[
   x^2 - x - 2 = 0
   \]
   Factor the quadratic:
   \[
   (x - 2)(x + 1) = 0
   \]
   So, the critical points are:
   \[
   x = 2 \quad \text{and} \quad x = -1
   \]

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### Step 2: Evaluate \( f(x) \) at critical points and endpoints  
We need to evaluate \( f(x) \) at the critical points \( x = 2 \) and \( x = -1 \), as well as at the endpoints of the interval \( [-2, 3] \).

1. **At \( x = -2 \):**
   \[
   f(-2) = 2(-2)^3 - 3(-2)^2 - 12(-2) + 1 = 2(-8) - 3(4) + 24 + 1 = -16 - 12 + 24 + 1 = -3
   \]

2. **At \( x = -1 \):**
   \[
   f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 1 = 2(-1) - 3(1) + 12 + 1 = -2 - 3 + 12 + 1 = 8
   \]

3. **At \( x = 2 \):**
   \[
   f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 1 = 2(8) - 3(4) - 24 + 1 = 16 - 12 - 24 + 1 = -19
   \]

4. **At \( x = 3 \):**
   \[
   f(3) = 2(3)^3 - 3(3)^2 - 12(3) + 1 = 2(27) - 3(9) - 36 + 1 = 54 - 27 - 36 + 1 = -8
   \]

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### Step 3: Identify the absolute maximum and minimum  
From the evaluations:
- \( f(-2) = -3 \)
- \( f(-1) = 8 \)
- \( f(2) = -19 \)
- \( f(3) = -8 \)

The **absolute maximum** is \( f(-1) = 8 \).  
The **absolute minimum** is \( f(2) = -19 \).

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### Final Answer:
- Absolute maximum: \( 8 \) at \( x = -1 \)
- Absolute minimum: \( -19 \) at \( x = 2 \)

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Would you like me to provide more details or have any questions?

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**5 Related Questions:**
1. How do we determine whether a critical point is a maximum or minimum?
2. Can absolute extrema occur at endpoints instead of critical points?
3. What happens if the interval is not closed? 
4. How do you solve cubic equations to find critical points if factoring is not possible?
5. How does the second derivative test help with extrema?

**Tip:** Always check both the critical points and endpoints when finding absolute extrema on a closed interval.

Find the absolute maximum and absolute minimum values of f(x) = 2x^3 - 3x^2 - 12x + 1 on [-2, 3]

Find the absolute maximum and minimum values of f(x) = 2x^3 - 3x^2 - 12x + 1 on the interval [-2, 3]. This calculus problem explores critical points and endpoint evaluation, ideal for students in Grades 11-12.

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