Find the absolute maxima and minima2x^3-15x^2+36x+1 in [1,5]

To find the absolute maxima and minima of the function \( f(x) = 2x^3 - 15x^2 + 36x + 1 \) on the interval \([1, 5]\), we need to follow these steps:

### Step 1: Find the derivative of the function
First, we compute the derivative of \( f(x) \):

\[
f'(x) = \frac{d}{dx}\left(2x^3 - 15x^2 + 36x + 1\right)
\]

\[
f'(x) = 6x^2 - 30x + 36
\]

### Step 2: Find the critical points
To find the critical points, we set the derivative equal to zero and solve for \( x \):

\[
6x^2 - 30x + 36 = 0
\]

Divide the entire equation by 6 to simplify:

\[
x^2 - 5x + 6 = 0
\]

Factor the quadratic equation:

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

So the critical points are \( x = 2 \) and \( x = 3 \).

### Step 3: Evaluate the function at the critical points and endpoints
Next, evaluate \( f(x) \) at the critical points and the endpoints of the interval \([1, 5]\).

1. \( f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 2 - 15 + 36 + 1 = 24 \)
2. \( f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 16 - 60 + 72 + 1 = 29 \)
3. \( f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 54 - 135 + 108 + 1 = 28 \)
4. \( f(5) = 2(5)^3 - 15(5)^2 + 36(5) + 1 = 250 - 375 + 180 + 1 = 56 \)

### Step 4: Identify the absolute maxima and minima
Compare the values obtained:

- \( f(1) = 24 \)
- \( f(2) = 29 \)
- \( f(3) = 28 \)
- \( f(5) = 56 \)

The **absolute maximum** value on \([1, 5]\) is \( 56 \) at \( x = 5 \).  
The **absolute minimum** value on \([1, 5]\) is \( 24 \) at \( x = 1 \).

Would you like further details or have any questions?

Here are 5 questions to consider:
1. How do you determine if a critical point is a maximum or minimum?
2. What happens if no critical points exist within the interval?
3. How do endpoints affect the determination of absolute extrema?
4. Can there be more than one absolute maximum or minimum?
5. How does the shape of the function's graph influence where maxima and minima occur?

**Tip:** When solving for extrema on a closed interval, always check the function's value at both critical points and the interval's endpoints.

Learn how to find the absolute maxima and minima of the function 2x^3 - 15x^2 + 36x + 1 on the interval [1, 5]. This problem covers calculus concepts, including derivatives and critical points. Suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation