Find the absolute maximum and minimum values of each of the following in the given intervals :
`f(x)=2x^(3)-15x^(2)+36x+1" in "[1, 5]`

Answer

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

### Step 1: Find the derivative of the function
We start by finding the derivative of the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2x^3 - 15x^2 + 36x + 1) \]
Using the power rule, we get:
\[ f'(x) = 6x^2 - 30x + 36 \]

### Step 2: Set the derivative to zero to find critical points
Next, we set the derivative equal to zero to find the critical points:
\[ 6x^2 - 30x + 36 = 0 \]
Dividing the entire equation by 6:
\[ x^2 - 5x + 6 = 0 \]
Now we can factor the quadratic:
\[ (x - 2)(x - 3) = 0 \]
Thus, the critical points are:
\[ x = 2 \quad \text{and} \quad x = 3 \]

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

1. **At \( x = 1 \)**:
\[ f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 2 - 15 + 36 + 1 = 24 \]

2. **At \( x = 2 \)**:
\[ f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 16 - 60 + 72 + 1 = 29 \]

3. **At \( x = 3 \)**:
\[ f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 54 - 135 + 108 + 1 = 28 \]

4. **At \( x = 5 \)**:
\[ f(5) = 2(5)^3 - 15(5)^2 + 36(5) + 1 = 250 - 375 + 180 + 1 = 56 \]

### Step 4: Compare the values
Now we compare the values obtained:
- \( f(1) = 24 \)
- \( f(2) = 29 \)
- \( f(3) = 28 \)
- \( f(5) = 56 \)

### Step 5: Determine the absolute maximum and minimum
From the values calculated:
- The **absolute maximum** value is \( 56 \) at \( x = 5 \).
- The **absolute minimum** value is \( 24 \) at \( x = 1 \).

### Final Answer:
- Absolute maximum value: \( 56 \) at \( x = 5 \)
- Absolute minimum value: \( 24 \) at \( x = 1 \)

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