The maximum and minimum values of `x^3-18x^2+96x` in interval `(0,9)` are

Options

A. 160, 0
B. 60, 0
C. 160, 128
D. 120, 28

Explanation

To find the maximum and minimum values of the function \( f(x) = x^3 - 18x^2 + 96x \) in the interval \( (0, 9) \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 - 18x^2 + 96x) \]
Using the power rule, we get:
\[ f'(x) = 3x^2 - 36x + 96 \]

### Step 2: Set the derivative to zero to find critical points
To find the critical points, we set the derivative equal to zero:
\[ 3x^2 - 36x + 96 = 0 \]
Dividing the entire equation by 3 simplifies it to:
\[ x^2 - 12x + 32 = 0 \]

### Step 3: Factor the quadratic equation
Next, we factor the quadratic:
\[ (x - 4)(x - 8) = 0 \]
This gives us the critical points:
\[ x = 4 \quad \text{and} \quad x = 8 \]

### Step 4: Evaluate the function at critical points and endpoints
Now we will evaluate \( f(x) \) at the critical points and the endpoints of the interval \( (0, 9) \). The endpoints are \( x = 0 \) and \( x = 9 \).

1. **At \( x = 0 \)**:
\[ f(0) = 0^3 - 18(0^2) + 96(0) = 0 \]

2. **At \( x = 4 \)**:
\[ f(4) = 4^3 - 18(4^2) + 96(4) = 64 - 288 + 384 = 160 \]

3. **At \( x = 8 \)**:
\[ f(8) = 8^3 - 18(8^2) + 96(8) = 512 - 1152 + 768 = 128 \]

4. **At \( x = 9 \)**:
\[ f(9) = 9^3 - 18(9^2) + 96(9) = 729 - 1458 + 864 = 135 \]

### Step 5: Determine the maximum and minimum values
Now we compare the values obtained:
- \( f(0) = 0 \)
- \( f(4) = 160 \)
- \( f(8) = 128 \)
- \( f(9) = 135 \)

From these values, we can conclude:
- The **maximum value** is \( 160 \) at \( x = 4 \).
- The **minimum value** is \( 0 \) at \( x = 0 \).

### Final Answer
- Maximum value: \( 160 \)
- Minimum value: \( 0 \)