The maximum and minimum values for the funtion f(x)=`3x^4-4x^3` on [-1,2] are

Options

A. 7,0
B. 0,-7
C. 16,-1
D. `-7,16`

Answer

To find the maximum and minimum values of the function \( f(x) = 3x^4 - 4x^3 \) on the interval \([-1, 2]\), we will follow these steps:

### Step 1: Find the derivative of the function
First, we need to calculate the derivative \( f'(x) \) to find the critical points.

\[ f'(x) = \frac{d}{dx}(3x^4 - 4x^3) = 12x^3 - 12x^2 \]

### Step 2: Set the derivative to zero
Next, we set the derivative equal to zero to find the critical points.

\[ 12x^3 - 12x^2 = 0 \]

Factoring out \( 12x^2 \):

\[ 12x^2(x - 1) = 0 \]

This gives us the critical points:

\[ x^2 = 0 \quad \Rightarrow \quad x = 0 \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

### Step 3: Evaluate the function at the critical points and endpoints
We need to evaluate \( f(x) \) at the critical points \( x = 0 \) and \( x = 1 \), as well as at the endpoints \( x = -1 \) and \( x = 2 \).

1. **At \( x = -1 \)**:
\[ f(-1) = 3(-1)^4 - 4(-1)^3 = 3(1) + 4(1) = 3 + 4 = 7 \]

2. **At \( x = 0 \)**:
\[ f(0) = 3(0)^4 - 4(0)^3 = 0 \]

3. **At \( x = 1 \)**:
\[ f(1) = 3(1)^4 - 4(1)^3 = 3(1) - 4(1) = 3 - 4 = -1 \]

4. **At \( x = 2 \)**:
\[ f(2) = 3(2)^4 - 4(2)^3 = 3(16) - 4(8) = 48 - 32 = 16 \]

### Step 4: Determine the maximum and minimum values
Now we compare the values obtained:

- \( f(-1) = 7 \)
- \( f(0) = 0 \)
- \( f(1) = -1 \)
- \( f(2) = 16 \)

From these values, we can see that:
- The **maximum value** is \( 16 \) at \( x = 2 \).
- The **minimum value** is \( -1 \) at \( x = 1 \).

### Final Result
Thus, the maximum and minimum values of the function \( f(x) = 3x^4 - 4x^3 \) on the interval \([-1, 2]\) are:
- Maximum value: \( 16 \)
- Minimum value: \( -1 \)

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