Find the image of interval `[-1,3]` under the mapping specified by the function `f(x)=4x^3-12 xdot`

Options

A. `[8,72]`
B. `[-8,72]`
C. `[0,8]`
D. `[8,-72]`

Explanation

To find the image of the interval \([-1, 3]\) under the mapping specified by the function \(f(x) = 4x^3 - 12x\), we will follow these steps:

### Step 1: Identify the function and the interval
The function given is:
\[ f(x) = 4x^3 - 12x \]
We need to find the image of the interval \([-1, 3]\).

### Step 2: Find the derivative of the function
To find the maximum and minimum values of \(f(x)\), we first need to calculate the derivative \(f'(x)\):
\[ f'(x) = \frac{d}{dx}(4x^3 - 12x) = 12x^2 - 12 \]

### Step 3: Set the derivative to zero to find critical points
Next, we set the derivative equal to zero to find the critical points:
\[ 12x^2 - 12 = 0 \]
\[ 12(x^2 - 1) = 0 \]
\[ x^2 - 1 = 0 \]
This gives us:
\[ x = 1 \quad \text{and} \quad x = -1 \]

### Step 4: Identify the boundary points
The boundary points of the interval are:
\[ x = -1 \quad \text{and} \quad x = 3 \]
Thus, the critical points and boundary points we need to evaluate are:
\[ x = -1, \quad x = 1, \quad x = 3 \]

### Step 5: Evaluate the function at these points
Now we will calculate \(f(x)\) at these points:

1. For \(x = -1\):
\[ f(-1) = 4(-1)^3 - 12(-1) = 4(-1) + 12 = -4 + 12 = 8 \]

2. For \(x = 1\):
\[ f(1) = 4(1)^3 - 12(1) = 4 - 12 = -8 \]

3. For \(x = 3\):
\[ f(3) = 4(3)^3 - 12(3) = 4(27) - 36 = 108 - 36 = 72 \]

### Step 6: Determine the minimum and maximum values
From our calculations, we have:
- \(f(-1) = 8\)
- \(f(1) = -8\)
- \(f(3) = 72\)

The minimum value is \(-8\) and the maximum value is \(72\).

### Step 7: Write the image of the interval
Thus, the image of the interval \([-1, 3]\) under the function \(f(x)\) is:
\[ [-8, 72] \]

### Final Answer
The image of the interval \([-1, 3]\) under the mapping specified by the function \(f(x) = 4x^3 - 12x\) is \([-8, 72]\).

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