The largest value of `2x^(3)-3x^(2)-12x+5` for `-2 le x le 2` occurs when

Options

A. `x=-2`
B. `x=-1`
C. `x=2`
D. `x=0`

Explanation

To find the largest value of the function \( f(x) = 2x^3 - 3x^2 - 12x + 5 \) for \( -2 \leq x \leq 2 \), we will evaluate the function at the endpoints of the interval and at any critical points within the interval.

### Step 1: Find the derivative of the function
First, we need to find the derivative of \( f(x) \):
\[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 5) \]
Using the power rule, we get:
\[ f'(x) = 6x^2 - 6x - 12 \]

### Step 2: Set the derivative to zero to find critical points
Next, we set the derivative equal to zero to find critical points:
\[ 6x^2 - 6x - 12 = 0 \]
Dividing the entire equation by 6 gives:
\[ x^2 - x - 2 = 0 \]
Now, we can factor the quadratic:
\[ (x - 2)(x + 1) = 0 \]
Thus, the critical points are:
\[ x = 2 \quad \text{and} \quad x = -1 \]

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

1. **At \( x = -2 \)**:
\[ f(-2) = 2(-2)^3 - 3(-2)^2 - 12(-2) + 5 \]
\[ = 2(-8) - 3(4) + 24 + 5 = -16 - 12 + 24 + 5 = 1 \]

2. **At \( x = -1 \)**:
\[ f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 5 \]
\[ = 2(-1) - 3(1) + 12 + 5 = -2 - 3 + 12 + 5 = 12 \]

3. **At \( x = 0 \)**:
\[ f(0) = 2(0)^3 - 3(0)^2 - 12(0) + 5 = 5 \]

4. **At \( x = 2 \)**:
\[ f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 5 \]
\[ = 2(8) - 3(4) - 24 + 5 = 16 - 12 - 24 + 5 = -15 \]

### Step 4: Compare the values
Now we compare the values we calculated:
- \( f(-2) = 1 \)
- \( f(-1) = 12 \)
- \( f(0) = 5 \)
- \( f(2) = -15 \)

The largest value occurs at \( x = -1 \) where \( f(-1) = 12 \).

### Conclusion
The largest value of \( f(x) \) for \( -2 \leq x \leq 2 \) occurs when \( x = -1 \).

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