The largest value of `2x^3 - 3x^2 - 12x + 5` for ` - 2 le x le 4` occurs at x =

Options

A. `-2`
B. `-1`
C. 2
D. 4

Answer

To find the largest value of the function \( f(x) = 2x^3 - 3x^2 - 12x + 5 \) for the interval \( -2 \leq x \leq 4 \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 5) \]

Using the power rule, we get:

\[ f'(x) = 6x^2 - 6x - 12 \]

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

\[ 6x^2 - 6x - 12 = 0 \]

Dividing the entire equation by 6 simplifies it:

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

### Step 3: Factor the quadratic equation
Next, we factor the quadratic equation:

\[ (x - 2)(x + 1) = 0 \]

### Step 4: Solve for \( x \)
Setting each factor to zero gives us the critical points:

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

### Step 5: Evaluate the function at the critical points and endpoints
Now we will evaluate the function at the critical points \( x = -1 \) and \( x = 2 \), as well as at the endpoints of the interval \( x = -2 \) and \( x = 4 \).

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

4. **At \( x = 4 \)**:
\[ f(4) = 2(4)^3 - 3(4)^2 - 12(4) + 5 = 2(64) - 3(16) - 48 + 5 = 128 - 48 - 48 + 5 = 37 \]

### Step 6: Compare the values
Now we compare the values obtained:

- \( f(-2) = 1 \)
- \( f(-1) = 12 \)
- \( f(2) = -15 \)
- \( f(4) = 37 \)

The largest value occurs at \( x = 4 \) with \( f(4) = 37 \).

### Conclusion
The largest value of \( 2x^3 - 3x^2 - 12x + 5 \) for \( -2 \leq x \leq 4 \) occurs at \( x = 4 \).

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