Maximum value of` f(x)=2x^(3)-3x^(2)-12x+6` is

Options

A. `12`
B. `11`
C. `10`
D. `13`

Answer

To find the maximum value of the function \( f(x) = 2x^3 - 3x^2 - 12x + 6 \), we will follow these steps:

### Step 1: Differentiate the function
We start by finding the first derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 6) \]

Using the power rule, we differentiate each term:

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

### Step 2: Set the first derivative to zero
To find the critical points, we set the first 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 critical points
Setting each factor equal 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: Determine the nature of critical points
To determine whether these critical points are maxima or minima, we need to find the second derivative of \( f(x) \).

\[ f''(x) = \frac{d}{dx}(6x^2 - 6x - 12) = 12x - 6 \]

Now we evaluate the second derivative at the critical points:

1. For \( x = 2 \):

\[ f''(2) = 12(2) - 6 = 24 - 6 = 18 \quad (\text{positive, so minimum}) \]

2. For \( x = -1 \):

\[ f''(-1) = 12(-1) - 6 = -12 - 6 = -18 \quad (\text{negative, so maximum}) \]

### Step 6: Calculate the maximum value
Now we find the maximum value by substituting \( x = -1 \) back into the original function \( f(x) \):

\[ f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 6 \]
\[ = 2(-1) - 3(1) + 12 + 6 \]
\[ = -2 - 3 + 12 + 6 \]
\[ = 13 \]

### Conclusion
The maximum value of \( f(x) \) is \( \boxed{13} \).

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