`f(x)=2x^(3)-3x^(2)-36x+24` has maximum value at x =

Options

A. `-2`
B. `-3`
C. 2
D. 3

Answer

To find the maximum value of the function \( f(x) = 2x^3 - 3x^2 - 36x + 24 \), we will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)
We start by differentiating the function with respect to \( x \).

\[ f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(3x^2) - \frac{d}{dx}(36x) + \frac{d}{dx}(24) \]
\[ f'(x) = 6x^2 - 6x - 36 \]

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

\[ 6x^2 - 6x - 36 = 0 \]

Dividing the entire equation by 6 gives:

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

### Step 3: Factor the quadratic equation
Now we factor the quadratic equation:

\[ x^2 - 3x + 2x - 6 = 0 \]
\[ (x - 3)(x + 2) = 0 \]

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

\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]

### Step 5: Find the second derivative \( f''(x) \)
Now we need to determine whether these critical points correspond to a maximum or minimum by finding the second derivative.

\[ f''(x) = \frac{d}{dx}(6x^2 - 6x - 36) = 12x - 6 \]

### Step 6: Evaluate the second derivative at the critical points
Now we evaluate the second derivative at each critical point.

1. For \( x = 3 \):

\[ f''(3) = 12(3) - 6 = 36 - 6 = 30 \quad (\text{which is } > 0) \]

This indicates that \( x = 3 \) is a local minimum.

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

\[ f''(-2) = 12(-2) - 6 = -24 - 6 = -30 \quad (\text{which is } < 0) \]

This indicates that \( x = -2 \) is a local maximum.

### Conclusion
Thus, the function \( f(x) \) has a maximum value at \( x = -2 \).