The maximum slope of curve y `=-x^(3)+3x^(2)+9x-27` is

Options

A. 0
B. 12
C. 16
D. 32

Explanation

To find the maximum slope of the curve given by the equation \( y = -x^3 + 3x^2 + 9x - 27 \), we will follow these steps:

### Step 1: Differentiate the function to find the slope
We start by differentiating the function \( y \) with respect to \( x \) to find the slope \( \frac{dy}{dx} \).

\[ \frac{dy}{dx} = \frac{d}{dx}(-x^3 + 3x^2 + 9x - 27) \]

Using the power rule of differentiation:

\[ \frac{dy}{dx} = -3x^2 + 6x + 9 \]

### Step 2: Find the critical points
To find the maximum slope, we need to find the critical points by setting the first derivative equal to zero.

\[ -3x^2 + 6x + 9 = 0 \]

Dividing the entire equation by -3 gives:

\[ x^2 - 2x - 3 = 0 \]

Now, we can factor this quadratic equation:

\[ (x - 3)(x + 1) = 0 \]

Thus, the critical points are:

\[ x = 3 \quad \text{and} \quad x = -1 \]

### Step 3: Determine the nature of the critical points
To determine whether these critical points correspond to a maximum or minimum, we differentiate the slope function again to find the second derivative.

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(-3x^2 + 6x + 9) \]

Calculating the second derivative:

\[ \frac{d^2y}{dx^2} = -6x + 6 \]

Now we evaluate the second derivative at the critical points:

1. For \( x = 3 \):

\[ \frac{d^2y}{dx^2} = -6(3) + 6 = -18 < 0 \quad \text{(indicating a maximum)} \]

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

\[ \frac{d^2y}{dx^2} = -6(-1) + 6 = 12 > 0 \quad \text{(indicating a minimum)} \]

### Step 4: Calculate the maximum slope
Since \( x = 3 \) is the point where the slope is maximized, we substitute \( x = 3 \) back into the first derivative to find the maximum slope.

\[ \frac{dy}{dx} \bigg|_{x=3} = -3(3^2) + 6(3) + 9 \]

Calculating this:

\[ = -3(9) + 18 + 9 \]
\[ = -27 + 18 + 9 \]
\[ = 0 \]

### Conclusion
The maximum slope of the curve \( y = -x^3 + 3x^2 + 9x - 27 \) occurs at \( x = 3 \) and the value of the maximum slope is:

\[ \text{Maximum slope} = 0 \]