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

Options

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

Answer

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 of the curve, we need to differentiate the function with respect to \( x \).

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

### Step 2: Find the critical points
Next, we need to find the critical points where the slope is either maximum or minimum. We do this by setting the first derivative equal to zero.

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

### Step 3: Solve the quadratic equation
We can simplify the equation by dividing by -3:

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

Now, we can factor the quadratic:

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

Thus, the critical points are:

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

### Step 4: Determine the nature of critical points
To determine whether these points are maxima or minima, we need to find the second derivative of the function.

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

Now, we will evaluate the second derivative at the critical points:

1. For \( x = 3 \):
\[ y''(3) = -6(3) + 6 = -18 + 6 = -12 \quad (\text{which is less than 0, indicating a maximum}) \]

2. For \( x = -1 \):
\[ y''(-1) = -6(-1) + 6 = 6 + 6 = 12 \quad (\text{which is greater than 0, indicating a minimum}) \]

### Step 5: Find the maximum slope
Since \( x = 3 \) is a point of maximum, we will substitute \( x = 3 \) back into the first derivative to find the maximum slope.

\[ y'(3) = -3(3^2) + 6(3) + 9 \]
\[ = -3(9) + 18 + 9 \]
\[ = -27 + 18 + 9 = 0 \]

Now, we also need to check the other critical point \( x = -1 \):

\[ y'(-1) = -3(-1)^2 + 6(-1) + 9 \]
\[ = -3(1) - 6 + 9 \]
\[ = -3 - 6 + 9 = 0 \]

### Step 6: Evaluate the maximum slope
Now, we will evaluate the slope at the critical points \( x = 1 \) (the midpoint between the two critical points) to find the maximum slope.

\[ y'(1) = -3(1^2) + 6(1) + 9 \]
\[ = -3 + 6 + 9 = 12 \]

Thus, the maximum slope of the curve is:

\[ \boxed{12} \]