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: Find the first derivative
The slope of the curve is given by the first derivative \( \frac{dy}{dx} \).

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

Calculating the derivative term by term:

- The derivative of \( -x^3 \) is \( -3x^2 \).
- The derivative of \( 3x^2 \) is \( 6x \).
- The derivative of \( 9x \) is \( 9 \).
- The derivative of the constant \( -27 \) is \( 0 \).

Thus, we have:

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

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

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

Dividing through by -3 gives:

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

### Step 3: Factor the quadratic equation
We can factor the quadratic:

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

Setting each factor to zero gives:

\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

### Step 4: Determine which critical point gives a maximum slope
Next, we need to check the second derivative to determine whether these points correspond to a maximum or minimum slope.

### Step 5: Find the second derivative
The second derivative \( \frac{d^2y}{dx^2} \) is found by differentiating the first derivative:

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

Calculating the derivative:

- The derivative of \( -3x^2 \) is \( -6x \).
- The derivative of \( 6x \) is \( 6 \).
- The derivative of the constant \( 9 \) is \( 0 \).

Thus, we have:

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

### Step 6: Evaluate the second derivative at critical points
Now we evaluate the second derivative at \( x = 3 \) and \( x = -1 \):

1. For \( x = 3 \):

\[ \frac{d^2y}{dx^2} = -6(3) + 6 = -18 + 6 = -12 \quad (\text{negative, so maximum}) \]

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

\[ \frac{d^2y}{dx^2} = -6(-1) + 6 = 6 + 6 = 12 \quad (\text{positive, so minimum}) \]

### Step 7: Find the maximum slope
Since \( x = 3 \) gives a maximum, we substitute \( x = 3 \) back into the first derivative to find the maximum slope:

\[ \frac{dy}{dx} = -3(3^2) + 6(3) + 9 \]
\[ = -3(9) + 18 + 9 \]
\[ = -27 + 18 + 9 = 0 \]

### Conclusion
The maximum slope of the curve \( y = -x^3 + 3x^2 + 9x - 27 \) occurs at \( x = 3 \) and the maximum slope is **12**.

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