The curve `y =-x^3 + 3x^2 + 2x - 27` has the maximum slope at

Options

A. `x = -1`
B. `x = 0`
C. `x = 1`
D. `x = 2`

Answer

To find the point at which the curve \( y = -x^3 + 3x^2 + 2x - 27 \) has the maximum slope, we will follow these steps:

### Step 1: Differentiate the function
We start by finding the first derivative of the function \( y \) with respect to \( x \). This derivative represents the slope of the curve.

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

Using the power rule for differentiation:

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

### Step 2: Find the second derivative
To find the maximum slope, we need to differentiate the first derivative to find the second derivative.

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

Differentiating again:

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

### Step 3: Set the second derivative to zero
To find the critical points where the slope is maximized, we set the first derivative equal to zero.

\[ -6x + 6 = 0 \]

Solving for \( x \):

\[ -6x = -6 \implies x = 1 \]

### Step 4: Verify that it is a maximum
To confirm that this point corresponds to a maximum slope, we check the second derivative at \( x = 1 \):

\[ \frac{d^2y}{dx^2} = -6(1) + 6 = 0 \]

Since the second derivative is negative, it indicates that the slope is indeed at a maximum at this point.

### Conclusion
Thus, the curve has the maximum slope at \( x = 1 \).

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