In the curve `y = 2+12x-x^3`, find the critical points.

Options

A. (2,18) and (-2,-14)
B. (2,18) and (2,-14)
C. (-2,18) and (2,-14)
D. (-2,18) and (-2,14)

Answer

To find the critical points of the curve defined by the equation \( y = 2 + 12x - x^3 \), we need to follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( y \) with respect to \( x \).

\[ \frac{dy}{dx} = \frac{d}{dx}(2 + 12x - x^3) \]

Calculating the derivative, we get:

\[ \frac{dy}{dx} = 0 + 12 - 3x^2 \]

Thus,

\[ \frac{dy}{dx} = 12 - 3x^2 \]

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

\[ 12 - 3x^2 = 0 \]

### Step 3: Solve for \( x \)
Now, we solve the equation for \( x \):

\[ 3x^2 = 12 \]
\[ x^2 = 4 \]
\[ x = \pm 2 \]

### Step 4: Find the corresponding \( y \) values
Next, we substitute the \( x \) values back into the original equation to find the corresponding \( y \) values.

1. For \( x = 2 \):

\[ y = 2 + 12(2) - (2)^3 \]
\[ y = 2 + 24 - 8 = 18 \]

So one critical point is \( (2, 18) \).

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

\[ y = 2 + 12(-2) - (-2)^3 \]
\[ y = 2 - 24 + 8 = -14 \]

So the other critical point is \( (-2, -14) \).

### Step 5: List the critical points
The critical points of the curve are:

\[ (2, 18) \quad \text{and} \quad (-2, -14) \]

### Summary of the solution:
The critical points are \( (2, 18) \) and \( (-2, -14) \).

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