If `y=x^(3)-3x`. Find the maximum value of y.

Options

A. 8
B. 6
C. 4
D. 2

Answer

To find the maximum value of the function \( y = x^3 - 3x \), we will follow these steps:

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

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

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

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

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

\[ 3x^2 = 3 \implies x^2 = 1 \implies x = 1 \quad \text{or} \quad x = -1 \]

### Step 4: Determine the nature of the critical points
To determine whether these critical points are maxima or minima, we will use the second derivative test. We first find the second derivative.

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

### Step 5: Evaluate the second derivative at the critical points
Now we evaluate the second derivative at the critical points \( x = 1 \) and \( x = -1 \).

1. For \( x = 1 \):
\[ \frac{d^2y}{dx^2} \bigg|_{x=1} = 6(1) = 6 \quad (\text{positive, indicates a local minimum}) \]

2. For \( x = -1 \):
\[ \frac{d^2y}{dx^2} \bigg|_{x=-1} = 6(-1) = -6 \quad (\text{negative, indicates a local maximum}) \]

### Step 6: Find the maximum value of \( y \)
Since \( x = -1 \) is a local maximum, we will substitute \( x = -1 \) back into the original function to find the maximum value of \( y \).

\[ y = (-1)^3 - 3(-1) = -1 + 3 = 2 \]

### Final Answer
The maximum value of \( y \) is \( 2 \).

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