Find the minimum and maximum values of the funciton `y=x^3-3x^2+6`. Also find the values of x at which these occur.

Options

A. `min = 2` , `max = 4`
B. `min = 4` , `max = 6`
C. `min = 1` , `max = 2`
D. `min = 2` , `max = 6`

Explanation

To find the minimum and maximum values of the function \( y = x^3 - 3x^2 + 6 \), we will follow these steps:

### Step 1: Find the derivative of the function
To find the critical points where the minimum and maximum values may occur, we first need to calculate the derivative of the function.

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

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

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

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

\[ 3x(x - 2) = 0 \]

### Step 4: Solve for \( x \)
Now, we solve the factored equation for \( x \):

\[ 3x = 0 \quad \Rightarrow \quad x = 0 \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

Thus, the critical points are \( x = 0 \) and \( x = 2 \).

### Step 5: Determine the second derivative
To classify these critical points as minimum or maximum, we need to find the second derivative of the function.

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

### Step 6: Evaluate the second derivative at the critical points
Now we evaluate the second derivative at the critical points:

1. For \( x = 0 \):
\[ y''(0) = 6(0) - 6 = -6 \quad (\text{Negative, so } x = 0 \text{ is a maximum}) \]

2. For \( x = 2 \):
\[ y''(2) = 6(2) - 6 = 6 \quad (\text{Positive, so } x = 2 \text{ is a minimum}) \]

### Step 7: Find the values of \( y \) at the critical points
Now we find the corresponding \( y \) values at these critical points:

1. For \( x = 0 \):
\[ y(0) = 0^3 - 3(0^2) + 6 = 6 \]

2. For \( x = 2 \):
\[ y(2) = 2^3 - 3(2^2) + 6 = 8 - 12 + 6 = 2 \]

### Conclusion
- The maximum value of \( y \) is \( 6 \) at \( x = 0 \).
- The minimum value of \( y \) is \( 2 \) at \( x = 2 \).

### Summary
- Maximum value: \( y = 6 \) at \( x = 0 \)
- Minimum value: \( y = 2 \) at \( x = 2 \)