Find the absolute minimum value of `y=x^(2)-3x` in `0lexle2.`

Answer

To find the absolute minimum value of the function \( y = x^2 - 3x \) in the interval \( [0, 2] \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function to find the critical points.

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

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

\[ 2x - 3 = 0 \]

### Step 3: Solve for \( x \)
Solving for \( x \):

\[ 2x = 3 \implies x = \frac{3}{2} = 1.5 \]

### Step 4: Check if the critical point is in the interval
The critical point \( x = 1.5 \) lies within the interval \( [0, 2] \).

### Step 5: Evaluate the function at the endpoints and the critical point
Now we will evaluate the function \( y \) at the endpoints \( x = 0 \) and \( x = 2 \), and at the critical point \( x = 1.5 \).

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

2. For \( x = 2 \):
\[ y(2) = 2^2 - 3(2) = 4 - 6 = -2 \]

3. For \( x = 1.5 \):
\[ y\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) = \frac{9}{4} - \frac{9}{2} = \frac{9}{4} - \frac{18}{4} = -\frac{9}{4} = -2.25 \]

### Step 6: Compare the values
Now we compare the values obtained:

- \( y(0) = 0 \)
- \( y(2) = -2 \)
- \( y(1.5) = -2.25 \)

### Step 7: Determine the absolute minimum value
The absolute minimum value of \( y \) in the interval \( [0, 2] \) is:

\[ \text{Absolute Minimum} = -2.25 \text{ at } x = 1.5 \]

### Summary
The absolute minimum value of the function \( y = x^2 - 3x \) in the interval \( [0, 2] \) is \( -2.25 \).

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