Find the absolute maximum and minimum values of each of the following in the given intervals :
`f(x)=(x-1)^(2)+3" in "[-3, 1]`

Answer

To find the absolute maximum and minimum values of the function \( f(x) = (x-1)^2 + 3 \) in the interval \([-3, 1]\), we will follow these steps:

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

\[ f'(x) = \frac{d}{dx}((x-1)^2 + 3) \]

Using the power rule:

\[ f'(x) = 2(x-1) \cdot \frac{d}{dx}(x-1) = 2(x-1) \cdot 1 = 2(x-1) \]

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

\[ 2(x-1) = 0 \]

Solving for \(x\):

\[ x - 1 = 0 \implies x = 1 \]

### Step 3: Evaluate the function at the critical points and endpoints
Now we need to evaluate the function \(f(x)\) at the critical point \(x = 1\) and at the endpoints of the interval, \(x = -3\) and \(x = 1\).

1. **Evaluate at \(x = -3\)**:

\[ f(-3) = (-3 - 1)^2 + 3 = (-4)^2 + 3 = 16 + 3 = 19 \]

2. **Evaluate at \(x = 1\)**:

\[ f(1) = (1 - 1)^2 + 3 = 0 + 3 = 3 \]

### Step 4: Compare the values
Now we compare the values obtained:

- \(f(-3) = 19\)
- \(f(1) = 3\)

### Conclusion
The absolute maximum value of \(f(x)\) on the interval \([-3, 1]\) is \(19\) at \(x = -3\), and the absolute minimum value is \(3\) at \(x = 1\).

**Final Answer:**
- Absolute Maximum: \(19\) at \(x = -3\)
- Absolute Minimum: \(3\) at \(x = 1\)