Find the range of
`f(x)=x^(2)-2|x|+3`

Answer

To find the range of the function \( f(x) = x^2 - 2|x| + 3 \), we will analyze it by considering the two cases for \( |x| \).

### Step 1: Define the function based on the absolute value
The function can be defined in two parts based on the value of \( x \):

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

2. For \( x < 0 \):
\[ f(x) = x^2 + 2x + 3 \]

### Step 2: Analyze the first case \( f(x) = x^2 - 2x + 3 \)
This is a quadratic function. To find its range, we will calculate the vertex of the parabola.

The vertex \( x \) coordinate is given by:
\[ x = -\frac{b}{2a} = -\frac{-2}{2 \cdot 1} = 1 \]

Now, we substitute \( x = 1 \) into the function to find the minimum value:
\[ f(1) = 1^2 - 2(1) + 3 = 1 - 2 + 3 = 2 \]

Since the parabola opens upwards (as the coefficient of \( x^2 \) is positive), the range for \( x \geq 0 \) is:
\[ [2, \infty) \]

### Step 3: Analyze the second case \( f(x) = x^2 + 2x + 3 \)
Again, this is a quadratic function. We will find the vertex here as well.

The vertex \( x \) coordinate is given by:
\[ x = -\frac{b}{2a} = -\frac{2}{2 \cdot 1} = -1 \]

Now, we substitute \( x = -1 \) into the function to find the minimum value:
\[ f(-1) = (-1)^2 + 2(-1) + 3 = 1 - 2 + 3 = 2 \]

Since this parabola also opens upwards, the range for \( x < 0 \) is:
\[ [2, \infty) \]

### Step 4: Combine the ranges
Both cases yield the same minimum value of 2, and since both ranges are \( [2, \infty) \), the overall range of the function \( f(x) \) is:
\[ \text{Range of } f(x) = [2, \infty) \]

### Final Answer
The range of the function \( f(x) = x^2 - 2|x| + 3 \) is:
\[ \boxed{[2, \infty)} \]