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

Answer

To find the range of the function \( f(x) = |x^2 - 3x + 2| \), we will follow these steps:

### Step 1: Find the roots of the quadratic function inside the absolute value.

The expression inside the absolute value is a quadratic function \( x^2 - 3x + 2 \). To find its roots, we can set it equal to zero:

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

Factoring the quadratic:

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

Thus, the roots are \( x = 1 \) and \( x = 2 \).

### Step 2: Determine the intervals for the quadratic function.

The quadratic \( x^2 - 3x + 2 \) opens upwards (since the coefficient of \( x^2 \) is positive). We can analyze the sign of the quadratic in the intervals defined by the roots:

1. For \( x < 1 \): Choose \( x = 0 \):
\[ f(0) = 0^2 - 3(0) + 2 = 2 \quad (\text{positive}) \]

2. For \( 1 < x < 2 \): Choose \( x = 1.5 \):
\[ f(1.5) = (1.5)^2 - 3(1.5) + 2 = 2.25 - 4.5 + 2 = -0.25 \quad (\text{negative}) \]

3. For \( x > 2 \): Choose \( x = 3 \):
\[ f(3) = 3^2 - 3(3) + 2 = 9 - 9 + 2 = 2 \quad (\text{positive}) \]

### Step 3: Analyze the absolute value.

Now, we can rewrite \( f(x) \) based on the intervals we found:

- For \( x < 1 \) and \( x > 2 \), \( f(x) = x^2 - 3x + 2 \).
- For \( 1 < x < 2 \), \( f(x) = -(x^2 - 3x + 2) = -x^2 + 3x - 2 \).

### Step 4: Find the maximum and minimum values.

1. **For \( x < 1 \) and \( x > 2 \)**:
- At \( x = 1 \):
\[ f(1) = |1^2 - 3(1) + 2| = |0| = 0 \]
- At \( x = 2 \):
\[ f(2) = |2^2 - 3(2) + 2| = |0| = 0 \]

As \( x \to -\infty \) or \( x \to \infty \), \( f(x) \to \infty \).

2. **For \( 1 < x < 2 \)**:
- The function \( f(x) = -x^2 + 3x - 2 \) is a downward-opening parabola. To find its vertex:
\[ x = -\frac{b}{2a} = -\frac{3}{2(-1)} = \frac{3}{2} \]
- Evaluate \( f\left(\frac{3}{2}\right) \):
\[ f\left(\frac{3}{2}\right) = -\left(\frac{3}{2}\right)^2 + 3\left(\frac{3}{2}\right) - 2 = -\frac{9}{4} + \frac{9}{2} - 2 = -\frac{9}{4} + \frac{18}{4} - \frac{8}{4} = \frac{1}{4} \]

### Step 5: Conclusion on the range.

From our analysis:
- The minimum value of \( f(x) \) is \( 0 \) (at \( x = 1 \) and \( x = 2 \)).
- The maximum value of \( f(x) \) in the interval \( 1 < x < 2 \) is \( \frac{1}{4} \).

Thus, the range of \( f(x) = |x^2 - 3x + 2| \) is:

\[ \text{Range} = [0, \infty) \]