Solution:

Given \( f(x) = -2x^2 + 4x + 1 \).

1. Understanding the Function:
The given quadratic function is \( f(x) = -2x^2 + 4x + 1 \). The function is in standard quadratic form \( ax^2 + bx + c \), where:

• \( a = -2 \)
• \( b = 4 \)
• \( c = 1 \)

2. Finding the Domain:
The domain of any quadratic function is all real numbers, as there are no restrictions preventing substitution of any real value into \( x \).
In interval notation, the domain is: \[ \text{Domain} = (-\infty, \infty) \]

3. Finding the Range:
To find the range, consider the direction of the parabola and the vertex (maximum or minimum point). The coefficient of \( x^2 \) (\( a \)) is negative (\( a = -2 \)), meaning the parabola opens downward, giving a maximum point.

4. Finding the Vertex:
The vertex formula for a quadratic function \( ax^2 + bx + c \) is: \[ x = -\frac{b}{2a}. \]
Substitute \( a = -2 \) and \( b = 4 \) into the formula: \[ x = -\frac{4}{2 \times -2} = 1 \]
This gives the x-coordinate of the vertex as \( x = 1 \).

5. Finding the Maximum Value:
Substitute \( x = 1 \) back into the function to find the maximum value (y-coordinate of the vertex): \[ f(1) = -2(1)^2 + 4 \times 1 + 1 = -2 + 4 + 1 \]= 3
Thus, the maximum value of \( f(x) \) is 3.

6. Conclusion - Range:
Since the parabola opens downward and has a maximum at 3, the range of the function is: \[ \text{Range} = (-\infty, 3] \]