Problem 45 - Precalculus Mathematics for Calculus

Question

Find the domain and range of the function. $$f(x)=-x^{2}+4 x-3$$

Step-by-Step Solution

Answer

Domain: all real numbers; Range: \((-\infty, 1]\).

Step 1: Determine the Domain

The domain of a function refers to all the possible input values (x-values) that the function can accept. Since the given function \( f(x) = -x^{2} + 4x - 3 \) is a quadratic polynomial, it is defined for all real numbers. Therefore, the domain is all real numbers \( \mathbb{R} \).

Step 2: Determine the Vertex

To find the range, begin by finding the vertex of the quadratic function. The vertex form of a quadratic function \( ax^2 + bx + c \) provides the vertex \( (h, k) \), where \( h = -\frac{b}{2a} \). For the function \(-x^{2} + 4x - 3\), \( a = -1 \), \( b = 4 \), and \( c = -3 \). Thus, \( h = -\frac{4}{2(-1)} = 2 \).

Step 3: Evaluate the Function at the Vertex

Substitute \( x = 2 \) back into the function to find the y-coordinate of the vertex. \( f(2) = -(2)^2 + 4(2) - 3 = -4 + 8 - 3 = 1 \). Thus, the vertex is \((2, 1)\).

Step 4: Determine the Range

Since the coefficient of \( x^2 \) in the function \( f(x) = -x^{2} + 4x - 3 \) is negative, the parabola opens downwards. The maximum value achieved by the function is at the vertex \( (2, 1) \), which gives the maximum y-value as 1. Therefore, the range of the function is \(( -\infty, 1 ]\).

Key Concepts

Understanding the Domain of a Function

The domain of a function is essentially the set of all possible input values, commonly referred to as the x-values, that the function can accept without any issues. For linear and quadratic functions, finding the domain can be straightforward. This is because these types of functions are typically defined for all real numbers. In the case of our quadratic function, \[ f(x) = -x^2 + 4x - 3 \]we are dealing with a polynomial. Polynomials, by nature, do not have any restrictions on their x-values. There are no square roots or divisions by zero to complicate things, so every real number is a valid input. Therefore, the domain of our given function is all real numbers, sometimes expressed as

• "the set of all real numbers"
• or \( \mathbb{R} \).

Exploring the Range of a Function

To find a function's range, we look at all possible output values, or y-values, that the function can produce. For quadratic functions, which form parabolas, the range is influenced by the orientation of the parabola. In our function,\[ f(x) = -x^2 + 4x - 3 \]the leading coefficient (the number in front of \( x^2 \)), is -1, indicating the parabola opens downwards. This means there is a maximum output value, which is found at the vertex of the parabola. After calculating the vertex of this quadratic, we found it to be

• \((2,1)\).

The vertex's y-coordinate, 1, is the highest value the function can reach. Therefore, the range of our function is all y-values up to and including this maximum, or symbolically,

• \(( -\infty, 1 ]\).

This tells us the function can produce values greater than negative infinity up to 1, the highest point.

The Vertex of a Parabola

The vertex of a parabola is a crucial part of understanding quadratic functions. It is the tip or the turning point of the parabola, where the curve changes direction. In the vertex form of a quadratic function \[ ax^2 + bx + c \],the vertex can be determined using the formula

• \( h = -\frac{b}{2a} \).

For our specific function, \[ f(x) = -x^2 + 4x - 3 \],we substitute \( a = -1 \) and \( b = 4 \) into the formula, yielding \( h = 2 \). To complete the picture, substitute \( h \) back into the function to find \( k \), the y-coordinate of the vertex:

• \( f(2) = -(2)^2 + 4(2) - 3 = 1 \).

Resulting in the vertex

• \((2, 1)\).

Remember, the vertex represents a peak (maximum) for downward-opening parabolas like ours. This point offers vital information about the behavior of the quadratic function.