Understanding Domain and Range of Linear and Quadratic Functions

Definitions

The domain of a function \( f(x) \) is the set of all possible input values (\( x \)) for which the function is defined. In simpler terms, it represents all the \( x \)-values that can go into the function.

The range of a function \( f(x) \) is the set of all possible output values (\( y \)) the function can produce. It depends on the behavior of the function and its domain.

1. Domain and Range of Linear Functions

A linear function is a function of the form \( f(x) = mx + b \), where \( m \) and \( b \) are constants. Linear functions are defined for all real numbers since there are no restrictions such as square roots or division by zero.

Key Points:

• The domain of a linear function is always \( (-\infty, \infty) \).
• The range of a linear function is also \( (-\infty, \infty) \) because the output can take any real value as \( x \) varies.

Examples of Linear Functions

Example 1: \( f(x) = 2x + 3 \)

• Domain: \( (-\infty, \infty) \)
• Range: \( (-\infty, \infty) \)

Example 2: \( f(x) = -x + 5 \)

• Domain: \( (-\infty, \infty) \)
• Range: \( (-\infty, \infty) \)

Word Problems for Linear Functions

Problem 1: A taxi company charges $2 per mile plus a flat fee of $5. Write the linear function for the cost, \( C(x) \), in terms of the number of miles, \( x \). Find the domain and range for a trip of up to 50 miles.

Solution: \( C(x) = 2x + 5 \)

• Domain: \( [0, 50] \) (since the trip is up to 50 miles)
• Range: \( [5, 105] \) (minimum cost is $5, maximum cost is \( 2(50) + 5 = 105 \))

Problem 2: A gym charges a membership fee of $20 and $10 for every hour spent with a personal trainer. Write the function for the total cost, \( C(x) \), where \( x \) is the number of hours spent with the trainer, and find the domain and range for up to 8 hours.

Solution: \( C(x) = 10x + 20 \)

• Domain: \( [0, 8] \)
• Range: \( [20, 100] \)

2. Domain and Range of Quadratic Functions

A quadratic function is a function of the form \( f(x) = ax^2 + bx + c \), where \( a \neq 0 \). The domain of a quadratic function is always \( (-\infty, \infty) \). The range, however, depends on whether the parabola opens upwards (\( a > 0 \)) or downwards (\( a < 0 \)).

Key Points:

• If the parabola opens upwards (\( a > 0 \)), the range is \( [\text{minimum value}, \infty) \).
• If the parabola opens downwards (\( a < 0 \)), the range is \( (-\infty, \text{maximum value}] \).

Examples of Quadratic Functions

Example 1: \( f(x) = 3x^2 - 5x + 2 \)

• Domain: \( (-\infty, \infty) \)
• Range: Find \( h = -\frac{b}{2a} = -\frac{-5}{2(3)} = \frac{5}{6} \).
  Substitute \( h \) into the function to find \( k = f\left(\frac{5}{6}\right) \):
  \( k = 3\left(\frac{5}{6}\right)^2 - 5\left(\frac{5}{6}\right) + 2 = -\frac{1}{12} \).
  \( \text{Range: } \left[-\frac{1}{12}, \infty\right) \).

Example 2: \( f(x) = -2x^2 + 4x + 1 \)

• Domain: \( (-\infty, \infty) \)
• Range: Find \( h = -\frac{b}{2a} = -\frac{4}{2(-2)} = 1 \).
  Substitute \( h \) into the function to find \( k = f(1) \):
  \( k = -2(1)^2 + 4(1) + 1 = 3 \).
  \( \text{Range: } (-\infty, 3] \).

Example 3: \( f(x) = x^2 + 6x + 5 \)

• Domain: \( (-\infty, \infty) \)
• Range: Find \( h = -\frac{b}{2a} = -\frac{6}{2(1)} = -3 \).
  Substitute \( h \) into the function to find \( k = f(-3) \):
  \( k = (-3)^2 + 6(-3) + 5 = -4 \).
  \( \text{Range: } [-4, \infty) \).

Constrained Optimization

Problem 1: A ball is thrown upward with an initial velocity of 20 m/s from a height of 5 m. Its height \( h(t) \) in meters after \( t \) seconds is given by \( h(t) = -5t^2 + 20t + 5 \). Find the maximum height of the ball.

Solution: \( t = -\frac{b}{2a} = -\frac{20}{2(-5)} = 2 \) seconds.
Substitute \( t = 2 \) into \( h(t) \):
\( h(2) = -5(2)^2 + 20(2) + 5 = 25 \).
The maximum height is 25 meters.

Problem 2: The profit from selling \( x \) units of a product is given by \( P(x) = -2x^2 + 8x + 5 \). Find the maximum profit and the number of units that produce it.

Solution: \( x = -\frac{b}{2a} = -\frac{8}{2(-2)} = 2 \) units.

A Fence Problem:

A farmer wants to build a rectangular pen for their livestock using 100 meters of fencing. One side of the pen will be against a barn, so only three sides need fencing. The area \( A(x) \) of the pen is modeled by the quadratic function:

\[ A(x) = 100x - 2x^2 \]

where \( x \) is the width of the pen in meters.

Questions

1. What is the domain of the function in this context?
2. What is the range of the function?
3. What is the maximum area the farmer can achieve, and what width corresponds to this maximum?

Solution

1. Domain:

The width \( x \) must be non-negative (\( x \geq 0 \)) and the total fencing used cannot exceed 100 meters. Since the pen uses \( x + 2x \) meters of fencing (the width and twice the length), we require:

\[ x + 2x \leq 100 \implies 3x \leq 100 \implies x \leq \frac{100}{3}. \]

Thus, the domain is: \[ [0, \frac{100}{3}] \quad \text{or approximately } [0, 33.33]. \]

2. Range:

The range depends on the maximum value of the quadratic function. First, find the width \( x \) that maximizes the area by using the vertex formula:

\[ x = -\frac{b}{2a} = -\frac{100}{2(-2)} = 25 \, \text{meters.} \]

Substitute \( x = 25 \) into the area function to find the maximum area:

\[ A(25) = 100(25) - 2(25)^2 = 2500 - 1250 = 1250 \, \text{square meters.} \]

The range is therefore: \[ [0, 1250] \, \text{square meters}. \]

3. Maximum Area and Corresponding Width:

The maximum area is \( 1250 \, \text{square meters} \), which occurs when the width is \( 25 \, \text{meters.} \)

Final Answer

• Domain: \( [0, \frac{100}{3}] \) or approximately \( [0, 33.33] \).
• Range: \( [0, 1250] \, \text{square meters.} \)
• Maximum Area: \( 1250 \, \text{square meters} \) at a width of \( 25 \, \text{meters.} \)