1-3 Additional Practice Piecewise Defined Functions

Mastering Piecewise Defined Functions: Three More Practice Problems

Piecewise defined functions, those intriguing mathematical creatures composed of distinct functions over different intervals, often present a challenge to students. Understanding them is key to succeeding in calculus and beyond. This article provides three additional practice problems focusing on different aspects of piecewise functions, complete with detailed solutions and explanations. We'll delve into evaluating functions, graphing them, and finding their domains and ranges, solidifying your understanding of this crucial concept.

Problem 1: Evaluating a Piecewise Function at Different Points

Let's consider the following piecewise function:

f(x) = {
    x² + 1,  if x < -1
    2x,      if -1 ≤ x ≤ 2
    x - 3,    if x > 2
}

Find the values of f(-2), f(-1), f(0), f(2), and f(3).

Solution:

This problem tests your ability to correctly identify which sub-function to use based on the input value.

• f(-2): Since -2 < -1, we use the first sub-function: f(-2) = (-2)² + 1 = 4 + 1 = 5

• f(-1): Since -1 satisfies -1 ≤ x ≤ 2, we use the second sub-function: f(-1) = 2(-1) = -2

• f(0): Since 0 satisfies -1 ≤ x ≤ 2, we use the second sub-function: f(0) = 2(0) = 0

• f(2): Since 2 satisfies -1 ≤ x ≤ 2, we use the second sub-function: f(2) = 2(2) = 4

• f(3): Since 3 > 2, we use the third sub-function: f(3) = 3 - 3 = 0

Key takeaway: Always carefully examine the conditions defining each interval to select the appropriate sub-function for evaluation.

Problem 2: Graphing a Piecewise Function and Determining its Domain and Range

Consider the piecewise function:

g(x) = {
    |x|,       if x < 1
    √(x - 1), if x ≥ 1
}

Graph the function g(x) and determine its domain and range.

Solution:

This problem requires us to graph each sub-function within its specified interval.

Graphing:

• For x < 1: The function is g(x) = |x|. This is the absolute value function, forming a V-shape with its vertex at (0,0). However, we only graph the portion for x < 1. There will be an open circle at (1,1) because x=1 is not included in this part of the function.

• For x ≥ 1: The function is g(x) = √(x - 1). This is a square root function, starting at (1,0) and increasing as x increases. There will be a closed circle at (1,0) because x=1 is included in this part.

By combining these two parts, we obtain the complete graph of g(x). The graph will show a V-shape for x<1 and a smoothly increasing square root curve for x≥1.

Domain: The domain is the set of all possible x-values. For g(x), x can be any real number. Therefore, the domain is (-∞, ∞).

Range: The range is the set of all possible y-values. Looking at the graph, the y-values start at 0 and extend to infinity. Therefore, the range is [0, ∞).

Key takeaway: Graphing piecewise functions involves graphing each sub-function separately within its defined interval, paying close attention to open and closed circles to indicate whether the endpoints are included or not. Carefully analyzing the graph helps to determine the domain and range accurately.

Problem 3: A Piecewise Function with a More Complex Definition

Let's analyze this piecewise function:

h(x) = {
    -x + 2,   if x ≤ 0
    x² - 2x, if 0 < x < 3
    5,       if x ≥ 3
}

1. Find the values of h(-1), h(0), h(1), h(3), and h(4).

2. Determine the intervals where h(x) is increasing and decreasing.

3. Identify any local maxima or minima.

Solution:

This problem combines evaluation, analysis of increasing/decreasing intervals, and identification of extrema – crucial aspects of function analysis.

1. Evaluating h(x):

• h(-1): Since -1 ≤ 0, h(-1) = -(-1) + 2 = 3
• h(0): Since x ≤ 0, h(0) = -(0) + 2 = 2
• h(1): Since 0 < 1 < 3, h(1) = (1)² - 2(1) = -1
• h(3): Since x ≥ 3, h(3) = 5
• h(4): Since x ≥ 3, h(4) = 5

2. Intervals of Increase and Decrease:

To determine where h(x) is increasing or decreasing, we analyze each sub-function individually within its defined interval:

• x ≤ 0: h(x) = -x + 2. This is a linear function with a negative slope, so it is decreasing on (-∞, 0].

• 0 < x < 3: h(x) = x² - 2x. This is a parabola. To find its vertex (which determines the minimum), we can complete the square or use the formula x = -b/2a. Here, a = 1 and b = -2, so x = 2/2 = 1. The parabola opens upwards, meaning it is decreasing on (0, 1) and increasing on (1, 3).

• x ≥ 3: h(x) = 5. This is a constant function, neither increasing nor decreasing.

3. Local Maxima and Minima:

• Local Minimum: The parabola has a minimum at x = 1, with a value of h(1) = -1.

• Local Maximum: There is no local maximum as the function increases to 5 at x=3. The function is constantly at y=5 for x>=3.

Key takeaway: Analyzing the behavior of a piecewise function often requires examining each piece separately. Understanding how to find intervals of increase/decrease and identify local extrema are important skills that are often tested in calculus.

This comprehensive exploration of three diverse piecewise function problems equips you with the tools to tackle similar challenges. Remember to always carefully consider the conditions defining each interval, and practice visualizing the graphs to enhance your understanding. Consistent practice is the key to mastering these complex yet crucial mathematical functions.