This guide provides a detailed answer key and explanation for evaluating piecewise functions, a crucial concept in algebra and pre-calculus. We'll cover various examples, common pitfalls, and strategies to master this topic. Understanding piecewise functions is essential for progressing to more advanced mathematical concepts.
What are Piecewise Functions?
A piecewise function is a function defined by multiple sub-functions, each applicable over a specified interval of the domain. Essentially, it's a function "broken" into pieces. Each piece is defined by a specific rule or equation, and the appropriate piece is selected based on the input value (x).
The general format looks like this:
f(x) = {
g(x), if a ≤ x < b
h(x), if b ≤ x ≤ c
k(x), if x > c
}
Where g(x), h(x), and k(x) are different functions, and a, b, and c are the boundaries defining the intervals.
Common Mistakes to Avoid
- Incorrect Interval Selection: The most frequent mistake is choosing the wrong sub-function based on the input value. Carefully examine the inequality conditions defining each interval.
- Ignoring Endpoint Behavior: Pay close attention to whether endpoints are included (≤ or ≥) or excluded (< or >) in each interval. This significantly impacts the function's value at those points.
- Order of Operations: Remember to follow the standard order of operations (PEMDAS/BODMAS) when evaluating each sub-function.
Example Problems and Solutions
Let's walk through some example problems, providing detailed solutions and explanations.
Example 1:
Evaluate the piecewise function:
f(x) = {
2x + 1, if x < 2
x² - 3, if x ≥ 2
}
for the following values of x:
- x = 1: Since 1 < 2, we use the first sub-function: f(1) = 2(1) + 1 = 3
- x = 2: Since 2 ≥ 2, we use the second sub-function: f(2) = 2² - 3 = 1
- x = 3: Since 3 ≥ 2, we use the second sub-function: f(3) = 3² - 3 = 6
Example 2:
Evaluate the piecewise function:
g(x) = {
|x| , if x ≤ 0
√x, if x > 0
}
for the following values of x:
- x = -3: Since -3 ≤ 0, we use the first sub-function: g(-3) = |-3| = 3
- x = 0: Since 0 ≤ 0, we use the first sub-function: g(0) = |0| = 0
- x = 4: Since 4 > 0, we use the second sub-function: g(4) = √4 = 2
Example 3: A More Complex Example
Evaluate:
h(x) = {
x³ - 2x + 5, if x < -1
2x² + 1, if -1 ≤ x ≤ 1
1/(x-1) if x >1
}
for x = -2, x = 0 and x = 2
- x = -2: Since -2 < -1, h(-2) = (-2)³ - 2(-2) + 5 = -8 + 4 + 5 = 1
- x = 0: Since -1 ≤ 0 ≤ 1, h(0) = 2(0)² + 1 = 1
- x = 2: Since 2 > 1, h(2) = 1/(2 - 1) = 1
Practice Problems
To solidify your understanding, try evaluating these piecewise functions for the given x values: (Solutions are provided at the end of the document – try them before checking!)
Problem 1:
f(x) = {
3x - 5, if x < 0
x² + 2, if x ≥ 0
}
Evaluate f(-2), f(0), f(3).
Problem 2:
g(x) = {
-x + 1, if x ≤ 1
2x - 1, if x > 1
}
Evaluate g(0), g(1), g(2).
Conclusion
Mastering piecewise functions involves careful attention to detail and a thorough understanding of interval notation. Practice is key to developing proficiency. Remember to always check the conditions that define each interval before applying the corresponding function. By consistently practicing and reviewing the examples, you can confidently tackle any piecewise function evaluation problem.
Solutions to Practice Problems:
Problem 1: f(-2) = -11, f(0) = 2, f(3) = 11
Problem 2: g(0) = 1, g(1) = 0, g(2) = 3
This comprehensive guide provides a solid foundation for understanding and evaluating piecewise functions. Remember to consult your textbook or instructor for further clarification if needed.
