Piecewise functions, those mathematical chameleons that change their behavior depending on the input, can seem daunting at first. But with a systematic approach and a solid understanding of their structure, evaluating them becomes straightforward. This guide will walk you through the process, providing clear explanations and example problems with a complete answer key. We'll cover various scenarios, including those involving absolute values and inequalities, ensuring you're well-equipped to tackle any piecewise function problem.
Understanding Piecewise Functions
A piecewise function is defined by multiple sub-functions, each applicable over a specific interval of the domain. The key is identifying which sub-function to use based on the input value (often denoted by 'x'). The general form looks like this:
f(x) = { f₁(x), if x ∈ I₁
f₂(x), if x ∈ I₂
f₃(x), if x ∈ I₃
...
}
Where:
f₁(x)
,f₂(x)
,f₃(x)
... are different functions.I₁
,I₂
,I₃
... are intervals defining the domains for each sub-function. These intervals can be open, closed, or half-open.
Step-by-Step Evaluation Process
To evaluate a piecewise function at a specific point, follow these steps:
- Identify the Interval: Determine which interval the input value falls into.
- Select the Correct Sub-function: Choose the sub-function associated with that interval.
- Substitute and Evaluate: Substitute the input value into the chosen sub-function and calculate the result.
Example Problems with Answer Key
Let's work through some examples to solidify your understanding.
Example 1:
Evaluate the following piecewise function at x = 2, x = 0, and x = -3:
f(x) = { x² + 1, if x > 1
2x, if -1 ≤ x ≤ 1
x - 3, if x < -1
}
Solution:
- f(2): Since 2 > 1, we use the first sub-function: f(2) = 2² + 1 = 5
- f(0): Since -1 ≤ 0 ≤ 1, we use the second sub-function: f(0) = 2(0) = 0
- f(-3): Since -3 < -1, we use the third sub-function: f(-3) = -3 - 3 = -6
Example 2 (Involving Absolute Value):
Evaluate the following piecewise function at x = 3, x = 0, and x = -2:
g(x) = { |x| + 2, if x ≥ 0
-x, if x < 0
}
Solution:
- g(3): Since 3 ≥ 0, we use the first sub-function: g(3) = |3| + 2 = 5
- g(0): Since 0 ≥ 0, we use the first sub-function: g(0) = |0| + 2 = 2
- g(-2): Since -2 < 0, we use the second sub-function: g(-2) = -(-2) = 2
Example 3 (More Complex Intervals):
Evaluate h(x) at x = 1, x = 4, and x = 7:
h(x) = { x/2, if 0 < x ≤ 2
x -1, if 2 < x ≤ 5
√(x-5), if x > 5
}
Solution:
- h(1): Since 0 < 1 ≤ 2, h(1) = 1/2 = 0.5
- h(4): Since 2 < 4 ≤ 5, h(4) = 4 - 1 = 3
- h(7): Since 7 > 5, h(7) = √(7-5) = √2
Answer Key Summary:
Function | Input (x) | Output |
---|---|---|
f(x) (Example 1) | 2 | 5 |
f(x) (Example 1) | 0 | 0 |
f(x) (Example 1) | -3 | -6 |
g(x) (Example 2) | 3 | 5 |
g(x) (Example 2) | 0 | 2 |
g(x) (Example 2) | -2 | 2 |
h(x) (Example 3) | 1 | 0.5 |
h(x) (Example 3) | 4 | 3 |
h(x) (Example 3) | 7 | √2 |
This guide provides a robust foundation for evaluating piecewise functions. Remember to always carefully examine the intervals defined for each sub-function before substituting your input value. Practice with different examples, and you'll master this essential mathematical concept in no time.