1. What is a Piecewise Function?
A **piecewise function** is a function that is defined by multiple sub-functions, each of which applies to a different interval of the main function's domain. Think of it as a set of rules, where you choose the right rule based on the value of the input.
Here is the general form of a piecewise function:
\[ f(x) = \begin{cases} g(x) & \text{if } x < c \\ h(x) & \text{if } x \geq c \end{cases} \]In this example, \( g(x) \) is the rule to follow for all \( x \) values less than \( c \), and \( h(x) \) is the rule for all \( x \) values greater than or equal to \( c \).
2. Limits and Continuity
Piecewise functions are a great way to explore the concepts of limits and continuity. The most important place to check for continuity is at the **critical point**, or the "break point" where the function's rule changes (like at \( x=c \) in the example above).
To determine if a piecewise function is continuous at a critical point, you must check the three conditions of continuity:
- Left-hand limit: Does the function approach a specific value as it comes from the left?
- Right-hand limit: Does the function approach a specific value as it comes from the right?
- Function value: Is the function defined at that point, and does it equal the limit?
If the left-hand limit and the right-hand limit are equal, and that value also equals the function value at the critical point, then the function is continuous there. Otherwise, it is discontinuous.
3. Check Yourself
Determine the continuity and evaluate the following piecewise functions.
Choosing the Correct Rule: When a problem asks you to evaluate a piecewise function at a specific point, always check the inequality symbols. The rule with the "or equal to" sign (\( \leq \) or \( \geq \)) is the one to use for the exact value of the critical point.
Question 1: Evaluate \( f(3) \) and determine if the function is continuous at \( x=3 \) for the function:
\[ f(x) = \begin{cases} x^2 & \text{if } x < 3 \\ 2x+3 & \text{if } x \geq 3 \end{cases} \]First, we evaluate the function at \( x=3 \) and the right-hand limit, since both use the same rule:
Function value: \( f(3) = 2(3)+3 = 9 \)
Right-hand limit: \( \lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (2x+3) = 2(3)+3 = 9 \)
Next, we check the left-hand limit to see if it matches:
Left-hand limit: \( \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x^2) = (3)^2 = 9 \)
Since the left-hand limit (\(9\)), the right-hand limit (\(9\)), and the function value (\(9\)) are all equal, the three conditions for continuity are met. The function is **continuous** at \( x=3 \).
Question 2: Find \( \lim_{x \to 2} g(x) \) and determine if \( g(x) \) is continuous at \( x=2 \) for the function:
\[ g(x) = \begin{cases} x^2+1 & \text{if } x < 2 \\ 3x-1 & \text{if } x \geq 2 \end{cases} \]We must check the left-hand and right-hand limits.
Left-hand limit: \( \lim_{x \to 2^-} g(x) = \lim_{x \to 2^-} (x^2+1) = (2)^2+1 = 5 \)
Right-hand limit: \( \lim_{x \to 2^+} g(x) = \lim_{x \to 2^+} (3x-1) = 3(2)-1 = 5 \)
Since the left-hand limit equals the right-hand limit, the overall limit exists and is 5. We also check the function value at \( x=2 \): \( g(2) = 3(2)-1 = 5 \).
Because \( \lim_{x \to 2} g(x) = g(2) \), the function is **continuous** at \( x=2 \).
Question 3: Find the value of the constant \( a \) that makes the function continuous for all real numbers.
\[ h(x) = \begin{cases} ax-1 & \text{if } x < 3 \\ 2x+5 & \text{if } x \geq 3 \end{cases} \]For the function to be continuous at the critical point \( x=3 \), the left-hand and right-hand limits must be equal.
\[ \lim_{x \to 3^-} (ax-1) = \lim_{x \to 3^+} (2x+5) \] \[ a(3)-1 = 2(3)+5 \] \[ 3a-1 = 6+5 \] \[ 3a-1 = 11 \] \[ 3a = 12 \] \[ a = 4 \]