Piecewise Functions

Defining and evaluating functions with different rules on different intervals

Piecewise Functions

What is a Piecewise Function?

A piecewise function is defined by different formulas on different parts of its domain.

General Form

\text{formula 1} & \text{if condition 1} \\ \text{formula 2} & \text{if condition 2} \\ \vdots & \vdots \end{cases}$$ ### Example $$f(x) = \begin{cases} x^2 & \text{if } x < 0 \\ 2x + 1 & \text{if } x \geq 0 \end{cases}$$ This means: - Use $f(x) = x^2$ when $x$ is negative - Use $f(x) = 2x + 1$ when $x$ is zero or positive ## Evaluating Piecewise Functions To evaluate $f(a)$ for a piecewise function: 1. Determine which condition $a$ satisfies 2. Use the formula for that piece 3. Calculate the value ### Example For the function above: - $f(-2) = (-2)^2 = 4$ (use first piece since $-2 < 0$) - $f(0) = 2(0) + 1 = 1$ (use second piece since $0 \geq 0$) - $f(3) = 2(3) + 1 = 7$ (use second piece since $3 \geq 0$) ## Continuity of Piecewise Functions A piecewise function is **continuous** at $x = a$ if: 1. $f(a)$ is defined 2. $\lim_{x \to a} f(x)$ exists 3. $\lim_{x \to a} f(x) = f(a)$ **Practical check:** The function is continuous if the pieces "connect" (no jump or gap). ### Checking Continuity at a Boundary For continuity at $x = c$ (a boundary point): - Evaluate the left piece at $x = c$: $\lim_{x \to c^-} f(x)$ - Evaluate the right piece at $x = c$: $\lim_{x \to c^+} f(x)$ - Check which piece includes $c$ to find $f(c)$ - If left limit = right limit = $f(c)$, the function is continuous ## Absolute Value as a Piecewise Function The absolute value function is a common piecewise function: $$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$ ## Domain and Range - **Domain**: Usually all real numbers, unless specified otherwise - **Range**: Combine the ranges from all pieces (may need to sketch or analyze) ## Graphing Piecewise Functions 1. Graph each piece on its specified interval 2. Use open circles ($\circ$) for excluded endpoints 3. Use closed circles ($\bullet$) for included endpoints 4. Check for continuity at boundary points

📚 Practice Problems

1Problem 1easy

❓ Question:

Evaluate the piecewise function f(x)={x2+1if x≤−13xif −1<x<25if x≥2f(x) = \begin{cases} x^2 + 1 & \text{if } x \leq -1 \\ 3x & \text{if } -1 < x < 2 \\ 5 & \text{if } x \geq 2 \end{cases} at x=−1x = -1, x=0x = 0, and x=3x = 3.

💡 Show Solution

Evaluate at each point:

For x=−1x = -1: Check which condition: −1≤−1-1 \leq -1 ✓ (first piece) f(−1)=(−1)2+1=1+1=2f(-1) = (-1)^2 + 1 = 1 + 1 = 2

For x=0x = 0: Check which condition: −1<0<2-1 < 0 < 2 ✓ (second piece) f(0)=3(0)=0f(0) = 3(0) = 0

For x=3x = 3: Check which condition: 3≥23 \geq 2 ✓ (third piece) f(3)=5f(3) = 5

Answers:

  • f(−1)=2f(-1) = 2
  • f(0)=0f(0) = 0
  • f(3)=5f(3) = 5

2Problem 2medium

❓ Question:

Determine whether the piecewise function g(x)={x+3if x<1x2+1if x≥1g(x) = \begin{cases} x + 3 & \text{if } x < 1 \\ x^2 + 1 & \text{if } x \geq 1 \end{cases} is continuous at x=1x = 1.

💡 Show Solution

Check continuity at x=1x = 1:

Step 1: Find g(1)g(1) (which piece includes x=1x = 1?) Since x≥1x \geq 1 includes 1, use second piece: g(1)=12+1=2g(1) = 1^2 + 1 = 2

Step 2: Find left-hand limit as x→1−x \to 1^- (use first piece): lim⁡x→1−g(x)=lim⁡x→1−(x+3)=1+3=4\lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} (x + 3) = 1 + 3 = 4

Step 3: Find right-hand limit as x→1+x \to 1^+ (use second piece): lim⁡x→1+g(x)=lim⁡x→1+(x2+1)=12+1=2\lim_{x \to 1^+} g(x) = \lim_{x \to 1^+} (x^2 + 1) = 1^2 + 1 = 2

Step 4: Compare

  • Left limit: 4
  • Right limit: 2
  • g(1)=2g(1) = 2

Since the left limit (44) ≠\neq right limit (22), the overall limit does not exist.

Answer: The function is NOT continuous at x=1x = 1 because there is a jump discontinuity (the two pieces don't connect).

3Problem 3hard

❓ Question:

Find the value of aa that makes h(x)={2x+aif x<3x2−4if x≥3h(x) = \begin{cases} 2x + a & \text{if } x < 3 \\ x^2 - 4 & \text{if } x \geq 3 \end{cases} continuous at x=3x = 3.

💡 Show Solution

For continuity at x=3x = 3, we need: lim⁡x→3−h(x)=lim⁡x→3+h(x)=h(3)\lim_{x \to 3^-} h(x) = \lim_{x \to 3^+} h(x) = h(3)

Step 1: Find h(3)h(3) (use second piece since 3≥33 \geq 3): h(3)=32−4=9−4=5h(3) = 3^2 - 4 = 9 - 4 = 5

Step 2: Find right-hand limit (use second piece): lim⁡x→3+h(x)=lim⁡x→3+(x2−4)=32−4=5\lim_{x \to 3^+} h(x) = \lim_{x \to 3^+} (x^2 - 4) = 3^2 - 4 = 5

Step 3: Find left-hand limit (use first piece with unknown aa): lim⁡x→3−h(x)=lim⁡x→3−(2x+a)=2(3)+a=6+a\lim_{x \to 3^-} h(x) = \lim_{x \to 3^-} (2x + a) = 2(3) + a = 6 + a

Step 4: Set left limit equal to right limit: 6+a=56 + a = 5 a=−1a = -1

Verify: With a=−1a = -1:

  • Left limit: 6+(−1)=56 + (-1) = 5 ✓
  • Right limit: 55 ✓
  • h(3)=5h(3) = 5 ✓

Answer: a=−1a = -1 makes the function continuous at x=3x = 3.

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