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 $f(x) = \begin{cases} x^2 + 1 & \text{if } x \leq -1 \\ 3x & \text{if } -1 < x < 2 \\ 5 & \text{if } x \geq 2 \end{cases}$ at $x = -1$ , $x = 0$ , and $x = 3$ .

💡 Show Solution

Evaluate at each point:

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

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

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

Answers:

$f(-1) = 2$
$f(0) = 0$
$f(3) = 5$

2Problem 2medium

❓ Question:

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

💡 Show Solution

Check continuity at $x = 1$ :

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

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

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

Step 4: Compare

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

Since the left limit ( $4$ ) $\neq$ right limit ( $2$ ), the overall limit does not exist.

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

3Problem 3hard

❓ Question:

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

💡 Show Solution

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

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

Step 2: Find right-hand limit (use second piece): $\lim_{x \to 3^+} h(x) = \lim_{x \to 3^+} (x^2 - 4) = 3^2 - 4 = 5$

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

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

Verify: With $a = -1$ :

Left limit: $6 + (-1) = 5$ ✓
Right limit: $5$ ✓
$h(3) = 5$ ✓

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

4Problem 4medium

❓ Question:

A function is defined as f(x) = { x² + 1, if x < 0; 2x + 1, if x ≥ 0 }. Evaluate f(-2), f(0), and f(3).

💡 Show Solution

Step 1: Evaluate f(-2): Is -2 < 0? Yes Use f(x) = x² + 1 f(-2) = (-2)² + 1 = 4 + 1 = 5

Step 2: Evaluate f(0): Is 0 < 0? No Is 0 ≥ 0? Yes Use f(x) = 2x + 1 f(0) = 2(0) + 1 = 1

Step 3: Evaluate f(3): Is 3 < 0? No Is 3 ≥ 0? Yes Use f(x) = 2x + 1 f(3) = 2(3) + 1 = 7

Answer: f(-2) = 5, f(0) = 1, f(3) = 7

5Problem 5hard

❓ Question:

For f(x) = { -x + 2, if x ≤ 1; x² - 2, if x > 1 }, determine if f is continuous at x = 1.

💡 Show Solution

Step 1: Find f(1) using the appropriate piece: Since 1 ≤ 1, use f(x) = -x + 2 f(1) = -1 + 2 = 1

Step 2: Find the left-hand limit as x approaches 1: lim(x→1⁻) f(x) = lim(x→1⁻) (-x + 2) = -1 + 2 = 1

Step 3: Find the right-hand limit as x approaches 1: lim(x→1⁺) f(x) = lim(x→1⁺) (x² - 2) = 1² - 2 = -1

Step 4: Check continuity conditions: For continuity at x = 1: • lim(x→1⁻) f(x) = 1 • lim(x→1⁺) f(x) = -1 • f(1) = 1

Step 5: Conclusion: Since lim(x→1⁻) f(x) ≠ lim(x→1⁺) f(x), the function is NOT continuous at x = 1. There is a jump discontinuity at x = 1.

Answer: Not continuous (jump discontinuity at x = 1)

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