Piecewise Functions

Functions defined by different formulas on different intervals

Piecewise Functions

Definition

A piecewise function uses different formulas for different parts of the domain.

Example: $f(x) = \begin{cases} x + 1 & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases}$

Evaluating

To find $f(a)$ :

Determine which condition $a$ satisfies
Use the corresponding formula

Example: For the function above:

$f(-2) = -2 + 1 = -1$ (use $x + 1$ since $-2 < 0$ )
$f(3) = 3^2 = 9$ (use $x^2$ since $3 \geq 0$ )

Graphing

Graph each piece on its domain
Use open circles for endpoints NOT included (< or >)
Use closed circles for endpoints included (≤ or ≥)

Common Types

Absolute Value: $|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$

Step Functions: Different constant values on intervals

Continuity

Check if pieces "connect" at boundary points.

Continuous if: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$

📚 Practice Problems

1Problem 1easy

❓ Question:

Evaluate $f(-3)$ and $f(2)$ for $f(x) = \begin{cases} 2x & \text{if } x < 0 \\ x + 5 & \text{if } x \geq 0 \end{cases}$

💡 Show Solution

For $f(-3)$ :

Since $-3 < 0$ , use the first formula: $f(-3) = 2(-3) = -6$

For $f(2)$ :

Since $2 \geq 0$ , use the second formula: $f(2) = 2 + 5 = 7$

Answer: $f(-3) = -6$ , $f(2) = 7$

2Problem 2medium

❓ Question:

Write the absolute value function $f(x) = |x - 3|$ as a piecewise function.

💡 Show Solution

The absolute value changes behavior at the point where the inside equals zero.

$x - 3 = 0$ when $x = 3$

When $x \geq 3$ : the inside is positive or zero $|x - 3| = x - 3$

When $x < 3$ : the inside is negative $|x - 3| = -(x - 3) = -x + 3$

Answer: $f(x) = \begin{cases} -x + 3 & \text{if } x < 3 \\ x - 3 & \text{if } x \geq 3 \end{cases}$

3Problem 3hard

❓ Question:

Is the function continuous at $x = 1$ ? $f(x) = \begin{cases} x^2 + 1 & \text{if } x \leq 1 \\ 3x - 1 & \text{if } x > 1 \end{cases}$

💡 Show Solution

Check if the function value and limits match at $x = 1$ .

From left: (use $x^2 + 1$ ) $\lim_{x \to 1^-} f(x) = 1^2 + 1 = 2$

From right: (use $3x - 1$ ) $\lim_{x \to 1^+} f(x) = 3(1) - 1 = 2$

Function value: (use $x^2 + 1$ since $1 \leq 1$ ) $f(1) = 1^2 + 1 = 2$

Since all three equal 2, the function is continuous at $x = 1$ .

Answer: Yes, continuous at $x = 1$

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