\(f(x) = \begin{cases} 2x + 1 & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases}\)
Find f(-3), f(0), and f(2).
Piecewise function: A function defined by multiple sub-functions, each applying to a certain interval of the domain.
- Identify which condition the input value satisfies
- Select the corresponding sub-function
- Evaluate the sub-function at the given input
Since -3 < 0, we use the first piece: f(x) = 2x + 1
f(-3) = 2(-3) + 1 = -6 + 1 = -5
Since 0 ≥ 0, we use the second piece: f(x) = x²
f(0) = 0² = 0
Since 2 ≥ 0, we use the second piece: f(x) = x²
f(2) = 2² = 4
f(-3) = -5, f(0) = 0, f(2) = 4
• Domain checking: Always verify which condition applies
• Function selection: Use the correct sub-function for the given input
• Substitution: Carefully substitute values to avoid calculation errors
\(g(x) = \begin{cases} -x + 3 & \text{if } x < 1 \\ 2 & \text{if } x = 1 \\ x - 1 & \text{if } x > 1 \end{cases}\)
Graphing piecewise functions: Plot each sub-function over its specified domain, being careful with open/closed circles at boundary points.
This is a line with slope -1 and y-intercept 3. Since x < 1, we draw it for values less than 1, ending with an open circle at (1, 2).
According to the definition, g(1) = 2, so we plot the point (1, 2) with a closed circle.
This is a line with slope 1 and y-intercept -1. We start drawing from x > 1, beginning with an open circle at (1, 0).
Left limit: lim(x→1⁻) g(x) = -1 + 3 = 2
Right limit: lim(x→1⁺) g(x) = 1 - 1 = 0
Since left limit ≠ right limit, the function has a jump discontinuity at x = 1.
The function has a jump discontinuity at x = 1. Domain: (-∞, ∞). Range: (-∞, 2] ∪ (0, ∞).
• Open vs Closed Circles: Open circle indicates exclusion, closed circle indicates inclusion
• Domain Restriction: Only graph each piece where its condition is satisfied
• Continuity Check: Compare left and right limits at boundary points
Real-world piecewise functions: Used to model situations where different rules apply to different ranges of input values.
For m ≤ 5: Cost = $2.50 × m
For m > 5: Cost = First 5 miles ($2.50 × 5 = $12.50) + remaining miles ($1.80 × (m-5))
C(m) = {2.50m if 0 ≤ m ≤ 5, 12.50 + 1.80(m-5) if m > 5}
Since 3 ≤ 5, use C(m) = 2.50m
C(3) = 2.50 × 3 = $7.50
Since 5 ≤ 5, use C(m) = 2.50m
C(5) = 2.50 × 5 = $12.50
Since 8 > 5, use C(m) = 12.50 + 1.80(m-5)
C(8) = 12.50 + 1.80(8-5) = 12.50 + 1.80(3) = 12.50 + 5.40 = $17.90
C(m) = {2.50m if 0 ≤ m ≤ 5, 12.50 + 1.80(m-5) if m > 5}
C(3) = $7.50, C(5) = $12.50, C(8) = $17.90
• Modeling: Break complex problems into simpler parts
• Boundary conditions: Be precise about which intervals include endpoints
• Unit consistency: Ensure all calculations use consistent units
Piecewise Function: A function defined by multiple sub-functions, each applying to a specific interval of the domain
Domain: The set of all possible input values for the function
Range: The set of all possible output values of the function
Continuity: A function is continuous at a point if the left limit equals the right limit equals the function value
- Identify the pieces: Determine all sub-functions and their domains
- Check boundaries: Note whether boundary points are included (≤, ≥) or excluded (<, >)
- Apply the correct rule: For any input, determine which condition it satisfies
- Verify continuity: At boundary points, check if left and right limits match
• Domain checking: Always determine which condition applies to your input
• Function evaluation: Substitute into the correct sub-function
• Graphing: Use closed circles for included points, open circles for excluded points
• Continuity: lim(x→a⁻) f(x) = lim(x→a⁺) f(x) = f(a) for continuity at point a
• Real-world applications: Interpret results in context of the problem
\(h(x) = \begin{cases} \sqrt{x} & \text{if } 0 \leq x \leq 4 \\ 2x - 4 & \text{if } x > 4 \end{cases}\)
Domain: Set of all possible input values. Range: Set of all possible output values.
For h(x) = √x on [0, 4]:
Domain contribution: [0, 4]
Range: When x = 0, h(0) = √0 = 0; when x = 4, h(4) = √4 = 2
Since √x is increasing, the range for this piece is [0, 2]
For h(x) = 2x - 4 on (4, ∞):
Domain contribution: (4, ∞)
Range: As x approaches 4 from the right, h(x) approaches 2(4) - 4 = 4
As x increases without bound, h(x) also increases without bound
So the range for this piece is (4, ∞)
Complete domain: [0, 4] ∪ (4, ∞) = [0, ∞)
Complete range: [0, 2] ∪ (4, ∞)
Note: There's a gap in the range from 2 to 4 (not inclusive)
Left limit: lim(x→4⁻) h(x) = √4 = 2
Right limit: lim(x→4⁺) h(x) = 2(4) - 4 = 4
Since left limit ≠ right limit, there's a jump discontinuity at x = 4
Domain: [0, ∞)
Range: [0, 2] ∪ (4, ∞)
Jump discontinuity at x = 4
• Domain combination: Union of all individual domains
• Range analysis: Consider behavior at interval endpoints
• Continuity check: Compare limits at boundary points
\(p(x) = \begin{cases} x^2 & \text{if } x < -1 \\ 2 & \text{if } -1 \leq x < 1 \\ x + 1 & \text{if } x \geq 1 \end{cases}\)
Find p(-2), p(-1), p(0.5), p(1), and p(3). Also, determine the domain and range.
Multi-piece function: A function with three or more sub-functions, each applying to distinct intervals.
Since -2 < -1, use p(x) = x²
p(-2) = (-2)² = 4
Since -1 ≤ -1 < 1, use p(x) = 2
p(-1) = 2
Since -1 ≤ 0.5 < 1, use p(x) = 2
p(0.5) = 2
Since 1 ≥ 1, use p(x) = x + 1
p(1) = 1 + 1 = 2
Since 3 ≥ 1, use p(x) = x + 1
p(3) = 3 + 1 = 4
All real numbers are covered: (-∞, -1) ∪ [-1, 1) ∪ [1, ∞) = ℝ
For x < -1: p(x) = x² produces values in (1, ∞)
For -1 ≤ x < 1: p(x) = 2 (constant)
For x ≥ 1: p(x) = x + 1 produces values in [2, ∞)
Combined range: {2} ∪ (1, ∞) = [1, ∞) (since 2 ∈ [2, ∞) and (1, ∞) ∪ [2, ∞) = [1, ∞))
Wait, let me recalculate: x² for x < -1 gives (1, ∞), constant 2 for [-1,1), and x+1 for [1,∞) gives [2,∞). Combined: {2} ∪ (1,∞) ∪ [2,∞) = [1,∞).
p(-2) = 4, p(-1) = 2, p(0.5) = 2, p(1) = 2, p(3) = 4
Domain: (-∞, ∞) or ℝ
Range: [1, ∞)
• Multiple pieces: Carefully identify which condition applies to each input
• Range determination: Analyze each piece separately then combine results
• Interval notation: Pay attention to open and closed brackets
Piecewise Function: A function defined by multiple sub-functions, each applying to a specific interval of the domain.
Domain: The complete set of input values for which the function is defined.
Range: The complete set of output values the function can produce.
Continuity: A function is continuous at a point if the left-hand limit equals the right-hand limit and both equal the function value.
- Analyze the function structure: Identify all pieces and their domains
- Check domain boundaries: Note whether endpoints are included or excluded
- Apply evaluation rules: For any input, determine which piece applies
- Verify continuity: At boundary points, compare left and right limits
- Determine domain and range: Combine results from all pieces
• Domain identification: Union of all individual piece domains
• Function evaluation: Always use the piece corresponding to the input value
• Continuity check: lim(x→a⁻) f(x) = lim(x→a⁺) f(x) = f(a)
• Graphing convention: Closed circle (●) for included, open circle (○) for excluded
• Range determination: Union of all individual piece ranges
Tax brackets: Different tax rates apply to different income levels
Shipping costs: Flat rate up to a weight threshold, then per-unit rate
Utility pricing: Tiered pricing based on consumption levels
Physics: Velocity changes at specific time intervals
- Read carefully: Identify all conditions and intervals
- Organize: List each piece with its domain
- Visualize: Sketch the function if possible
- Verify: Check your answers make sense