Let f(x) = {3x + 1 if x < 2; x² − 1 if x ≥ 2}. What is f(2)?
A7 — substituting x = 2 into 3x + 1 gives 3(2) + 1 = 7
B3 — substituting x = 2 into x² − 1 gives 2² − 1 = 3
CBoth 7 and 3 — the function has two values at boundary points
DUndefined — x = 2 sits exactly on the boundary between pieces
The domain restriction x ≥ 2 means the second piece owns the boundary point x = 2. The first piece, x < 2, does not include x = 2 (strict inequality). So f(2) = 2² − 1 = 3. Option A is the classic error — applying the first formula without checking the inequality. Always determine which interval contains your input value before evaluating.
Question 2 Multiple Choice
A piecewise function is defined as f(x) = {x + 1 if x ≤ 1; 2x if x > 1}. A student argues the function is discontinuous at x = 1 because two different formulas are defined there. Which response is most accurate?
ACorrect — any piecewise function is discontinuous at its boundary points
BIncorrect — both pieces give the same output at x = 1 (both equal 2), so the function is continuous there
CIncorrect — only the first piece applies at x = 1, so continuity is determined solely by that formula
DCorrect — a closed circle and an open circle at the same x-value always indicate a discontinuity
Continuity at a boundary depends on whether the pieces agree in value there — not on whether multiple formulas exist. First piece at x = 1: f(1) = 1 + 1 = 2 (closed circle, x ≤ 1). Second piece limit from the right: lim_{x→1+} 2x = 2 (open circle). Both equal 2, so the graph has no jump and the function is continuous at x = 1. The student's reasoning is the most common misconception: the mere existence of two formulas does not cause discontinuity. You must check whether they agree at the boundary.
Question 3 True / False
Nearly every piecewise-defined function has a jump discontinuity at each boundary point between its pieces.
TTrue
FFalse
Answer: False
Many piecewise functions are perfectly continuous. The absolute value function |x| = {x if x ≥ 0; −x if x < 0} is continuous at x = 0 because both pieces give y = 0 there. Continuity at a boundary requires that the left-hand limit, right-hand limit, and function value all agree. The piecewise structure introduces the possibility of discontinuity, but whether one actually occurs depends on the specific formulas.
Question 4 True / False
On the graph of a piecewise function, a closed circle at a boundary point means that piece includes that endpoint value.
TTrue
FFalse
Answer: True
Graphical convention: a closed (filled) circle means the endpoint is included in the piece — the function actually equals that y-value at that x. An open (hollow) circle means the piece approaches but does not include that endpoint. When one piece ends with an open circle and another begins with a closed circle at the same x, exactly one piece owns the point, and the function is well-defined there (not necessarily continuous, but unambiguous).
Question 5 Short Answer
Explain how to evaluate a piecewise function at a specific input value, and describe what must be true at a boundary point for the function to be continuous there.
Think about your answer, then reveal below.
Model answer: To evaluate a piecewise function at x = a: first determine which interval a belongs to by checking the domain conditions (the inequalities), then apply the formula for that interval. At a boundary point, the function is continuous if the output values from the piece to the left and the piece to the right both equal the same number — the graph has no jump. If they differ, there is a jump discontinuity.
The two-step process — check the condition, then apply the formula — is the whole algorithm. The continuity condition formalizes the intuition of a 'connected' graph: the left limit equals the right limit equals the function value. Piecewise functions can fail continuity in various ways: a jump (different left and right limits), a removable hole (limits agree but function value differs), or the function being undefined at the boundary.