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Quizzes > Mathematics

Piecewise Function Quiz: Absolute Value and Domain Breaks

Quick, focused practice to test writing absolute value as piecewise. Instant results.

Editorial: Review CompletedCreated By: Ltcol C. Nicole JohansenUpdated Aug 24, 2025
Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art illustration for quiz on absolute value, piecewise functions, discontinuous domain challenges on coral background.

This piecewise function quiz helps you write absolute value as piecewise, check split cases, and spot domain breaks on graphs. Solve short questions with instant feedback. For more practice, try the absolute value function quiz, a focused absolute value practice quiz, or build graph sense with graphing functions practice.

How can the function f(x) = |x| be written as a piecewise function?
f(x) = -x for x ? 0; f(x) = x for x < 0
f(x) = x for x > 0; f(x) = -x for x ? 0
f(x) = -x for x > 0; f(x) = x for x ? 0
f(x) = x for x ? 0; f(x) = -x for x < 0
The absolute value function is defined as x when x ? 0 and -x when x < 0, ensuring non-negative outputs for all real inputs. This piecewise definition captures its 'V' shape around the origin. By splitting at x = 0, we cover both positive and negative inputs correctly.
What is the value of f(-3) for f(x) = |x|?
3
-3
Undefined
0
By definition, the absolute value function returns the non-negative magnitude of its input. Since -3 is negative, |?3| = 3. This ensures the output is always zero or positive.
What is the domain of the function f(x) = |x - 2| + 3?
All real numbers
x ? 2
x ? 2
x > 2
Absolute value expressions and constant shifts are defined for every real x, so there is no restriction on x. Adding or subtracting inside or outside the absolute value does not remove any real input. Therefore, the domain remains all real numbers. Domain of a Function - Purplemath
Which description best matches the graph of y = |x|?
A U-shaped parabola opening upward
A straight line with slope 1
A W-shaped graph with two minima
A V-shaped graph with vertex at (0, 0) opening upward
The graph of y = |x| forms a 'V' because it mirrors the positive and negative parts of x. It has a sharp corner, or vertex, at the origin, and both branches rise linearly away from it. Unlike a parabola, its slopes are constant on each side of x = 0.
Solve the equation |x - 1| = 5.
x = 4 or x = -6
x = 5 or x = -5
x = 6 or x = -4
x = 1 or x = -1
Split into two cases: x - 1 = 5 gives x = 6, and x - 1 = -5 gives x = -4. Both satisfy the original equation once plugged back in. This approach works for any equation of the form |A| = B where B ? 0.
Express g(x) = 2|x + 3| - 4 as a piecewise function.
g(x) = 2(x - 3) - 4 for x ? 3; g(x) = 2(-x + 3) - 4 for x < 3
g(x) = 2(x + 3) - 4 for x ? -3; g(x) = 2(-x - 3) - 4 for x < -3
g(x) = 2(x + 3) - 4 for x > -3; g(x) = 2(-x - 3) - 4 for x ? -3
g(x) = 2(-x - 3) - 4 for x ? -3; g(x) = 2(x + 3) - 4 for x < -3
The break occurs where the expression inside the absolute value is zero, x + 3 = 0 so x = -3. For x ? -3, |x + 3| = x + 3; for x < -3, |x + 3| = -(x + 3). Distribute the 2 and subtract 4 in each case.
Find the solution set of the inequality |2x - 4| > 6.
x < -1 or x > 5
x > -1 and x < 5
x ? -1 or x ? 5
-1 < x < 5
Solve 2x - 4 > 6 to get x > 5, and 2x - 4 < -6 to get x < -1. Combining gives x < -1 or x > 5. This is the standard approach for |A| > B.
What is the domain of the function h(x) = |x| / x?
All real x except x = 0
x > 0 only
All real x including x = 0
x < 0 only
The expression divides by x, so x cannot be zero. For any other real x, |x|/x is defined and equals 1 if x > 0 or -1 if x < 0. Points where the denominator is zero are excluded from the domain. Sign Function - Wikipedia
Is the function f(x) = |x| differentiable at x = 0?
Yes, with derivative 0
No, because it's not defined at x = 0
No, because there is a cusp at x = 0
Yes, with derivative 1
The graph of y = |x| has a sharp corner (cusp) at the origin, so the left-hand derivative (-1) and right-hand derivative (1) do not match. Differentiability requires a single slope from both sides. Thus f is not differentiable at x = 0.
Solve the equation |x - 2| + |x + 1| = 5.
x = 1 or x = -4
x = 2 or x = -1
x = 4 or x = -3
x = 3 or x = -2
Break into intervals at x = -1 and x = 2. Testing each region yields two valid solutions: x = 3 in the region x ? 2 and x = -2 in the region x < -1. The middle region gives no solution. Absolute-Value Functions - Lamar University
Express h(x) = |x| + |x - 3| as a piecewise function.
h(x) = -x - 3 for x < 0; h(x) = x for 0 ? x < 3; h(x) = x - 3 for x ? 3
h(x) = 2x - 3 for x < 0; h(x) = 3 for 0 ? x < 3; h(x) = -2x + 3 for x ? 3
h(x) = -2x + 3 for x < 0; h(x) = 3 for 0 ? x < 3; h(x) = 2x - 3 for x ? 3
h(x) = 2x + 3 for x < 0; h(x) = 3 for 0 ? x < 3; h(x) = -2x - 3 for x ? 3
Breakpoints are x = 0 and x = 3. For x < 0 both inside values are negative, giving -(x) + -(x-3) = -2x + 3. For 0 ? x < 3, x ? 0 but x - 3 < 0, yielding x + -(x-3) = 3. For x ? 3 both are nonnegative, so x + (x - 3) = 2x - 3.
Consider f(x) = |x² - 9|/(x - 3) for x ? 3. Which piecewise definition is correct and what type of discontinuity occurs at x = 3?
f(x) = x + 3 if x > 3; f(x) = -(x + 3) if x < 3; removable discontinuity at x = 3
f(x) = -(x + 3) if x > 3; f(x) = x + 3 if x < 3; jump discontinuity at x = 3
f(x) = x + 3 if x > 3; f(x) = x + 3 if x < 3; removable discontinuity at x = 3
f(x) = x + 3 if x > 3; f(x) = -(x + 3) if x < 3; jump discontinuity at x = 3
Factor |x² - 9| = |(x - 3)(x + 3)| = |x - 3||x + 3| and divide by (x - 3) for x ? 3. This yields f(x) = (x + 3) when x > 3 and f(x) = -(x + 3) when x < 3. Left and right limits are -6 and 6, respectively, so there's a jump at x = 3.
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Study Outcomes

  1. Understand Piecewise Decomposition -

    Analyze how absolute value expressions can be rewritten as piecewise functions by defining separate cases for positive and negative inputs.

  2. Apply Piecewise Construction -

    Construct piecewise expressions for a variety of absolute value functions and confirm their equivalence to the original expressions.

  3. Graph Piecewise Functions -

    Plot absolute value and discontinuous piecewise functions accurately, identifying key points and intervals on the graph.

  4. Evaluate Discontinuous Domains -

    Determine and handle discontinuities in the domain of piecewise functions, interpreting open and closed intervals correctly.

  5. Assess Problem-Solving Strategies -

    Use instant feedback from the quiz to refine methods for solving piecewise and absolute value problems effectively.

Cheat Sheet

  1. Piecewise Definition of Absolute Value -

    According to MIT OpenCourseWare, any absolute value |x − h| splits into x − h for x ≥ h and −(x − h) for x < h. For example, |x−3| becomes {x−3 if x ≥ 3; 3−x if x < 3}. Mastering this conversion is the foundation for acing your piecewise functions quiz.

  2. Graphing with Slope and Intercept -

    Khan Academy shows that each piece of an absolute value graph is a line with slope +1 or −1. Identify the "vertex" (h,k) and plot two rays, using y=k±(x−h). This trick makes graphing piecewise functions feel like drawing a confident V for victory.

  3. Handling Discontinuous Domains -

    University of California resources highlight how domain restrictions create breaks in piecewise graphs. Always check whether endpoints are open ( ◯ ) or closed ( ◝ ) on interval notation. That attention to detail ensures you nail a piecewise function with a discontinuous domain worksheet answers.

  4. Boundary-Checking Strategy -

    Purdue University recommends plugging boundary values back into each piece to verify continuity or jumps. Label each case clearly when x equals a critical point like h or k. This boundary-checking habit boosts your confidence during absolute value functions practice.

  5. Targeted Practice and Feedback -

    Use free online quizzes such as our piecewise functions quiz for instant grading and hints (Wolfram MathWorld endorses continuous self-testing). Tackle varied problems - graphing piecewise functions, solving equations, and domain puzzles - to reinforce learning. Continuous practice is key to transform theory into quick, accurate skills.

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