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Piecewise Functions Practice Problems

Thirteen hand-built problems on piecewise functions, each fully worked. They drill the habit that decides every one of these questions — check the condition before you substitute — then move on to boundary values, absolute value rewritten as two rules, gaps in the domain, and a tiered pricing model.

  1. 1.

    For \(f(x) = x + 4\) when \(x < 1\), and \(f(x) = 3x\) when \(x \ge 1\), find \(f(-2)\).

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    Is \(-2 < 1\)? Yes, so use the first rule: \(f(-2) = -2 + 4 = 2\).

  2. 2.

    For \(f(x) = x + 4\) when \(x < 1\), and \(f(x) = 3x\) when \(x \ge 1\), find \(f(4)\).

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    \(4 \ge 1\), so use \(3x\): \(f(4) = 12\).

  3. 3.

    For \(f(x) = x + 4\) when \(x < 1\), and \(f(x) = 3x\) when \(x \ge 1\), find \(f(1)\).

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    The boundary. The condition \(x \ge 1\) includes 1, so the second rule owns it: \(f(1) = 3\).

  4. 4.

    For \(f(x) = x^2\) when \(x \le 0\), and \(f(x) = 3x\) when \(x > 0\), find \(f(-2)\).

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    \(-2 \le 0\), so \(f(-2) = (-2)^2 = 4\).

  5. 5.

    For \(f(x) = x^2\) when \(x \le 0\), and \(f(x) = 3x\) when \(x > 0\), find \(f(0)\).

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    \(0 \le 0\) is true, so the first rule applies: \(f(0) = 0\). Here both rules happen to agree, so the graph joins up.

  6. 6.

    For \(f(x) = 5\) when \(x < 1\), and \(f(x) = 2x + 1\) when \(x \ge 1\), find \(f(1)\).

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    \(x \ge 1\) includes 1, so \(f(1) = 2(1) + 1 = 3\). The first rule approaches 5 from the left, so the graph jumps down at \(x = 1\).

  7. 7.

    For \(g(x) = x^2\) when \(x \le 2\), and \(g(x) = 7\) when \(x > 2\), what is the size of the jump at \(x = 2\)?

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    The left branch gives \(g(2) = 4\); the right branch approaches 7. The jump is \(7 - 4 = 3\).

  8. 8.

    Write \(\lvert x - 3 \rvert\) as a piecewise rule. For \(x < 3\), what is the rule? Give it in terms of \(x\).

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    For \(x < 3\) the inside is negative, so the absolute value negates it: \(-(x - 3) = 3 - x\). For \(x \ge 3\) the rule is simply \(x - 3\).

  9. 9.

    For \(f(x) = 2\) when \(x < 0\), and \(f(x) = x + 2\) when \(0 \le x \le 3\), find the domain in interval notation.

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    The union of \((-\infty,\ 0)\) and \([0,\ 3]\) is \((-\infty,\ 3]\). Nothing above 3 is covered by any condition, so it is outside the domain.

  10. 10.

    For that same function, find the range in interval notation.

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    The first branch produces only \(\{2\}\). The second runs from \(f(0) = 2\) to \(f(3) = 5\). The union is \([2,\ 5]\).

  11. 11.

    For \(f(x) = x\) when \(x < -1\), and \(f(x) = x^2\) when \(x > 1\), find the domain in interval notation.

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    No condition covers \(-1 \le x \le 1\), so that whole stretch is excluded and the domain comes in two pieces.

  12. 12.

    A courier charges \(\pounds 4\) for parcels up to 2 kg, then \(\pounds 4\) plus \(\pounds 1.50\) per extra kilogram. What does a 5 kg parcel cost, in pounds?

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    Above the threshold the rule is \(4 + 1.5(w - 2)\). At \(w = 5\): \(4 + 1.5(3) = \pounds 8.50\). The surcharge applies only to the weight above 2 kg.

  13. 13.

    A gym charges \(\pounds 20\) a month for up to 8 visits, then \(\pounds 2\) per extra visit. What does a month with 12 visits cost, in pounds?

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    \(C(v) = 20 + 2(v - 8)\) for \(v > 8\). At \(v = 12\): \(20 + 2(4) = \pounds 28\).
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Piecewise Functions Practice Problems

The Piecewise Functions lesson covers the notation and the graphing conventions; this set is thirteen problems to practise on, each fully solved.

How to use this set

One habit decides almost every problem here: check the condition first, then substitute. Ask which interval the input falls in, pick that branch, and only then do the arithmetic.

Pay particular attention to problems 3, 5 and 6 — all three evaluate at a boundary, which is where marks are actually lost.

Common mistakes to watch for

Substituting before checking. Decide the branch first.

Getting the boundary branch wrong. owns the value 1; does not.

Assuming the branches join up. Many real piecewise models jump, and that is a feature, not an error.

Reading as negative. For a negative , comes out positive — that is exactly how the absolute value rule works.

Forgetting uncovered values. If no condition includes a number, it is outside the domain.

Where to go next

Function Notation is the evaluation skill underneath, Domain and Range goes further on the interval work, and Absolute Value Equations applies the two-case split.

Frequently Asked Questions

What is a piecewise function?+

One defined by two or more rules, each applying to a different part of the domain. Which rule you use depends on which interval the input falls in.

How do you evaluate one?+

Check the conditions first to see which interval the input belongs to, then substitute into that branch only. Deciding the branch before touching the algebra prevents nearly every error on this topic.

Which rule owns the boundary value?+

Whichever condition uses or . For and , the input 1 belongs to the second rule.

What do the filled and open dots mean?+

A filled dot means the endpoint is included, matching or . An open circle means it is excluded, matching or . At any boundary in the domain exactly one dot is filled.

How do I find the domain?+

Take the union of the intervals the branches are defined on. If some value is covered by no condition at all, it is not in the domain — which is how a piecewise function can have a gap in the middle.

Is absolute value piecewise?+

Yes. is when and when . Rewriting it that way is how absolute value equations and inequalities get solved.

More practice problems

Function Notation Practice Problems Domain and Range Practice Problems Absolute Value Equations Practice Problems