Lesson 1.5 · Unit 1 · F-IF.7b

Functions with different rules on different intervals

Tax brackets, postage, phone plans and the absolute value itself all change their rule at a boundary. A piecewise function states the rule for each interval and says where the switch happens. The absolute value turns out to be the simplest example, and understanding it as a distance makes equations and inequalities with it straightforward.

The method

A piecewise function lists a rule for each interval, with a condition saying which rule applies.
To evaluate, find which condition the input satisfies, then use only that rule.
At a boundary, exactly one piece includes the endpoint. A filled dot marks that piece; an open dot marks the other.
$ |x| $ is the distance from $ x $ to 0, and $ |x - a| $ is the distance from $ x $ to $ a $.
As a piecewise rule: $ |x| = x $ when $ x \ge 0 $, and $ -x $ when $ x \lt 0 $.
$ |A| = c $ (with $ c \gt 0 $) means $ A = c $ or $ A = -c $. Solve both and check.
$ |A| \lt c $ means $ -c \lt A \lt c $; $ |A| \gt c $ means $ A \lt -c $ or $ A \gt c $.
Equations with the variable outside as well must be checked, since a distance cannot equal a negative number.

Where students lose marks: forgetting the second case. The equation $ |x - 3| = 2 $ has two solutions, 1 and 5, both at distance 2 from 3. Writing only $ x = 5 $ answers half the question.

Worked example

The problem. (a) For $ f(x) = \begin{cases} x + 2 & x \lt 1 \\ x^2 & x \ge 1 \end{cases} $, find $ f(-3) $, $ f(1) $ and $ f(4) $, and describe the graph at $ x = 1 $. (b) Solve $ |2x - 5| = 9 $. (c) Solve $ |x - 3| \lt 4 $ and $ |x - 3| \ge 4 $. (d) A tax is 10 percent of the first \$10,000 and 12 percent of the amount above that, up to \$40,000. Write the tax function and find the tax on \$25,000.

Step one: evaluate (a). For $ f(-3) $: $ -3 \lt 1 $, so use $ x + 2 $, giving $ -1 $. For $ f(1) $: $ 1 \ge 1 $, so use $ x^2 $, giving 1. For $ f(4) $: $ 4^2 = 16 $.

Step two: describe the boundary. As $ x $ approaches 1 from the left the value approaches $ 1 + 2 = 3 $ but never reaches it, so there is an open dot at $ (1, 3) $. At $ x = 1 $ the function equals 1, a filled dot at $ (1, 1) $. The graph jumps by 2 at $ x = 1 $: it cannot be drawn without lifting the pencil.

Step three: solve (b) in two cases. $ 2x - 5 = 9 $ gives $ 2x = 14 $, so $ x = 7 $. $ 2x - 5 = -9 $ gives $ 2x = -4 $, so $ x = -2 $. Check: $ |2(7) - 5| = 9 $ and $ |2(-2) - 5| = |-9| = 9 $. Both correct.

Step four: interpret (c) as distance. $ |x - 3| \lt 4 $ says $ x $ is within 4 of 3, so $ -4 \lt x - 3 \lt 4 $, which gives $ -1 \lt x \lt 7 $. The complement, $ |x - 3| \ge 4 $, is at least 4 away: $ x \le -1 $ or $ x \ge 7 $. Test $ x = 0 $: distance 3, inside the first set. Test $ x = 8 $: distance 5, in the second.

Step five: set up the tax function for (d). On the first \$10,000, $ T(x) = 0.10x $. Above \$10,000, the first \$10,000 still costs \$1,000, and the excess is taxed at 12 percent: $ T(x) = 1000 + 0.12(x - 10000) $.

Step six: write it piecewise. $ T(x) = \begin{cases} 0.10x & 0 \le x \le 10000 \\ 1000 + 0.12(x - 10000) & 10000 \lt x \le 40000 \end{cases} $.

Step seven: compute and check continuity. $ T(25000) = 1000 + 0.12(15000) = 1000 + 1800 = 2800 $. At the boundary the two pieces agree: $ 0.10(10000) = 1000 $ and the second piece gives $ 1000 + 0 = 1000 $. A well-designed tax schedule has no jump.

Step eight: read the slopes. The slope of each piece is its marginal rate, 0.10 then 0.12. A common belief is that moving into a higher bracket taxes all income at the higher rate; the function shows that only the excess is taxed at the higher rate, and the graph has a corner at \$10,000 and no jump.

Practice · 10 problems

Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.

For $ f(x) = \begin{cases} 2x & x \lt 0 \\ x + 1 & x \ge 0 \end{cases} $, find $ f(-2) $, $ f(0) $ and $ f(3) $.
Show the full solution

$ f(-2) = 2(-2) = -4 $. At 0 the second rule applies: $ f(0) = 1 $. $ f(3) = 3 + 1 = 4 $. $ -4, 1, 4 $
Solve $ |x| = 7 $.
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Two numbers are at distance 7 from 0. $ x = 7 $ or $ x = -7 $
Solve $ |x - 2| = 5 $.
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$ x - 2 = 5 $ gives 7, and $ x - 2 = -5 $ gives $ -3 $. $ x = 7 $ or $ x = -3 $
Solve $ |x + 1| \lt 3 $.
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$ -3 \lt x + 1 \lt 3 $, so subtract 1 throughout. $ -4 \lt x \lt 2 $
Evaluate $ |-8| - |3 - 10| $.
Show the full solution

$ 8 - |-7| = 8 - 7 $. 1
Solve $ |3x - 1| = 11 $.
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$ 3x - 1 = 11 $ gives $ x = 4 $. $ 3x - 1 = -11 $ gives $ x = -\dfrac{10}{3} $. Check the second: $ 3\left(-\dfrac{10}{3}\right) - 1 = -11 $, and $ |-11| = 11 $. $ x = 4 $ or $ x = -\dfrac{10}{3} $
Solve $ |2x - 4| \ge 6 $.
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$ 2x - 4 \ge 6 $ gives $ x \ge 5 $. $ 2x - 4 \le -6 $ gives $ x \le -1 $. Test $ x = 0 $: $ |-4| = 4 $, not at least 6, so 0 is correctly outside. $ x \le -1 $ or $ x \ge 5 $
Solve $ |x - 3| = 2x $, and identify any extraneous solution.
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Case 1: $ x - 3 = 2x $ gives $ x = -3 $. Case 2: $ 3 - x = 2x $ gives $ x = 1 $. Check $ x = -3 $: the left side is $ |-6| = 6 $, the right side is $ 2(-3) = -6 $. They are not equal, so $ x = -3 $ is extraneous: a distance cannot equal a negative number. Check $ x = 1 $: $ |-2| = 2 $ and $ 2(1) = 2 $. Correct. $ x = 1 $ only
Write $ g(x) = |x - 4| + |x + 1| $ as a piecewise function and check it at $ x = -2 $.
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The pieces change at $ x = -1 $ and $ x = 4 $. For $ x \lt -1 $: $ (4 - x) + (-x - 1) = 3 - 2x $. For $ -1 \le x \le 4 $: $ (4 - x) + (x + 1) = 5 $. For $ x \gt 4 $: $ (x - 4) + (x + 1) = 2x - 3 $. Check at $ x = -2 $: directly $ |-6| + |-1| = 7 $, and $ 3 - 2(-2) = 7 $. Correct. $ g(x) = 3 - 2x $ for $ x \lt -1 $; $ 5 $ for $ -1 \le x \le 4 $; $ 2x - 3 $ for $ x \gt 4 $
Extend the tax function of the worked example: above \$40,000 the rate is 22 percent on the excess. Find the tax on \$50,000 and explain why the function is continuous.
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At \$40,000 the tax is $ 1000 + 0.12(30000) = 1000 + 3600 = 4600 $. For $ x \gt 40000 $: $ T(x) = 4600 + 0.22(x - 40000) $. $ T(50000) = 4600 + 0.22(10000) = 4600 + 2200 = 6800 $. Continuity: at \$40,000 the third piece gives $ 4600 + 0 $, matching the second. Each new piece starts from the total accumulated so far, which is why the function has corners and no jumps. \$6,800

Functions with different rules on different intervals

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