From Verbal Conditions to a Function Rule
A verbal description may specify one calculation for inputs in one range and a different calculation for inputs in another. To write this as a piecewise definition, two decisions must be made carefully: which formula applies in each range, and exactly which inputs belong to that range.
The second decision matters especially at a transition value. If one rule applies below \(2\) and another applies at \(2\) or above, then the first condition is \(x<2\) and the second is \(x\ge 2\). The boundary input \(2\) belongs to exactly one rule.
Earlier material introduced piecewise function notation and endpoint conventions. Here the emphasis is on building the conditions from verbal statements. The goal is not merely to list formulas, but to encode the verbal rule without leaving out any allowed input or assigning an input to two different formulas.
Translate Boundary Words Precisely
Boundary phrases indicate whether the boundary value is included. For a real input \(x\) and a boundary \(c\), the common translations are:
| Verbal phrase | Inequality |
|---|---|
| Less than \(c\), below \(c\), or before \(c\) | \(x<c\) |
| At most \(c\), no greater than \(c\), or up to and including \(c\) | \(x\le c\) |
| Greater than \(c\), above \(c\), or after \(c\) | \(x>c\) |
| At least \(c\), no less than \(c\), or \(c\) and above | \(x\ge c\) |
For example, “from \(1\) up to but not including \(4\)” means \(1\le x<4\). The lower endpoint is included; the upper endpoint is excluded. The words do not describe a vague region around the endpoints: they determine exact inequalities.
When a description says only “up to \(4\)” or “from \(4\),” it may not be clear whether \(4\) is included. Do not silently guess if the context does not settle the meaning. State the interpretation being used, or seek clarification. A piecewise rule needs an unambiguous assignment at every boundary.
A Reliable Construction Method
Begin by identifying the domain and the transition values. Then translate each verbal range into inequalities, keeping the formula attached to its corresponding condition. Finally, check both coverage and uniqueness: every allowed input must satisfy a condition, and no allowed input should satisfy two different conditions.
Write down which inputs are allowed, including any stated lower or upper limit.
These are the numbers at which the verbal description changes from one rule to another.
Use strict or inclusive inequalities according to the boundary words.
Keep each formula with the range for which the verbal description specifies it.
Test the transition values and confirm that the conditions cover the domain exactly once.
Worked Example: A Rule with Two Transition Values
Worked Example: A Rule with Two Transition Values
A calibration rule accepts inputs \(x\) from \(-2\) through \(6\). It says: use \(2x+4\) for inputs below \(1\); use \(x^2-1\) for inputs from \(1\) up to but not including \(4\); and use \(3x-5\) for inputs of \(4\) or greater. Build the piecewise definition.
Step 1: Record the domain. “From \(-2\) through \(6\)” includes both endpoints, so the domain is \([-2,6]\).
Step 2: Translate the first range. “Below \(1\)” gives \(x<1\). Within the stated domain, this range is \(-2\le x<1\). Its formula is \(2x+4\).
Step 3: Translate the middle range. “From \(1\) up to but not including \(4\)” gives \(1\le x<4\). Its formula is \(x^2-1\).
Step 4: Translate the last range. “Of \(4\) or greater” gives \(x\ge4\). The domain ends at \(6\), so this condition is \(4\le x\le6\). Its formula is \(3x-5\).
Combining the ranges and formulas gives:
Check the transitions. At \(x=1\), the first condition fails because \(1<1\) is false, while the second holds because \(1\le1<4\). The assigned value is \(1^2-1=0\). At \(x=4\), the middle condition fails because \(4<4\) is false, while the last holds because \(4\le4\le6\). The assigned value is \(3(4)-5=12-5=7\).
As an interior check, \(x=0\) satisfies \(-2\le0<1\), so \(f(0)=2(0)+4=4\). The ranges cover the full domain: they meet at \(1\) and \(4\), with each transition included in exactly one neighboring range. Thus the definition follows the verbal conditions on the entire stated domain.
Why These Conditions Form a Proper Partition
The basic structure can be justified for any finite list of increasing transition values. Suppose the allowed inputs form the interval \([a,b]\), and the transition values satisfy \(a<c_1<c_2<\cdots<c_n<b\). One consistent assignment is to include each transition in the range to its right, using half-open intervals until the last range.
These conditions cover \([a,b]\) without overlap. To see why, take any \(x\in[a,b]\). If \(x<c_1\), it belongs to the first interval. Otherwise, move along the increasing transition values until reaching the first interval whose upper endpoint exceeds \(x\), or the final interval if there is no such earlier interval. This places \(x\) in one of the listed intervals. At a transition \(x=c_i\), the interval ending at \(c_i\) excludes it, while the next interval begins at \(c_i\) and includes it. Therefore no transition is omitted.
For uniqueness, suppose \(x\) belonged to two different listed intervals. Those intervals are ordered and meet only at a transition value. The interval on the left excludes that transition, while the interval on the right includes it. So \(x\) cannot belong to both. Every input is covered, and no input is assigned twice.
This particular left-inclusive convention is useful, but it is not compulsory. Verbal conditions may instead assign a transition to the interval on its left. For example, “at most \(c\)” includes \(c\) in the range ending at \(c\). The essential requirement is to translate the words accurately and assign each allowed input once, not to use one endpoint convention in every problem.
Worked Example: Reading Three Ranges on a Time Domain
Worked Example: Reading Three Ranges on a Time Domain
A sensor’s reported value \(S(t)\), where \(t\) is time in minutes, is defined for \(0\le t\le9\). Its calibration description says: before \(2\) minutes, report \(2t+5\); from \(2\) minutes up to but not including \(6\) minutes, report \(3t+3\); from \(6\) minutes through \(9\) minutes, report \(40-t\). Write the rule and check the transition inputs.
“Before \(2\)” means \(t<2\), and the domain supplies \(t\ge0\), giving \(0\le t<2\). “From \(2\) up to but not including \(6\)” means \(2\le t<6\). “From \(6\) through \(9\)” includes both endpoints, giving \(6\le t\le9\). Thus:
At \(t=2\), the first condition excludes the input and the second includes it. Therefore \(S(2)=3(2)+3=6+3=9\). At \(t=6\), the second condition excludes the input and the third includes it. Therefore \(S(6)=40-6=34\).
Check an interior input in each range: \(t=1\) gives \(S(1)=2(1)+5=7\); \(t=4\) gives \(S(4)=3(4)+3=12+3=15\); and \(t=8\) gives \(S(8)=40-8=32\). These checks confirm that each sample is evaluated using the formula attached to its own condition, and that both transition inputs have an assigned formula.
Worked Example: When the Boundary Belongs to the Left
Worked Example: When the Boundary Belongs to the Left
Suppose a rule is defined for \(-5\le x\le5\). It uses \(x+6\) for inputs no greater than \(-2\), uses \(2-x\) for inputs greater than \(-2\) and no greater than \(3\), and uses \(x^2\) for inputs greater than \(3\). Translate the description into a piecewise definition.
“No greater than \(-2\)” means \(x\le-2\), which becomes \(-5\le x\le-2\) after applying the domain. “Greater than \(-2\) and no greater than \(3\)” means \(-2<x\le3\). “Greater than \(3\)” means \(x>3\), which becomes \(3<x\le5\). The conditions deliberately include \(-2\) and \(3\) in the ranges to their left.
At \(x=-2\), the first condition applies, giving \(g(-2)=-2+6=4\). The second condition does not apply because \(-2<-2\) is false. At \(x=3\), the second condition applies, giving \(g(3)=2-3=-1\); the third condition does not apply because \(3>3\) is false. An interior check at \(x=4\) gives \(g(4)=4^2=16\). The intervals cover the domain from \(-5\) to \(5\), and each transition is assigned once in accordance with the words.
A Common Mistake: Giving a Boundary to Both Rules
A frequent error is to make both neighboring ranges inclusive at the same transition. For instance, a mistaken attempt to divide inputs at \(x=2\) might use \(x\le2\) for one formula and \(x\ge2\) for the next. Then \(x=2\) satisfies both conditions.
If the first formula is \(3x\) and the second is \(x+10\), the two instructions at \(x=2\) give different results:
That does not define a single function value at \(2\). Correct the conditions by assigning the boundary to one side, in a way that matches the verbal description. If the wording is “below \(2\)” and “at least \(2\),” the correct conditions are \(x<2\) and \(x\ge2\). Then \(x=2\) belongs only to the second range, which gives \(2+10=12\).
The opposite error is a gap: using \(x<2\) and \(x>2\) leaves \(x=2\) out of both ranges. Checking the transition itself catches either problem. At each boundary, ask: Which condition includes this value? Does any other condition include it too?
Why Careful Translation Matters
A piecewise definition is a compact mathematical record of a conditional rule. If the inequalities are wrong, the definition may fail exactly where the input range changes, even when every formula is correct. This can lead to conflicting outputs, missing cases, or a domain that differs from the one described.
Before accepting a constructed rule, make three checks:
- Formula check: Does each formula match the verbal instruction for its range?
- Coverage check: Does every input in the stated domain satisfy at least one condition?
- Boundary check: Does every transition input satisfy exactly one condition?
These checks separate two tasks that are easy to confuse: calculating with a formula and deciding which formula applies. A reliable piecewise rule does both unambiguously.
Check Your Understanding
For each question, pay particular attention to what happens at the transition value.
- Translate “less than \(5\)” and “at least \(5\)” into inequalities. Which condition contains \(5\)?
- A rule applies “from \(-1\) through \(3\), but not including \(3\).” Write its input condition.
- For a domain \([0,8]\), a description says “below \(4\)” and “\(4\) or greater.” Write conditions that cover the domain exactly once.
- What is the problem with using \(x\le c\) and \(x\ge c\) for two neighboring ranges when the formulas give different values at \(c\)?
- A rule says “no greater than \(2\),” then “greater than \(2\).” Which range contains the boundary, and why?
