Separate the input, rule, and output
The expression f(3) asks for the output when the input is 3. The rule may be given by a formula, a table, or several formulas with conditions. For f(x) = x² − 4x, evaluating f(−2) gives (−2)² − 4(−2) = 4 + 8 = 12. Parentheses preserve the sign of the input through both substitutions. If the rule is a table, locate the input first and read across to its output. An output appearing in another row does not determine where to start.
Worked example at a rule boundary
Suppose a service calculates a balance adjustment using p(t) = t + 4 when t ≤ 0 and p(t) = 2t + 1 when t > 0. At zero the first condition includes equality, so p(0) = 4. At one the second condition applies, giving p(1) = 3. The formulas do not have to give the same value at the boundary because the conditions choose exactly one rule there. Substituting into both and averaging would invent an output that neither condition specifies.
Keep a restriction after cancellation
The expression (x − 5)(x + 1)/(x + 1) can be simplified to x − 5 only when x + 1 is nonzero. At x = −1, the original quotient has denominator zero, so it is undefined. A new function defined by x − 5 for every real input would have an output there. A calculation using the shortened rule must carry the excluded input along with it. Write the restriction before canceling to avoid accidentally changing which function you are studying.
Separate square-root and denominator conditions
A real square root can have a radicand equal to zero. In the expression √(2x − 6), that allows x = 3. If the entire square root is a denominator, as in 1/√(2x − 6), then x = 3 would cause division by zero. In that case require 2x − 6 > 0 and conclude x > 3. When an expression has several restrictions, the allowed inputs must satisfy all of them. Test a boundary directly to see which inequality belongs in the answer.
Transfer check with a square root in the denominator
Find the domain of (x + 8)/√(x − 1). Check x = 1 in the original expression before deciding whether to include it. The radicand must be positive, so x > 1. Zero is allowed in a real square root on its own, but here it would put zero in the denominator. As a final check, x = 2 gives a real, nonzero denominator. Any input below one gives a negative radicand. The numerator adds no restriction.
Try it, then explain it.
Complete a short set without looking at the example. After each answer, explain which operation preserved the relationship and why the result makes sense.
Sources & authorship
Original OpenExam explanations and examples. References support the concepts; they are not the source of copied questions. Independent educator review is pending.
College Board ACCUPLACER Math test contentOpenStax Intermediate Algebra 2eOpenStax College Algebra 2e