ORIGINAL WORKED EXAMPLE
Apply the idea
If f(t)=2t²−3t+1, find f(−2).
- Substitute t=−2 in every occurrence.
- 2(−2)²−3(−2)+1=8+6+1.
- Add the terms.
Answer 15.
GMAT · Algebra and constraints · TOPIC LESSON
Functions, sequences and defined operations
Original GMAT Problem Solving instruction: reason from constraints, calculate efficiently without a calculator, and check the requested quantity.
BY THE END, YOU CAN
- Explain and apply function notation and evaluation.
- Explain and apply compositions.
- Explain and apply domain restrictions.
- Explain and apply defined operations.
- Explain and apply arithmetic sequences.
- Explain and apply geometric patterns.
- Explain and apply recursive rules.
- Explain and apply algebraic pattern inference.
BEFORE YOU START
- Substitution, powers, fractions and simple equations.
Original educational lesson. Work through the examples before revealing the practice answers. This original lesson supports current GMAT study through an editorial skill map. It does not reproduce official questions, predict exam content or guarantee a score.
LEARN THROUGH ORIGINAL PRACTICE
Original GMAT-style worked examples and practice
All passages, examples and exercises here are original. No official GMAC questions, choices or explanations are reproduced. This independent preparation resource is not affiliated with or endorsed by GMAC. GMAT is a trademark of the Graduate Management Admission Council.
01 / UNDERSTAND & APPLY
A function is a rule applied to an input
Function notation f(x) names the output of a rule, not multiplication by f. Replace every occurrence of the input variable with the complete supplied input. If f(x)=x²−2x, then f(a+1)=(a+1)²−2(a+1), with parentheses protecting grouping.
An equation f(x)=c asks which inputs produce an output; evaluating f(c) asks for the output at a given input. These are different tasks. Functions can send multiple inputs to the same output without violating the definition of a function.
ORIGINAL PRACTICE · 1
If f(x)=x²+1, solve f(x)=10 over the reals.
Reveal answer & reasoning
x=−3 or x=3.
Set x²+1=10, giving x²=9. Both signed inputs give output 10. Computing f(10)=101 answers a different question.
02 / UNDERSTAND & APPLY
Composition applies the inner rule first
The composition f(g(x)) means first compute g(x), then use that output as the input to f. The order usually matters. Write the inner result in parentheses before substitution to avoid applying the outer rule to only part of it.
The input must belong to the domain of the inner function, and the inner output must belong to the domain of the outer function. A simplified composition may hide one of these original restrictions.
ORIGINAL WORKED EXAMPLE
Apply the idea
Let f(x)=3x−1 and g(x)=x². Find f(g(−2)) and g(f(−2)).
- g(−2)=4, so f(g(−2))=f(4)=11.
- f(−2)=−7, so g(f(−2))=g(−7)=49.
- The different answers show composition is not generally commutative.
Answer 11 and 49, respectively.
ORIGINAL PRACTICE · 1
If f(x)=x+4 and g(x)=2x, find f(g(x))−g(f(x)).
Reveal answer & reasoning
−4.
f(g(x))=2x+4, while g(f(x))=2(x+4)=2x+8. Subtracting gives −4. Assuming the order does not matter would give zero.
03 / UNDERSTAND & APPLY
The domain contains only valid inputs
For real-valued rules, denominators must be nonzero and even-root radicands must be nonnegative. When both occur, intersect the restrictions. A root in a denominator must be strictly positive because zero is excluded there.
Domain restrictions come from the original rule and any contextual requirements. A formula might accept negative numbers algebraically while a model for the number of customers requires a nonnegative integer. Do not infer extra restrictions merely because a variable has a familiar name.
ORIGINAL WORKED EXAMPLE
Apply the idea
Find the real domain of f(x)=√(x−2)/(x−5).
- The radical requires x−2≥0, so x≥2.
- The denominator requires x≠5.
- Both conditions must hold simultaneously.
Answer x≥2 with x≠5.
ORIGINAL PRACTICE · 1
What is the real domain of g(x)=1/√(4−x)?
Reveal answer & reasoning
x<4.
The square root must exist and cannot be zero in the denominator. Thus 4−x>0. Including x=4 would leave division by zero.
04 / UNDERSTAND & APPLY
A new symbol has only its stated definition
A problem may define a custom operation such as a★b=2a+b². The symbol does not inherit commutativity, associativity or ordinary multiplication rules unless the definition implies them. Replace the symbol with its rule each time.
For nested operations, follow parentheses and apply the definition separately at each stage. If the definition contains a denominator or root, carry the corresponding restrictions. A cleverly chosen counterexample can disprove a claimed property.
ORIGINAL WORKED EXAMPLE
Apply the idea
Define a★b=2a−b. Calculate (3★4)★5 and 3★(4★5).
- 3★4=6−4=2, so (3★4)★5=4−5=−1.
- 4★5=8−5=3, so 3★(4★5)=6−3=3.
- The different results show the operation is not associative.
Answer −1 and 3, respectively.
ORIGINAL PRACTICE · 1
Define a◇b=a²+b. Is 2◇3 equal to 3◇2?
Reveal answer & reasoning
No: 7 versus 11.
2◇3=4+3=7 and 3◇2=9+2=11. The custom symbol does not automatically share the commutative property of addition.
05 / UNDERSTAND & APPLY
A constant difference leads to a linear nth term
An arithmetic sequence adds a fixed difference d each step. If the first term is a₁, then aₙ=a₁+(n−1)d because reaching term n requires n−1 changes. The first term is indexed 1 unless the problem says otherwise.
The sum of n arithmetic terms is n(first+last)/2, since symmetrically placed pairs have equal sums. This formula requires equally spaced terms; do not use it for arbitrary increasing lists.
ORIGINAL WORKED EXAMPLE
Apply the idea
An arithmetic sequence begins 7,11,15,… . Find its twentieth term and the sum of its first twenty terms.
- The common difference is 4. a₂₀=7+19×4=83.
- The average of first and last terms is (7+83)/2=45.
- Multiply by 20 terms.
Answer Twentieth term 83; sum 900.
ORIGINAL PRACTICE · 1
The fifth term of an arithmetic sequence is 18 and its difference is 3. What is the first term?
Reveal answer & reasoning
6.
Four steps separate the first and fifth terms, so a₁=18−4×3=6. Subtracting five differences creates an index error.
06 / UNDERSTAND & APPLY
A constant ratio leads to exponential change
A geometric sequence multiplies by a fixed ratio r each step: aₙ=a₁r^(n−1). A negative ratio alternates signs; a positive ratio between 0 and 1 shrinks positive terms toward zero. A constant difference and a constant ratio describe different models.
For a finite sum, small sequences can be listed or paired strategically. When r≠1, the sum is a₁(1−r^n)/(1−r), derived by subtracting r times the sum from the sum. For r=1, every term equals a₁ and the sum is na₁.
ORIGINAL WORKED EXAMPLE
Apply the idea
A geometric sequence begins 3,6,12,… . Find its sixth term and the sum of its first six terms.
- The ratio is 2, so a₆=3×2⁵=96.
- List the six terms: 3,6,12,24,48,96.
- Their sum is 3(2⁶−1)=3×63.
Answer Sixth term 96; sum 189.
ORIGINAL PRACTICE · 1
A geometric sequence has first term 80 and ratio 1/2. What is its fifth term?
Reveal answer & reasoning
5.
There are four ratio applications: 80×(1/2)⁴=80/16=5. Applying five halvings would give the sixth term instead.
07 / UNDERSTAND & APPLY
A recurrence needs a starting value
A recursive rule defines a term from earlier terms. Compute in order, retaining the given initial values and index convention. A rule involving two previous terms needs enough initial values to start it.
A recurrence need not be arithmetic or geometric. The rule aₙ₊₁=2aₙ+1 doubles and then adds, so successive ratios are not constant. Sometimes shifting the terms reveals structure, but verify any proposed closed form against both the rule and the initial term.
ORIGINAL WORKED EXAMPLE
Apply the idea
Let a₁=2 and aₙ₊₁=2aₙ+1. Find a₄.
- a₂=2×2+1=5.
- a₃=2×5+1=11.
- a₄=2×11+1=23.
Answer 23.
ORIGINAL PRACTICE · 1
Let b₁=1, b₂=3 and bₙ=bₙ₋₁+bₙ₋₂ for n≥3. Find b₅.
Reveal answer & reasoning
11.
b₃=3+1=4, b₄=4+3=7, and b₅=7+4=11. Reusing only the first term in each calculation would violate the two-term recurrence.
08 / UNDERSTAND & APPLY
Finite observations do not uniquely define a rule
A finite list can fit many different continuation rules. If a question explicitly states that a sequence is arithmetic, geometric or obeys a recurrence, use that condition. Without a rule, observing the first few terms alone cannot prove a unique next term.
When deriving an algebraic pattern, test the starting index and the transition. A formula that matches two terms can still fail later. In a well-specified Problem Solving question, exploit the supplied rule rather than inventing one based only on appearance.
ORIGINAL WORKED EXAMPLE
Apply the idea
A sequence is defined by aₙ=n²+n for positive integers n. Find aₙ₊₁−aₙ and the fifth term.
- aₙ₊₁=(n+1)²+(n+1)=n²+3n+2.
- Subtract n²+n to get 2n+2.
- a₅=25+5=30.
Answer Difference 2n+2; fifth term 30.
ORIGINAL PRACTICE · 1
A list starts 2,4,8. With no further rule, must the next term be 16?
Reveal answer & reasoning
No.
Doubling gives 16, but the rule aₙ=n²−n+2 gives the same first three terms and then 14. The displayed values alone do not establish a unique continuation; an explicit model is required.
KEEP THE BIG IDEAS
Quick revision
- Function notation gives a rule, not multiplication.
- Apply inner functions first and preserve original domains.
- Count n−1 changes from the first term to term n.
- Distinguish a stated rule from an unsupported pattern guess.
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