Lesson 10 of 35 · Quantitative Reasoning: algebra

Functions, sequences and custom operations

Evaluate and compose functions, apply operations defined in the question, and work with arithmetic, geometric and recursive sequences.

About 22 minutes

In this lesson

  • Evaluate f(x)f(x) for numbers and expressions, and compose two functions in the right order.
  • Apply a symbol defined in the question exactly as written, including nested uses.
  • Find terms and sums of arithmetic sequences, and terms of geometric and recursive ones.
  • Use repeating patterns, such as units digits, to reach far-off terms.

The idea

A function is a rule for substitution

f(x)=2x2−3f(x) = 2x^2 - 3 is an instruction: whatever goes in the brackets replaces every xx. So f(3)=2(9)−3=15f(3) = 2(9) - 3 = 15, and f(a+1)=2(a+1)2−3f(a + 1) = 2(a + 1)^2 - 3. When the input is an expression, put it in brackets before you substitute; most errors here come from writing 2a+122a + 1^2 instead of 2(a+1)22(a + 1)^2.

Function questions in the problem-solving format usually ask one of four things: evaluate at a number, evaluate at an expression, solve f(x)=kf(x) = k for xx, or compare ff at two inputs. All four are substitution followed by ordinary algebra.

Composition: inside first

f(g(x))f(g(x)) means: apply gg first, then apply ff to the result. Work from the inside out. With f(x)=2x2−3f(x) = 2x^2 - 3 and g(x)=x+4g(x) = x + 4:

f(g(−1))=f(3)=15,g(f(−1))=g(−1)=3f(g(-1)) = f(3) = 15, \qquad g(f(-1)) = g(-1) = 3

The two orders give different answers, which is normal. Composition is not commutative, so read the brackets carefully.

Custom operations: the question defines the rules

Some questions invent a symbol, for instance "for all numbers aa and bb, let a⋆b=ab−a+ba \star b = ab - a + b". The symbol has no meaning beyond that definition. Treat it as a function of two inputs: the first number replaces aa, the second replaces bb.

Three habits help:

  • Copy the definition and substitute slowly. 5⋆2=10−5+2=75 \star 2 = 10 - 5 + 2 = 7, while 2⋆5=10−2+5=132 \star 5 = 10 - 2 + 5 = 13. Order matters unless you can show it does not.
  • Nested symbols go inside out, like composition. In 3⋆(2⋆5)3 \star (2 \star 5), compute 2⋆52 \star 5 first.
  • Questions about properties ("for which xx does x⋆3=3⋆xx \star 3 = 3 \star x?") become equations once you write both sides from the definition.

Arithmetic sequences

In an arithmetic sequence each term adds the same common difference dd:

an=a1+(n−1)da_n = a_1 + (n - 1)d

The (n−1)(n - 1) is the point people miss: from term 1 to term nn there are n−1n - 1 steps. Two known terms give dd directly: if a4=17a_4 = 17 and a10=41a_{10} = 41, six steps cover 2424, so d=4d = 4.

The sum of the first nn terms is the number of terms times the average of the first and last:

Sn=n(a1+an)2S_n = \frac{n(a_1 + a_n)}{2}

That formula works because the terms pair up from the outside in, each pair adding to a1+ana_1 + a_n. It also gives sums of runs of consecutive integers: 1+2+⋯+50=50×512=12751 + 2 + \dots + 50 = \frac{50 \times 51}{2} = 1275.

Geometric sequences

Each term multiplies by a common ratio rr: an=a1r n−1a_n = a_1 r^{\,n-1}. A sequence 3, 6, 12, 24, … has r=2r = 2, and its 8th term is 3⋅27=3843 \cdot 2^7 = 384. To find rr from two terms, divide them: if a2=6a_2 = 6 and a5=162a_5 = 162, then r3=27r^3 = 27 and r=3r = 3.

Recursive definitions

A recursive rule defines each term from the previous one, such as a1=3a_1 = 3 and an+1=3an−4a_{n+1} = 3a_n - 4. There is usually no shortcut: write the terms one by one (3,5,11,29,…3, 5, 11, 29, \dots) and look for a pattern before computing many.

Patterns that repeat

Units digits of powers repeat in short cycles. The units digits of 71,72,73,747^1, 7^2, 7^3, 7^4 are 7,9,3,17, 9, 3, 1, and then the cycle starts again. To find the units digit of 7437^{43}, divide the exponent by the cycle length 4: 43=4×10+343 = 4 \times 10 + 3, so 7437^{43} ends like 737^3, in 3. The same idea, list until it repeats then use the remainder, settles many questions about the 100th term of a repeating pattern.

Worked examples

Example 1: Composition in both orders

If f(x)=2x2−3f(x) = 2x^2 - 3 and g(x)=x+4g(x) = x + 4, find f(g(−1))f(g(-1)) and g(f(−1))g(f(-1)).

  1. Inside first for f(g(−1))f(g(-1)): g(−1)=−1+4=3g(-1) = -1 + 4 = 3, then f(3)=2(9)−3=15f(3) = 2(9) - 3 = 15.

  2. Inside first for g(f(−1))g(f(-1)): f(−1)=2(1)−3=−1f(-1) = 2(1) - 3 = -1, then g(−1)=3g(-1) = 3.

  3. The results differ because each order applies the rules to different inputs.

Answer

f(g(−1))=15f(g(-1)) = 15 and g(f(−1))=3g(f(-1)) = 3. Computing f(−1)f(-1) and g(−1)g(-1) and multiplying them (−1×3=−3-1 \times 3 = -3) is a misreading of composition.

Example 2: A nested custom operation

For all numbers aa and bb, a⋆b=ab−a+ba \star b = ab - a + b. What is 3⋆(2⋆5)3 \star (2 \star 5)?

  1. Innermost first: 2⋆5=2(5)−2+5=132 \star 5 = 2(5) - 2 + 5 = 13.

  2. Then 3⋆13=3(13)−3+13=39−3+13=493 \star 13 = 3(13) - 3 + 13 = 39 - 3 + 13 = 49.

  3. Note that 5⋆2=10−5+2=75 \star 2 = 10 - 5 + 2 = 7, not 13: the operation is not commutative, so swapping the inner inputs would change the answer.

Answer

3⋆(2⋆5)=493 \star (2 \star 5) = 49. Using 5⋆2=75 \star 2 = 7 by mistake gives 3⋆7=21−3+7=253 \star 7 = 21 - 3 + 7 = 25.

Example 3: Two terms, then a sum

In an arithmetic sequence, the 4th term is 17 and the 10th term is 41. What is the sum of the first 20 terms?

  1. From term 4 to term 10 there are 6 steps: 6d=41−17=246d = 41 - 17 = 24, so d=4d = 4.

  2. Step back three times from the 4th term: a1=17−3(4)=5a_1 = 17 - 3(4) = 5.

  3. The 20th term: a20=5+19(4)=81a_{20} = 5 + 19(4) = 81.

  4. Sum: S20=20(5+81)2=10×86=860S_{20} = \frac{20(5 + 81)}{2} = 10 \times 86 = 860. Check dd with the 10th term: 5+9(4)=415 + 9(4) = 41.

Answer

The sum is 860. Using a20=a1+20d=85a_{20} = a_1 + 20d = 85 instead gives 900, the n−1n - 1 slip.

Example 4: A units digit far out

What is the units digit of 7437^{43}?

  1. List units digits: 71→77^1 \to 7, 72=49→97^2 = 49 \to 9, 73=343→37^3 = 343 \to 3, 74=2401→17^4 = 2401 \to 1, then 75→77^5 \to 7 again. The cycle has length 4.

  2. Divide the exponent by 4: 43=4×10+343 = 4 \times 10 + 3, remainder 3.

  3. Remainder 3 means the same position as 737^3, whose units digit is 3.

Answer

The units digit is 3. A remainder of 0 would match 747^4 (units digit 1), not 707^0; mixing that up is the usual slip.

Common traps

  • Substituting an expression without brackets

    For f(x)=x2−2xf(x) = x^2 - 2x, f(a+1)f(a + 1) is (a+1)2−2(a+1)(a + 1)^2 - 2(a + 1), not a2+1−2a+1a^2 + 1 - 2a + 1. Write the brackets first, then expand.

  • Assuming a custom operation is commutative

    Most invented operations are not symmetric. a⋆ba \star b and b⋆ab \star a must each be computed from the definition; test with numbers before swapping.

  • Using nn steps instead of n−1n - 1

    The nnth term of an arithmetic sequence is a1+(n−1)da_1 + (n - 1)d. Count gaps, not terms: from term 3 to term 8 is 5 steps.

  • Forgetting to halve in the sum formula

    Sn=n(a1+an)2S_n = \frac{n(a_1 + a_n)}{2}: the sum is the number of terms times the average. Without the 2 you have doubled the sum.

Check yourself

Answer each question before you open its answer.

Question 1: A function of an expression

Problem Solving: algebra

If f(x)=x2−2xf(x) = x^2 - 2x, which of the following is equal to f(a+1)f(a + 1)?

  1. A.a2−2a−1a^2 - 2a - 1
  2. B.a2−2a+1a^2 - 2a + 1
  3. C.a2−1a^2 - 1
  4. D.a2+1a^2 + 1
  5. E.a2+3a^2 + 3
Show the answer

Answer: C. a2−1a^2 - 1

Working

f(a+1)=(a+1)2−2(a+1)=a2+2a+1−2a−2=a2−1f(a + 1) = (a + 1)^2 - 2(a + 1) = a^2 + 2a + 1 - 2a - 2 = a^2 - 1. Check with a=2a = 2: f(3)=9−6=3f(3) = 9 - 6 = 3, and 22−1=32^2 - 1 = 3.

Why the other options are wrong

  • A. This squares a+1a + 1 as a2+1a^2 + 1, losing the middle term 2a2a.
  • B. This is f(a)+1f(a) + 1: the 1 was added to the output instead of the input.
  • D. This multiplies only the aa by −2-2, writing −2(a+1)-2(a + 1) as −2a-2a, so the −2-2 goes missing.
  • E. This writes −2(a+1)-2(a + 1) as −2a+2-2a + 2, a sign slip on the constant.

What carries over

Bracket the input, substitute it for every xx, expand, then test with a number.

Question 2: When does the order not matter?

Problem Solving: algebra

For all numbers aa and bb, a⊕b=a2−2ba \oplus b = a^2 - 2b. If x⊕3=3⊕xx \oplus 3 = 3 \oplus x and x≠3x \neq 3, what is the value of xx?

  1. A.−5-5
  2. B.−3-3
  3. C.33
  4. D.55
  5. E.Any value of xx
Show the answer

Answer: A. −5-5

Working

From the definition, x⊕3=x2−6x \oplus 3 = x^2 - 6 and 3⊕x=9−2x3 \oplus x = 9 - 2x. Setting them equal: x2+2x−15=0x^2 + 2x - 15 = 0, so (x+5)(x−3)=0(x + 5)(x - 3) = 0 and x=−5x = -5 or x=3x = 3. The question excludes 3, so x=−5x = -5. Check: 25−6=1925 - 6 = 19 and 9+10=199 + 10 = 19.

Why the other options are wrong

  • B. This comes from moving −2x-2x across without changing its sign, giving x2−2x−15=0x^2 - 2x - 15 = 0, whose roots are 5 and −3-3.
  • C. x=3x = 3 does satisfy the equation trivially, but the question excludes it.
  • D. This is the other root of the sign-slipped equation x2−2x−15=0x^2 - 2x - 15 = 0; check: 25−6=1925 - 6 = 19 but 9−10=−19 - 10 = -1.
  • E. This assumes the operation is commutative. It is not: x⊕3x \oplus 3 and 3⊕x3 \oplus x are different expressions.

What carries over

Write both sides from the definition, then solve the ordinary equation that results.

Question 3: Sum of the first ten terms

Problem Solving: algebra

In an arithmetic sequence, the 3rd term is 11 and the 8th term is 31. What is the sum of the first 10 terms?

  1. A.170
  2. B.210
  3. C.230
  4. D.390
  5. E.420
Show the answer

Answer: B. 210

Working

Five steps from the 3rd to the 8th term: 5d=205d = 20, so d=4d = 4 and a1=11−2(4)=3a_1 = 11 - 2(4) = 3. Then a10=3+9(4)=39a_{10} = 3 + 9(4) = 39 and S10=10(3+39)2=210S_{10} = \frac{10(3 + 39)}{2} = 210. Check by listing: 3,7,11,15,19,23,27,31,35,393, 7, 11, 15, 19, 23, 27, 31, 35, 39, which add to 210.

Why the other options are wrong

  • A. This finds the first term as 11−3d=−111 - 3d = -1, stepping back three times from the 3rd term instead of twice.
  • C. This takes the 10th term as a1+10d=43a_1 + 10d = 43 instead of a1+9d=39a_1 + 9d = 39.
  • D. This multiplies the number of terms by the last term, 10×3910 \times 39, instead of by the average term.
  • E. This is 10×(3+39)10 \times (3 + 39): the sum formula without dividing by 2.

What carries over

Count steps between terms, then sum as number of terms times the average of first and last.

Key points

  • Bracket every input before substituting it into a function or a defined operation.

  • Composition and nested operations go from the inside out, and order matters.

  • Arithmetic: an=a1+(n−1)da_n = a_1 + (n - 1)d and Sn=n(a1+an)2S_n = \frac{n(a_1 + a_n)}{2}.

  • For far-off terms of a repeating pattern, find the cycle length and use the remainder.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

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