If f(x) = 2x + 3 and (f∘g)(x) = 6x − 1, then g(x) is
- (f∘g)(x) = f(g(x)) = 2g(x) + 3.
- Set this equal to 6x − 1: 2g(x) + 3 = 6x − 1, so 2g(x) = 6x − 4.
- Divide by 2: g(x) = 3x − 2. Check: 2(3x − 2) + 3 = 6x − 1.
Answer: g(x) = 3x − 2
Sets and functions
NUST NET Maths · Maths · Numbers, sets and functions. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.
The idea
Set questions are counting questions. A Venn diagram shows the union A ∪ B (in A or B or both), the intersection A ∩ B (in both) and the complement A′ (not in A). When two groups overlap, the overlap is counted twice unless you subtract it once.
A function gives each input exactly one output. Its domain is the set of allowed inputs and its range is the set of outputs it actually produces. To find a domain, look for what must be avoided: a zero denominator, or a negative number under a square root.
A composite f∘g means g acts first, then f. An inverse undoes the function and exists only when the function is one-one.
Rules to remember
- n(A ∪ B) = n(A) + n(B) − n(A ∩ B)
- A set with n elements has 2n subsets, of which 2n − 1 are proper subsets
- De Morgan's laws: (A ∪ B)′ = A′ ∩ B′ and (A ∩ B)′ = A′ ∪ B′
- Domain: denominator ≠ 0; expression under a square root ≥ 0 (strictly > 0 if the root is in a denominator)
- (f∘g)(x) = f(g(x)): work out g first, then put the result into f
- Inverse: write y = f(x), swap x and y, solve for y; domain of f−1 = range of f
- Even: f(−x) = f(x). Odd: f(−x) = −f(x), so an odd function defined at 0 has f(0) = 0
Common mistake
Working out f∘g in the wrong order. (f∘g)(x) means f(g(x)), so g is applied first; f(x) × g(x) and g(f(x)) are different things.
Worked example
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