Grade 7-12 Math Review · Grade 11-12 · Gen Math: Functions
Operations on Functions and Composition
Composition is not multiplication. The inner function runs first.
Evaluate
f(x) = 2x − 1, g(x) = x². Find (f ∘ g)(3)
(f ∘ g)(3) means f(g(3)). Inside first: g(3) = 9. Then f(9) = 2(9) − 1 = 17. C does it backwards: g(f(3)) = g(5) = 25. D multiplies the two outputs, 5 × 9 = 45, which is (f · g)(3), a different operation. A stops at f(3) and never uses g at all.
1
Add, subtract and multiply by combining the rules.
Work out both outputs, then do the arithmetic. Nothing new to learn.
(f + g)(x)= f(x) + g(x)add the two rules
(f − g)(x)= f(x) − g(x)watch the sign on every term
(f · g)(x)= f(x) × g(x)multiply the outputs
f(x) = x + 4, g(x) = x² − 6
find (f − g)(x)
= (x + 4) − (x² − 6)
bracket the whole of g
= x + 4 − x² + 6 = −x² + x + 10
the minus reaches both terms
Bracket the second function before subtracting
Writing x + 4 − x² − 6 loses the sign on the −6 and gives −x² + x − 2. The bracket
is the only thing standing between those two answers.
2
Division throws away part of the domain.
f(x) = x + 1, g(x) = x − 4
find (fg)(x)
(fg)(x) = x + 1x − 4
the rule is easy
x ≠ 4
where g is zero
Both f and g are fine at x = 4 on their own. It is the division that breaks, so the
restriction belongs to the quotient and has to be stated with it.
3
Composition feeds one function into the other.
(f ∘ g)(x) = f(g(x)). Read it inside out: g runs first.
The symbol ∘ is not a multiplication dot. (f · g)(3) is 5 × 9 = 45; (f ∘ g)(3) is 17.
They are different questions with different answers.
4
The order almost never commutes.
f(x) = 3x + 2, g(x) = x²
build both compositions
(f ∘ g)(x) = f(x²) = 3x² + 2
square, then triple and add 2
(g ∘ f)(x) = g(3x + 2) = (3x + 2)²
triple and add 2, then square
= 9x² + 12x + 4
not the same function
Put a shirt on, then a jacket. Reverse the order and you get a different result — and
in maths, usually a different formula.
5
Variations: a function inside itself, a pair that undoes, two holes in the domain.
A function composed with itself: f(x) = 2x + 3, find (f ∘ f)(x).
(f ∘ f)(x) = f(2x + 3)
the inner f goes in whole
= 2(2x + 3) + 3
double it, add 3
= 4x + 9
expand and tidy
at x = 1: f(1) = 5, f(5) = 13, and 4 + 9 = 13 ✓
check with a number
f ∘ f is not f², the square of f. It means "do f, then do f again".
A pair that undoes each other: f(x) = 3x − 6 and g(x) = x3 + 2, find f(g(x)).
f(g(x)) = 3(x3 + 2) − 6
g goes in whole
= x + 6 − 6
3 times x3 is x
= x
everything cancels
When a composition gives back plain x, each function undoes the other. That is the idea behind inverse functions, two lessons on.
Two holes in the domain: f(x) = x + 5 and g(x) = x² − 4, state the domain of (fg)(x).
(fg)(x) = x + 5x² − 4
the rule
x² − 4 = 0 when x = 2 or x = −2
where the bottom is zero
all real x except 2 and −2
state both
A squared denominator can vanish at two places. Missing the negative one is the usual slip.
6
Composition is how two rules stack in real life.
d(x) = 0.85x, v(x) = 1.12x
discount, then VAT
v(d(1200)) = v(1020) = 1142.40
discount first
d(v(1200)) = d(1344) = 1142.40
VAT first
₱1,142.40 either way
These two DO commute, because both are multiplications: 0.85 × 1.12 = 1.12 × 0.85.
That is the exception, not the rule. Change the discount to a flat ₱200 off and the
order starts to matter a great deal.
On the exam
Periodical exams ask for a composition at a number, a composition as a formula, or the value excluded from a quotient; entrance tests like to swap the order of f and g to see who notices. At a number, work inside out — the function on the right first. To check a formula, put x = 1 into it and into f(g(1)) worked out the long way: the two must match.
Remember the flow
Combine rules for + − × ÷ · Exclude where the bottom is zero · For ∘, work inside out · Check the order
∘ is composition, · is multiplication. They are never interchangeable.
Now you try
Practice problem
Evaluate
f(x) = x + 4, g(x) = 3x. Find (g ∘ f)(2)
(g ∘ f)(2) = g(f(2)). Inside first: f(2) = 6. Then g(6) = 3(6) = 18. B is the reverse composition, f(g(2)) = f(6) = 10. D multiplies instead of composing: (f · g)(2) = 6 × 6 = 36. A stops at f(2) and never reaches g. The function on the right of the ∘ always runs first.
Hint: the function written on the right of the ∘ is the one
that runs first. Here that is f, so work out f(2) before you touch g.
