Let f(x) = x² and g(x) = x + 3. A student claims f(g(x)) = g(f(x)) because 'the same two functions are involved either way.' What is the correct evaluation of each at x = 2?
Af(g(2)) = 25, g(f(2)) = 25 — they are equal at x = 2
Bf(g(2)) = 25, g(f(2)) = 7 — they differ because composition is not commutative
Cf(g(2)) = 10, g(f(2)) = 10 — composition is commutative for polynomials
Df(g(2)) = 7, g(f(2)) = 25 — the inner function always produces the larger result
f(g(2)) = f(5) = 25; g(f(2)) = g(4) = 7. The student's claim is wrong — composition is generally NOT commutative. The inner function executes first, and swapping which function is 'inner' changes the computation entirely. Always track the order carefully: f(g(x)) means apply g first, then f to whatever g produced.
Question 2 Multiple Choice
Let g(x) = √x and f(x) = 1/(x − 4). What values must be excluded from the domain of f(g(x))?
AOnly x = 4, because f is undefined there
BOnly x < 0, because g is undefined for negative inputs
Cx < 0 and x = 4, because both sources of restriction apply
Dx < 0 and x = 16, because g(x) = 4 when x = 16, making f undefined
The domain of f(g(x)) has two sources of restriction: (1) g(x) = √x requires x ≥ 0; (2) f(u) = 1/(u − 4) is undefined when u = 4, and g(x) = 4 only when x = 16. So x = 16 must also be excluded. The trap in option B is forgetting to check whether the inner function's output lands in a gap in the outer function's domain — both sources of trouble must be caught.
Question 3 True / False
For f(x) = x² and g(x) = x + 1, the composition f(g(x)) equals x² + 2x + 1.
TTrue
FFalse
Answer: True
f(g(x)) = f(x + 1) = (x + 1)² = x² + 2x + 1. This is correct. The inner function g produces (x + 1), and the outer function f squares whatever it receives. Note that g(f(x)) = g(x²) = x² + 1 — a completely different function — which illustrates why knowing the order of composition matters.
Question 4 True / False
If f(g(x)) is defined at x = 3, then g(f(x)) is necessarily defined at x = 3 as well.
TTrue
FFalse
Answer: False
The two compositions can have entirely different domains. f(g(x)) requires x to be in the domain of g and g(x) to be in the domain of f. g(f(x)) requires x to be in the domain of f and f(x) to be in the domain of g — a completely separate set of conditions. A value that satisfies the first pair of conditions need not satisfy the second pair.
Question 5 Short Answer
Why does mastering function composition matter for calculus, and what specific skill should you practice now to prepare for it?
Think about your answer, then reveal below.
Model answer: The chain rule — the most-used differentiation rule — says that if h(x) = f(g(x)), then h'(x) = f'(g(x)) · g'(x). To apply it, you must identify which part of a complex expression is the 'outer' function f and which is the 'inner' function g. For example, in h(x) = sin(x²), f(u) = sin(u) is outer and g(x) = x² is inner. The skill to practice now is decomposition: given a complicated expression, name its layers.
Function composition is the conceptual engine behind the chain rule. Students who struggle with the chain rule in calculus almost always struggle because they can't identify which function is inside which — they treat the whole expression as a single flat object. Practicing decomposition (given h, find f and g such that h = f ∘ g) directly trains the skill the chain rule demands.