12 multiple-choice questions, progressively harder.
If fff and ggg are functions, what does (g∘f)(x)(g \circ f)(x)(g∘f)(x) mean?
Solution
Correct answer: D
The composition g∘fg \circ fg∘f applies fff first, then feeds the result to ggg.
(g∘f)(x)=g(f(x))(g \circ f)(x) = g(f(x))(g∘f)(x)=g(f(x))
The right function acts first, so fff runs before ggg; f(g(x))f(g(x))f(g(x)) would be the reverse composition f∘gf \circ gf∘g.
In the composition g∘fg \circ fg∘f, which function is applied first?
Correct answer: C
Reading (g∘f)(x)=g(f(x))(g \circ f)(x) = g(f(x))(g∘f)(x)=g(f(x)), the inner function is fff.
(g∘f)(x)=g(f(x)⏟first)(g \circ f)(x) = g(\underbrace{f(x)}_{\text{first}})(g∘f)(x)=g(firstf(x))
The function on the right, fff, acts first; the circle is composition, not multiplication.
Let f(x)=x+2f(x) = x + 2f(x)=x+2 and g(x)=3xg(x) = 3xg(x)=3x. What is (g∘f)(1)(g \circ f)(1)(g∘f)(1)?
Work from the inside out: apply fff first, then ggg.
f(1)=1+2=3,g(3)=3⋅3=9f(1) = 1 + 2 = 3, \qquad g(3) = 3 \cdot 3 = 9f(1)=1+2=3,g(3)=3⋅3=9
So (g∘f)(1)=9(g \circ f)(1) = 9(g∘f)(1)=9. The value 555 is the other order, (f∘g)(1)=f(3)(f \circ g)(1) = f(3)(f∘g)(1)=f(3).
Let f(x)=2xf(x) = 2xf(x)=2x and g(x)=x+5g(x) = x + 5g(x)=x+5. Which expression is (g∘f)(x)(g \circ f)(x)(g∘f)(x)?
Substitute f(x)=2xf(x) = 2xf(x)=2x into ggg.
(g∘f)(x)=g(2x)=2x+5(g \circ f)(x) = g(2x) = 2x + 5(g∘f)(x)=g(2x)=2x+5
The reverse order gives (f∘g)(x)=2(x+5)=2x+10(f \circ g)(x) = 2(x + 5) = 2x + 10(f∘g)(x)=2(x+5)=2x+10.
Let f(x)=x2f(x) = x^2f(x)=x2 and g(x)=x+1g(x) = x + 1g(x)=x+1. Which expression is (f∘g)(x)(f \circ g)(x)(f∘g)(x)?
Correct answer: A
In f∘gf \circ gf∘g the inner function is ggg, so substitute g(x)=x+1g(x) = x + 1g(x)=x+1 into fff.
(f∘g)(x)=f(x+1)=(x+1)2(f \circ g)(x) = f(x + 1) = (x + 1)^2(f∘g)(x)=f(x+1)=(x+1)2
The reverse order gives (g∘f)(x)=x2+1(g \circ f)(x) = x^2 + 1(g∘f)(x)=x2+1.
The composition f∘gf \circ gf∘g is the function that does which of these?
The notation f∘gf \circ gf∘g places ggg on the right, and the right function acts first.
(f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x))(f∘g)(x)=f(g(x))
So ggg runs first and fff second.
Let f(x)=x−4f(x) = x - 4f(x)=x−4 and g(x)=x+4g(x) = x + 4g(x)=x+4. What is (g∘f)(x)(g \circ f)(x)(g∘f)(x)?
Apply fff first, then ggg.
(g∘f)(x)=g(x−4)=(x−4)+4=x(g \circ f)(x) = g(x - 4) = (x - 4) + 4 = x(g∘f)(x)=g(x−4)=(x−4)+4=x
The added and subtracted 444 cancel, leaving xxx.
In this lesson, the notation f2f^2f2 stands for which function?
The superscript on a function name counts compositions, not a product.
f2=f∘ff^2 = f \circ ff2=f∘f
So f2(x)=f(f(x))f^2(x) = f(f(x))f2(x)=f(f(x)), which is generally different from the output square (f(x))2(f(x))^2(f(x))2.
Let f(x)=xf(x) = \sqrt{x}f(x)=x and g(x)=x+1g(x) = x + 1g(x)=x+1. What is the domain of g∘fg \circ fg∘f?
Correct answer: B
You need x∈domfx \in \operatorname{dom} fx∈domf, and g(x)=x+1g(x) = x + 1g(x)=x+1 accepts every real number.
x needs x≥0\sqrt{x} \text{ needs } x \ge 0x needs x≥0
So the only restriction is x≥0x \ge 0x≥0; the output x+1\sqrt{x} + 1x+1 places no further limit.
Let f(x)=2xf(x) = 2xf(x)=2x. Using f2=f∘ff^2 = f \circ ff2=f∘f, what is f2(3)f^2(3)f2(3)?
f2f^2f2 means f∘ff \circ ff∘f, so apply fff twice starting at 333.
f(3)=6,f(6)=12f(3) = 6, \qquad f(6) = 12f(3)=6,f(6)=12
So f2(3)=12f^2(3) = 12f2(3)=12; the value 363636 is the output square (f(3))2=62(f(3))^2 = 6^2(f(3))2=62, a different meaning of the superscript.
Let f(x)=x2f(x) = x^2f(x)=x2 and g(x)=x−3g(x) = x - 3g(x)=x−3. What is (g∘f)(x)(g \circ f)(x)(g∘f)(x)?
(g∘f)(x)=g(x2)=x2−3(g \circ f)(x) = g(x^2) = x^2 - 3(g∘f)(x)=g(x2)=x2−3
The reverse order gives (f∘g)(x)=(x−3)2(f \circ g)(x) = (x - 3)^2(f∘g)(x)=(x−3)2.
For the identity function id(x)=x\operatorname{id}(x) = xid(x)=x, both f∘idf \circ \operatorname{id}f∘id and id∘f\operatorname{id} \circ fid∘f equal which function?
The identity contributes nothing on either side of a composition.
f∘id=fandid∘f=ff \circ \operatorname{id} = f \quad\text{and}\quad \operatorname{id} \circ f = ff∘id=fandid∘f=f
So the identity is the neutral element for composition, like 000 for addition.
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