AP Precalculus: Composition of Functions (Drill 1)

Question 1. The tables below give selected values of f and g.

x	1	2	3	4
f(x)	3	1	4	2

x	1	2	3	4
g(x)	4	3	2	1

What is the value of f(g(2))?

• A) 1
• B) 2
• C) 3
• D) 4 ✓

Explanation: Choice D is correct. To evaluate f(g(2)), work from the inside out. First find g(2): from the g-table, g(2) = 3. Then find f(3): from the f-table, f(3) = 4. So f(g(2)) = 4. Choice A is incorrect because 1 = f(2); the student applied f directly to 2 without first evaluating g(2). Choice B is incorrect because 2 = f(4) = f(g(1)); the student used x = 1 as the input instead of x = 2. Choice C is incorrect because 3 = g(2); the student correctly evaluated the inner function but stopped there and never applied f.

Question 2. Let f(x) = 2x − 1 and g(x) = x² + 3. Which of the following is an expression for (f ∘ g)(x)?

• A) 2x² + 5 ✓
• B) 4x² − 4x + 4
• C) 2x² + 2
• D) 2x³ + 6x − 1

Explanation: Choice A is correct. (f ∘ g)(x) = f(g(x)). Substitute g(x) = x² + 3 into f: f(x² + 3) = 2(x² + 3) − 1 = 2x² + 6 − 1 = 2x² + 5. Choice B is incorrect because 4x² − 4x + 4 = (2x − 1)² + 3 = g(f(x)), not f(g(x)); the student reversed the composition order. Choice C is incorrect because 2x² + 2 results from computing 2(x² + 3) − 4; the student subtracted 4 instead of 1 from the last step. Choice D is incorrect because 2x³ + 6x − 1 = f(x) · g(x), the product of f and g; the student multiplied the functions instead of composing them.

Question 3. Let f(x) = √x − 4 and g(x) = x². What is the domain of (f ∘ g)(x)?

• A) x ≥ 0
• B) x ≥ 4
• C) x ≤ −2 or x ≥ 2 ✓
• D) x ≥ 2

Explanation: Choice C is correct. (f ∘ g)(x) = f(g(x)) = f(x²) = √(x² − 4). For the square root to be defined, the radicand must be non-negative: x² − 4 ≥ 0, so x² ≥ 4, which means |x| ≥ 2. This gives x ≤ −2 or x ≥ 2. Choice A is incorrect because x ≥ 0 ignores the additional restriction imposed by f; it assumes the only domain constraint comes from g, but the square root in f requires the output of g to be at least 4. Choice B is incorrect because x ≥ 4 would apply if the inner function were x rather than x²; the student applied the domain restriction of f directly to x without first substituting g(x) = x². Choice D is incorrect because x ≥ 2 captures only the positive branch; since x² ≥ 4 is also satisfied for all x ≤ −2, the student incorrectly ignored the symmetric negative solution.

Question 4. A company's monthly profit P (in thousands of dollars) depends on the number of units sold n according to P(n) = 0.4n − 8. The number of units sold depends on the price per unit p (in dollars) according to n(p) = 200 − 5p. Which of the following expresses the monthly profit as a function of price?

• A) P = 0.4(200 − 5p − 8)
• B) P = 80 − 2p
• C) P = 72 − 2p ✓
• D) P = 0.4(200) − 5p − 8

Explanation: Choice C is correct. Substitute n(p) = 200 − 5p into P(n) = 0.4n − 8: P = 0.4(200 − 5p) − 8 = 80 − 2p − 8 = 72 − 2p. Choice A is incorrect because it places the −8 inside the argument to be multiplied by 0.4 rather than subtracting it afterward; the student incorrectly treated −8 as part of n rather than as the separate constant in P(n). Choice B is incorrect because 80 − 2p omits the −8 term entirely; the student correctly distributed 0.4 across n(p) but dropped the additive constant from P(n). Choice D is incorrect because it distributes 0.4 only to the constant 200 and then subtracts 5p separately, showing a failure to multiply the entire expression (200 − 5p) by 0.4.

Question 5. Functions f and g are defined by f(x) = \(\dfrac{1}{x-2}\) and g(x) = \(\dfrac{3}{x}\). For what value(s) of x is (f ∘ g)(x) undefined?

• A) x = 2 only
• B) x = 0 only
• C) x = 0 and x = 2
• D) x = 0 and x = (3/2) ✓

Explanation: Choice D is correct. (f ∘ g)(x) = f(g(x)) = f(3/x) = 1/((3/x) − 2). This expression is undefined in two cases. First, g(x) = 3/x is undefined when x = 0. Second, f is undefined when its input equals 2, so set g(x) = 2: 3/x = 2, giving x = 3/2. Therefore (f ∘ g)(x) is undefined at x = 0 and x = 3/2. Choice A is incorrect because x = 2 makes f(x) undefined when x is the direct input to f, but in this composition the input to f is g(x) = 3/x; when x = 2, g(2) = 3/2 ≠ 2, so f(g(2)) is defined. Choice B is incorrect because x = 0 is only one of the two restrictions; it identifies the domain gap of g but ignores the additional value of x that causes g(x) to equal 2. Choice C is incorrect because x = 2 is not a restriction in this composition (see above), while x = 3/2 is; the student correctly identified one gap from g but substituted x = 2 directly into f rather than solving g(x) = 2.