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AP Precalculus Quiz

AP Precalculus Quiz: Composition Of Functions

Practice Composition Of Functions in AP Precalculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 12

0 of 12 answered

A sound sensor models intensity by g(x)=10xg(x)=10^xg(x)=10x and converts to decibels with f(I)=10log⁡10(I)f(I)=10\log_{10}(I)f(I)=10log10​(I). Based on the scenario, what is f(g(3))f(g(3))f(g(3))?

Select an answer to continue

What this quiz covers

This quiz focuses on Composition Of Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A sound sensor models intensity by g(x)=10xg(x)=10^xg(x)=10x and converts to decibels with f(I)=10log⁡10(I)f(I)=10\log_{10}(I)f(I)=10log10​(I). Based on the scenario, what is f(g(3))f(g(3))f(g(3))?

  1. 303030 (correct answer)
  2. 300300300
  3. 10log⁡10(3)10\log_{10}(3)10log10​(3)
  4. 10310^3103

Explanation: This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential functions and logarithmic decibel conversion. Composition requires applying one function to the output of another, where g(x) = 10^x models intensity and f(I) = 10log₁₀(I) converts to decibels. In this scenario, we need to find f(g(3)), which means first calculating g(3) = 10³ = 1000, then applying f to get the decibel level. Choice A is correct because f(g(3)) = f(1000) = 10log₁₀(1000) = 10·3 = 30 decibels. Choice D is incorrect because it gives g(3) = 1000 instead of f(g(3)), confusing the intensity value with the decibel measurement. To help students: Draw diagrams showing the flow from input through each function, practice with the decibel formula, and emphasize that log₁₀(10^n) = n is a key logarithm property.

Question 2

A culture grows as g(t)=500⋅2t/3g(t)=500\cdot 2^{t/3}g(t)=500⋅2t/3 and analysis uses f(x)=ln⁡(x)f(x)=\ln(x)f(x)=ln(x). Using the function defined, what is f(g(6))f(g(6))f(g(6))?​

  1. ln⁡(2000)\ln(2000)ln(2000) (correct answer)
  2. ln⁡(1000)\ln(1000)ln(1000)
  3. ln⁡(500)+2\ln(500)+2ln(500)+2
  4. ln⁡(500)+ln⁡(2)\ln(500)+\ln(2)ln(500)+ln(2)

Explanation: This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and natural logarithm functions. Composition f(g(x)) means we evaluate g(x) first, then apply f to that result, requiring careful attention to function order and properties. In this scenario, a culture grows as g(t) = 500·2^(t/3) and analysis uses f(x) = ln(x), so we need to find f(g(6)). Choice A is correct because g(6) = 500·2^(6/3) = 500·2² = 500·4 = 2000, so f(g(6)) = ln(2000). Choice B is incorrect because it represents ln(1000), which might come from miscalculating the exponential growth. To help students: Emphasize evaluating the inner function completely before applying the outer function, practice with specific numerical values, and reinforce properties of exponential functions like 2^(6/3) = 2².

Question 3

A population grows as g(t)=300e0.04tg(t)=300e^{0.04t}g(t)=300e0.04t and analysis uses f(x)=ln⁡(x/300)f(x)=\ln(x/300)f(x)=ln(x/300). Using the function defined, what is f(g(t))f(g(t))f(g(t))?

  1. ln⁡(300)+0.04t\ln(300)+0.04tln(300)+0.04t
  2. 0.04t0.04t0.04t (correct answer)
  3. ln⁡(300e0.04t)\ln(300e^{0.04t})ln(300e0.04t)
  4. ln⁡(t)0.04\dfrac{\ln(t)}{0.04}0.04ln(t)​

Explanation: This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential growth with base e and natural logarithm transformations. Composition requires substituting g(t) into f(x) and simplifying using logarithm properties, particularly ln(e^x) = x. In this scenario, population grows as g(t) = 300e^(0.04t) and f(x) = ln(x/300) normalizes relative to initial population, requiring careful algebraic manipulation. Choice B is correct because f(g(t)) = ln(300e^(0.04t)/300) = ln(e^(0.04t)) = 0.04t, using the fundamental property that ln and e are inverse functions. Choice A is incorrect because it fails to simplify ln(e^(0.04t)) = 0.04t, keeping the unnecessary ln(300) term. To help students: Emphasize the inverse relationship between ln and e, practice simplifying expressions like ln(ae^x/a) = ln(e^x) = x, and reinforce that f removes the initial population factor.

Question 4

A sound meter models intensity by I(d)=106−0.2dI(d)=10^{6-0.2d}I(d)=106−0.2d and converts to decibels with L(x)=10log⁡10(x)L(x)=10\log_{10}(x)L(x)=10log10​(x). Based on the scenario, what is (L∘I)(10)(L\circ I)(10)(L∘I)(10)?

  1. 404040 (correct answer)
  2. 808080
  3. 444
  4. 10410^{4}104

Explanation: This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. Composition involves applying one function to the result of another, denoted as (L∘I)(10) = L(I(10)). In this scenario, sound intensity decreases exponentially with distance, and the logarithmic function converts intensity to decibels. To find (L∘I)(10), first calculate I(10) = 10^(6-0.2×10) = 10^(6-2) = 10^4, then apply L: L(10^4) = 10log₁₀(10^4) = 10×4 = 40. Choice A is correct because it accurately computes the composition, yielding 40 decibels. Choice D might tempt students who stop at I(10) = 10^4 without applying the second function. To help students: emphasize the order of operations in composition, practice substituting step-by-step, and reinforce that log₁₀(10^n) = n.

Question 5

A population grows as P(t)=500⋅3tP(t)=500\cdot 3^tP(t)=500⋅3t and f(x)=log⁡3(x)f(x)=\log_3(x)f(x)=log3​(x). Using the function defined, what transformation occurs in (f∘P)(t)(f\circ P)(t)(f∘P)(t)?

  1. Vertical stretch by factor 500500500
  2. Vertical shift up by log⁡3(500)\log_3(500)log3​(500) (correct answer)
  3. Horizontal shift right by 500500500
  4. Reflection across the xxx-axis

Explanation: This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. The composition (f∘P)(t) = f(P(t)) = log₃(500×3^t) requires understanding logarithmic properties. In this population growth scenario, applying the logarithm to the exponential function reveals the transformation. Using logarithm properties: log₃(500×3^t) = log₃(500) + log₃(3^t) = log₃(500) + t. Choice B is correct because this represents a vertical shift up by log₃(500) units from the basic function y = t. Choice A incorrectly suggests a vertical stretch, but logarithms convert multiplication to addition. To help students: review logarithm properties like log(ab) = log(a) + log(b), graph both the original exponential and the composition, and recognize that log₃(3^t) = t creates a linear function with a vertical shift.

Question 6

A radioactive sample has N(t)=80(12)t/4N(t)=80\left(\tfrac12\right)^{t/4}N(t)=80(21​)t/4 grams and h(x)=log⁡1/2(x/80)h(x)=\log_{1/2}(x/80)h(x)=log1/2​(x/80). Based on the scenario, what is (h∘N)(12)(h\circ N)(12)(h∘N)(12)?

  1. −3-3−3
  2. 333 (correct answer)
  3. 121212
  4. log⁡1/2(80)−3\log_{1/2}(80)-3log1/2​(80)−3

Explanation: This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. Composition requires evaluating (h∘N)(12) = h(N(12)), applying h to the output of N. In this scenario, radioactive decay follows an exponential model, and the logarithmic function extracts the time information. First calculate N(12) = 80×(1/2)^(12/4) = 80×(1/2)^3 = 80×(1/8) = 10, then apply h: h(10) = log₁/₂(10/80) = log₁/₂(1/8) = log₁/₂((1/2)^3) = 3. Choice B is correct because log_b(b^x) = x applies here. Choice A might result from a sign error, as some students confuse logs with base less than 1. To help students: review that log₁/₂(1/8) asks 'what power of 1/2 gives 1/8?', practice with fractional bases, and verify answers by checking (1/2)^3 = 1/8.

Question 7

An investment grows as A(t)=500⋅20.5tA(t)=500\cdot 2^{0.5t}A(t)=500⋅20.5t dollars, and f(x)=log⁡2(x/500)f(x)=\log_2(x/500)f(x)=log2​(x/500) analyzes growth. Using the function defined, what is (f∘A)(6)(f\circ A)(6)(f∘A)(6)?

  1. 333 (correct answer)
  2. 666
  3. 222
  4. log⁡2(500)+3\log_2(500)+3log2​(500)+3

Explanation: This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. Composition requires applying f to the result of A, written as (f∘A)(6) = f(A(6)). In this scenario, an investment grows exponentially, and the logarithmic function analyzes the growth factor relative to the initial amount. First calculate A(6) = 500×2^(0.5×6) = 500×2^3 = 500×8 = 4000, then apply f: f(4000) = log₂(4000/500) = log₂(8) = log₂(2^3) = 3. Choice A is correct because it properly evaluates the composition. Choice B might result from confusing the input value 6 with the output. To help students: draw function diagrams showing the flow from input to output, practice identifying which function to apply first, and use the property that log_b(b^x) = x.

Question 8

A bacteria culture follows P(t)=200e0.3tP(t)=200e^{0.3t}P(t)=200e0.3t and g(x)=ln⁡(x/200)g(x)=\ln(x/200)g(x)=ln(x/200) recovers elapsed time. Based on the scenario, what is (g∘P)(5)(g\circ P)(5)(g∘P)(5)?

  1. 1.51.51.5 (correct answer)
  2. 555
  3. e1.5e^{1.5}e1.5
  4. ln⁡(200)+1.5\ln(200)+1.5ln(200)+1.5

Explanation: This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. Composition means applying g to the output of P, denoted (g∘P)(5) = g(P(5)). In this scenario, bacteria grow exponentially, and the logarithmic function recovers the elapsed time from the population count. Calculate P(5) = 200e^(0.3×5) = 200e^1.5, then apply g: g(200e^1.5) = ln(200e^1.5/200) = ln(e^1.5) = 1.5. Choice A is correct because it uses the fundamental property that ln(e^x) = x. Choice C shows e^1.5, which is P(5)/200, not the final composition result. To help students: emphasize that ln and e are inverse functions, practice recognizing when expressions simplify, and work through the composition step-by-step to avoid shortcuts.

Question 9

In a greenhouse, temperature is T(t)=20+15(1.1)tT(t)=20+15(1.1)^tT(t)=20+15(1.1)t and f(x)=log⁡1.1 ⁣(x−2015)f(x)=\log_{1.1}\!\left(\frac{x-20}{15}\right)f(x)=log1.1​(15x−20​). Using the function defined, what is (f∘T)(4)(f\circ T)(4)(f∘T)(4)?

  1. 444 (correct answer)
  2. 1.141.1^41.14
  3. log⁡1.1(20)+4\log_{1.1}(20)+4log1.1​(20)+4
  4. log⁡1.1(15)+4\log_{1.1}(15)+4log1.1​(15)+4

Explanation: This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. Composition means evaluating (f∘T)(4) = f(T(4)), where f is applied to the temperature at time 4. In this greenhouse scenario, temperature grows exponentially from a baseline, and the logarithmic function extracts the time from temperature readings. Calculate T(4) = 20 + 15(1.1)^4, then apply f: f(20 + 15(1.1)^4) = log₁.₁((20 + 15(1.1)^4 - 20)/15) = log₁.₁((1.1)^4) = 4. Choice A is correct because the functions are designed to be inverses of each other. Choice B shows (1.1)^4 without the logarithm applied. To help students: recognize when functions undo each other, practice identifying inverse relationships, and verify by checking that f recovers the original time input.

Question 10

Radioactive mass is N(t)=200e−0.4tN(t)=200e^{-0.4t}N(t)=200e−0.4t and g(x)=ln⁡(x/200)g(x)=\ln(x/200)g(x)=ln(x/200). Based on the scenario, solve for ttt in (g∘N)(t)=−2(g\circ N)(t)=-2(g∘N)(t)=−2.

  1. t=0.8t=0.8t=0.8
  2. t=5t=5t=5 (correct answer)
  3. t=−5t=-5t=−5
  4. t=2.5t=2.5t=2.5

Explanation: This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential and logarithmic functions. We need to solve (g∘N)(t) = -2, which means g(N(t)) = -2. In this radioactive decay scenario, the logarithmic function extracts time information from the remaining mass. Setting up: ln(200e^(-0.4t)/200) = -2, which simplifies to ln(e^(-0.4t)) = -2, giving -0.4t = -2, so t = 5. Choice B is correct because solving -0.4t = -2 yields t = 5. Choice A shows 0.8, which might result from arithmetic errors. To help students: use the property ln(e^x) = x to simplify, check the answer by substituting back (N(5) = 200e^(-2) and g(200e^(-2)) = ln(e^(-2)) = -2), and practice solving equations involving compositions of exponentials and logarithms.

Question 11

A sample decays by g(t)=100(12)tg(t)=100\left(\tfrac12\right)^tg(t)=100(21​)t and analysis uses f(x)=log⁡10(x)f(x)=\log_{10}(x)f(x)=log10​(x). Using the function defined, what is f(g(2))f(g(2))f(g(2))?​

  1. log⁡10(25)\log_{10}(25)log10​(25) (correct answer)
  2. log⁡10(50)\log_{10}(50)log10​(50)
  3. 2log⁡10(50)2\log_{10}(50)2log10​(50)
  4. log⁡10(100)−log⁡10(2)\log_{10}(100)-\log_{10}(2)log10​(100)−log10​(2)

Explanation: This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential decay and logarithmic analysis. Composition requires evaluating g(t) first to find the remaining amount, then applying f(x) = log₁₀(x) to analyze the scale. In this scenario, a sample decays by g(t) = 100(½)^t and we need to find f(g(2)), which means first calculating g(2) then applying the logarithm. Choice A is correct because g(2) = 100(½)² = 100(¼) = 25, so f(g(2)) = log₁₀(25). Choice B is incorrect because it gives log₁₀(50), which would be g(1) not g(2), confusing the decay rate. To help students: Emphasize evaluating functions step-by-step, practice with powers of ½ like (½)² = ¼, and verify answers by checking that log₁₀(25) ≈ 1.398.

Question 12

A greenhouse warms as g(t)=20+15(1.08)tg(t)=20+15(1.08)^tg(t)=20+15(1.08)t and uses f(x)=ln⁡(x−20)f(x)=\ln(x-20)f(x)=ln(x−20) to isolate growth. Using the function defined, what is f(g(t))f(g(t))f(g(t))?​

  1. \ln\!big(15(1.08)^t\u00big) (correct answer)
  2. \ln\!big(20+15(1.08)^t\u00big)
  3. \ln\!big(15+20(1.08)^t\u00big)
  4. \ln\!big(15(1.08)\u00big)^t

Explanation: This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential growth with a vertical shift and logarithmic transformation. Composition f(g(t)) requires substituting the entire expression g(t) into f(x), being careful with domain restrictions and algebraic manipulation. In this scenario, greenhouse temperature is g(t) = 20 + 15(1.08)^t and f(x) = ln(x - 20) isolates the growth component by removing the baseline temperature. Choice A is correct because f(g(t)) = ln(g(t) - 20) = ln((20 + 15(1.08)^t) - 20) = ln(15(1.08)^t). Choice B is incorrect because it includes the constant 20 inside the logarithm, failing to subtract it as required by f(x) = ln(x - 20). To help students: Emphasize careful substitution and simplification, highlight how f(x) = ln(x - 20) specifically removes the baseline value, and practice with functions that have domain restrictions.