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ACT Math: Functions (Drill 2)

Drill 2 · Math · Functions

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About This Drill

ACT Math: Functions (Drill 2) is a Math practice drill covering Functions. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.

This ACT Functions drill covers evaluating piecewise functions, modeling exponential growth in context, solving for an input given a specified output, evaluating logarithmic expressions, and performing algebraic composition of two functions.

Questions & Explanations

Question 1. A function f(x) is defined as follows:
f(x) = 3x + 1, if x < 2
f(x) = x2 − 3, if x ≥ 2

What is f(−1) + f(4)?

  • A) 9
  • B) 11 ✓
  • C) 13
  • D) 15

Explanation: Since −1 < 2, use the first rule: f(−1) = 3(−1) + 1 = −3 + 1 = −2. Since 4 ≥ 2, use the second rule: f(4) = 42 − 3 = 16 − 3 = 13. Sum: −2 + 13 = 11. Choice A (9) results from a sign error in the first rule, computing 3(−1) − 1 = −4 instead of −3 + 1 = −2, giving −4 + 13 = 9. Choice C (13) results from applying the second rule incorrectly to f(−1): (−1)2 − 3 = −2, and then making an arithmetic error on f(4) to get 15, giving −2 + 15 = 13. Choice D (15) results from correctly evaluating f(4) = 13 but making a sign error on f(−1) that produces +2 instead of −2, giving 2 + 13 = 15.

Question 2. The number of users on a new app is modeled by N(t) = 1,200 · (1.5)t, where t is the number of months since launch. What does the value 1,200 represent in this model?

  • A) The number of users added each month
  • B) The number of users after 1.5 months
  • C) The number of users at launch ✓
  • D) The rate at which users increase per month

Explanation: In an exponential model N(t) = a · bt, the coefficient a is the initial value, the output when t = 0. N(0) = 1,200 · (1.5)0 = 1,200 · 1 = 1,200. This is the number of users at launch (t = 0). Choice A confuses the coefficient with an additive rate of change, that would describe a linear model. Choice B confuses the coefficient with an output at t = 1.5. Choice D confuses the coefficient with the growth factor; the growth rate is represented by 1.5 (a 50% increase per month), not 1,200.

Question 3. If h(x) = 4x − 9 and h(a) = 19, what is the value of a?

  • A) 2.5
  • B) 5
  • C) 7 ✓
  • D) 10

Explanation: Set h(a) = 19: 4a − 9 = 19. Add 9 to both sides: 4a = 28. Divide by 4: a = 7. Choice A (2.5) results from subtracting 9 from 19 instead of adding it: 4a = 19 − 9 = 10, then 10 ÷ 4 = 2.5. Choice B (5) results from adding only 1 instead of 9 to both sides: 4a = 19 + 1 = 20, then 20 ÷ 4 = 5. Choice D (10) results from correctly reaching 4a = 28 but then making an arithmetic error in the final division, recording 28 ÷ 4 = 10 instead of 7.

Question 4. If f(x) = log2(x + 4), what is f(28)?

  • A) 3
  • B) 4
  • C) 5 ✓
  • D) 8

Explanation: Substitute x = 28: f(28) = log2(28 + 4) = log2(32). Ask: 2 to what power equals 32? Since 25 = 32, log2(32) = 5. Choice A results from computing log2(28) ≈ log2(23 · 3.5) ≈ 3, forgetting to add 4 first. Choice B results from computing log2(16) = 4, perhaps substituting x = 12 instead of x = 28. Choice D results from computing log2(28 + 4) = log2(32) but then reporting 32 ÷ 4 = 8 rather than evaluating the logarithm.

Question 5. Let f(x) = 2x + 1 and g(x) = x2 − 4. Which of the following is equivalent to f(g(x))?

  • A) 2x2 − 7
  • B) 2x2 − 3 ✓
  • C) 4x2 + 4x − 3
  • D) 2x2 + x − 4

Explanation: f(g(x)) means substitute g(x) into f: f(g(x)) = 2(g(x)) + 1 = 2(x2 − 4) + 1 = 2x2 − 8 + 1 = 2x2 − 7. Choice B results from computing 2(x2 − 4) + 5 = 2x2 − 3, adding 5 instead of 1, a common error when students substitute g(x) into f(x) = 2x + 5 by mistake. Choice C results from computing g(f(x)) instead of f(g(x)): g(f(x)) = (2x + 1)2 − 4 = 4x2 + 4x + 1 − 4 = 4x2 + 4x − 3. Choice D results from distributing only partially: 2(x2) + 1 − 4 = 2x2 + x − 4, adding x incorrectly.