ACT Math: Functions (Drill 4)

Question 1. A linear function f passes through the points (0, −4) and (2, 0). What is f(6)?

• A) 4
• B) 6
• C) 8 ✓
• D) 10

Explanation: The slope is (0 − (−4)) / (2 − 0) = 4/2 = 2. The y-intercept is −4 (given by the point (0, −4)). So f(x) = 2x − 4. Evaluate at x = 6: f(6) = 2(6) − 4 = 12 − 4 = 8. Choice A results from computing f(4) = 2(4) − 4 = 4, evaluating at x = 4 instead of x = 6. Choice B results from using slope 1 instead of 2: f(x) = x − 4, f(6) = 2... or from computing 6 + 0 = 6 without using the function. Choice D results from computing f(6) = 2(6) − 2 = 10, using the x-intercept (2) as the y-intercept.

Question 2. A cell phone plan charges $30 per month for up to 5 gigabytes (GB) of data, plus $10 for each additional GB beyond 5. Which of the following piecewise functions gives the monthly cost C(g), in dollars, for g gigabytes of data used?

• A) C(g) = 10g, if g ≤ 5; C(g) = 30 + 10g, if g > 5
• B) C(g) = 30, if g ≤ 5; C(g) = 30 + 10g, if g > 5
• C) C(g) = 30, if g ≤ 5; C(g) = 30 + 10(g − 5), if g > 5 ✓
• D) C(g) = 30 + 10g, if g ≤ 5; C(g) = 30 + 10(g − 5), if g > 5

Explanation: For g ≤ 5 GB, the cost is the flat rate: C(g) = 30. For g > 5 GB, the extra charge applies only to the gigabytes beyond 5: C(g) = 30 + 10(g − 5). Check: g = 7: 30 + 10(7 − 5) = 30 + 20 = $50 ✓. Choice A charges $10 per GB even within the 5 GB allowance, ignoring the flat rate structure entirely. Choice B correctly uses $30 for g ≤ 5, but for g > 5 it charges $10 for every gigabyte used (including the first 5 that are already covered): C(7) = 30 + 10(7) = $100 instead of $50. Choice D incorrectly adds overage charges even for usage within the 5 GB plan.

Question 3. The table below gives values for functions f and g.
x        1    2    3    4    5
f(x)    3    5    2    1    4
g(x)    4    1    5    2    3
What is the value of f(g(3))?

• A) 2
• B) 3
• C) 4 ✓
• D) 5

Explanation: Work from the inside out. First find g(3): from the table, g(3) = 5. Then find f(5): from the table, f(5) = 4. So f(g(3)) = f(5) = 4. Choice A results from reversing the composition: g(f(3)) = g(2) = 1... that gives 1, not 2. Actually Choice A (2) results from computing f(3) = 2 and stopping, without applying g first. Choice B results from reading g(f(3)): f(3) = 2, g(2) = 1... that is 1, not 3. Choice D results from reporting g(3) = 5 as the final answer without evaluating f of that result.

Question 4. The graph of f(x) is transformed to produce the graph of g(x) = −3f(x) + 2. Which of the following describes the transformation?

• A) Reflected across the x-axis, stretched vertically by a factor of 3, and shifted up 2 units ✓
• B) Stretched vertically by a factor of 3 and shifted right 2 units
• C) Reflected across the y-axis, stretched vertically by a factor of 3, and shifted up 2 units
• D) Reflected across the x-axis and shifted up 2 units, with no stretch

Explanation: In g(x) = −3f(x) + 2, three transformations are applied to f(x): the factor of −3 multiplies f(x) by 3 (vertical stretch by factor 3) and the negative sign reflects the graph across the x-axis; the +2 shifts the graph up 2 units. The x-axis reflection comes from the negative coefficient, not a change inside the function argument. Choice B omits the reflection and misidentifies +2 as a horizontal shift, adding a constant outside the function always shifts vertically, not horizontally. Choice C incorrectly identifies the reflection as across the y-axis; a reflection across the y-axis would come from f(−x), not −f(x). Choice D correctly identifies the reflection and vertical shift but ignores the factor of 3 entirely.

Question 5. If f(x) = 3^x and g(x) = log₃(x), what is f(g(81))?

• A) 3
• B) 4
• C) 27
• D) 81 ✓

Explanation: Work from the inside out. First evaluate g(81) = log₃(81). Since 3⁴ = 81, log₃(81) = 4. Then evaluate f(4) = 3⁴ = 81. So f(g(81)) = 81. This illustrates that f and g are inverse functions, f(g(x)) = x for all valid x. In general, a^log_a(x) = x. Choice A results from stopping at g(81) = 4 and then computing f(4) = 3 by subtracting 1 from the exponent. Choice B results from stopping at g(81) = 4 and reporting that as the final answer without evaluating f(4). Choice C results from computing 3³ = 27, using the wrong exponent of 3 instead of 4.