Bluebook SAT Math: Advanced Functions Questions and Answers Flashcards

Let f(x) = (x + 1)/(x − 1) and g(x) = 1/x. Which of the following is equivalent to f(g(x)), for all x where the expression is defined?

Answer: (x + 1)/(1 − x)

Substitute g(x) = 1/x into f: f(1/x) = (1/x + 1)/(1/x − 1). Multiply numerator and denominator by x to get (1 + x)/(1 − x), which equals (x + 1)/(1 − x). Choice B is f(x) itself; choice A is 1/f(x); choice D is the negative reciprocal — all common errors when substituting rational expressions.

If f(x) = (2x − 3)/(x + 1), which of the following correctly expresses f⁻¹(x)?

Answer: (x + 3)/(2 − x)

Set y = (2x − 3)/(x + 1) and solve for x: y(x + 1) = 2x − 3 → xy + y = 2x − 3 → x(y − 2) = −(y + 3) → x = (y + 3)/(2 − y). Swapping variables gives f⁻¹(x) = (x + 3)/(2 − x). Choice D differs by a sign in the denominator — a subtle but critical error that students often miss when handling the negative.

A piecewise function is defined as f(x) = √(x + 4) for x ≥ −4, and f(x) = |x| − 2 for x < −4. What is the value of f(f(−7))?

Answer: 3

First, evaluate the inner function: f(−7). Since −7 < −4, use the second piece: f(−7) = |−7| − 2 = 7 − 2 = 5. Then evaluate the outer function: f(5). Since 5 ≥ −4, use the first piece: f(5) = √(5 + 4) = √9 = 3. Choice A (5) is the result of stopping after the first evaluation; choice D (√13) comes from incorrectly computing √(7 + 4 + 2) without following the correct order.

The graph of y = f(x) passes through the point (3, −2). Which point must lie on the graph of y = −f(x − 1) + 4?

Answer: (4, 6)

Apply each transformation in sequence. The substitution (x − 1) shifts the graph right by 1, so the x-coordinate moves from 3 to 4. The outer transformation −f(·) + 4 reflects the y-value and shifts it up: −(−2) + 4 = 2 + 4 = 6. The required point is (4, 6). Choice A reflects a leftward shift; choice C correctly shifts x but misapplies the vertical transformation; choice D forgets the horizontal shift entirely.

The function f(x) = ax² + bx + c has its vertex at (2, −3) and passes through the point (0, 5). What is the value of a + b + c?

Answer: −1

Write f in vertex form: f(x) = a(x − 2)² − 3. Use the point (0, 5): f(0) = a(0 − 2)² − 3 = 4a − 3 = 5, so a = 2. Thus f(x) = 2(x − 2)² − 3. Note that a + b + c = f(1) = 2(1 − 2)² − 3 = 2(1) − 3 = −1. Choice A (−3) is the vertex y-value; choice B (5) is f(0); both are plausible traps if students substitute the wrong x-value.

For what value(s) of k does the parabola f(x) = x² − kx + k² intersect the line y = x at exactly one point?

Answer: k = −1/3 or k = 1

Set x² − kx + k² = x, giving x² − (k + 1)x + k² = 0. For exactly one intersection point, the discriminant equals zero: (k + 1)² − 4k² = 0 → k² + 2k + 1 − 4k² = 0 → −3k² + 2k + 1 = 0 → 3k² − 2k − 1 = 0 → (3k + 1)(k − 1) = 0. So k = −1/3 or k = 1. Choice A flips the signs; choice C only gives one solution; choice D comes from incorrectly expanding the discriminant.