Bluebook SAT Math: Advanced Functions Questions and Answers Flashcards

Let f(x) = (x + 1)/(x − 1) for x ≠ 1. What is f(f(f(x)))?

Answer: (x + 1)/(x − 1)

First compute f(f(x)): substituting u = (x+1)/(x−1) into f gives f(u) = (u+1)/(u−1) = [(x+1)/(x−1) + 1] / [(x+1)/(x−1) − 1] = [2x/(x−1)] / [2/(x−1)] = x. So f(f(x)) = x, meaning f is its own inverse (an involution). Therefore f(f(f(x))) = f(x) = (x+1)/(x−1).

Let g(x) = x² − 4x + 7 with domain x ≥ 2. What is g⁻¹(3)?

Answer: 2

Completing the square: g(x) = (x − 2)² + 3. On the domain x ≥ 2, the minimum value is 3, achieved exactly at x = 2. So g(2) = 3, which means g⁻¹(3) = 2. The domain restriction x ≥ 2 is essential — without it, g would not be one-to-one and the inverse would not exist.

A polynomial p(x) has zeros at x = −2 (multiplicity 2), x = 1 (multiplicity 1), and x = 3 (multiplicity 1). If p(0) = −12, what is the leading coefficient of p(x)?

Answer: −1

From the zeros and multiplicities: p(x) = a(x + 2)²(x − 1)(x − 3). Substituting x = 0: p(0) = a(2)²(−1)(−3) = a · 4 · 1 · 3 = 12a. Setting this equal to −12 gives 12a = −12, so a = −1.

Which statement about the graph of h(x) = (x² − 9)/(x² − x − 6) is true?

Answer: It has a hole at x = 3 and a vertical asymptote at x = −2

Factor completely: x² − 9 = (x − 3)(x + 3) and x² − x − 6 = (x − 3)(x + 2). The factor (x − 3) cancels, giving h(x) = (x + 3)/(x + 2) for x ≠ 3. The canceled factor creates a removable discontinuity (hole) at x = 3, while the remaining denominator factor (x + 2) creates a vertical asymptote at x = −2.

Let g be an odd function and h be an even function, both defined for all real numbers. Which of the following must be an odd function?

Answer: g(g(x))

For g(g(x)): evaluate at −x → g(g(−x)) = g(−g(x)) [since g is odd] = −g(g(x)) [since g is odd again]. So g(g(−x)) = −g(g(x)), confirming g∘g is odd. By contrast: h(g(−x)) = h(−g(x)) = h(g(x)) [h is even], so h∘g is even. Similarly g(h(−x)) = g(h(x)) [h is even], so g∘h is even. And g(x) + h(x) is neither odd nor even in general.

A function f satisfies f(x) + f(1 − x) = 1 for all real numbers x. If f(1/3) = 2/5, what is the value of f(2/3)?

Answer: 3/5

Substitute x = 1/3 into the functional equation: f(1/3) + f(1 − 1/3) = 1, which gives f(1/3) + f(2/3) = 1. Since f(1/3) = 2/5, we get 2/5 + f(2/3) = 1, so f(2/3) = 3/5. The functional equation creates a pairing between x and (1 − x), forcing their outputs to sum to 1.