Bluebook SAT Test Math: Advanced Functions 4

Question 1

The piecewise function f(x) = { ax² + b, for x ≤ 1; 3x + 2, for x > 1 } is both continuous and differentiable at x = 1. What is the value of a + b?

Accepted Answer

5

Answer

Continuity at x = 1 requires a(1)² + b = 3(1) + 2 = 5, so a + b = 5 directly. The differentiability condition (equating the derivatives: 2a = 3) gives a = 3/2 and b = 7/2, but their sum is still 5. The key insight is that a + b is fully determined by the continuity condition alone — differentiability yields individual values but doesn't change their sum.

Question 2

Let f(x) = 1/(x − 3) and g(x) = √x. Which of the following correctly describes the domain of the composite function (f ∘ g)(x)?

Accepted Answer

x ≥ 0 and x ≠ 9

Answer

(f ∘ g)(x) = f(g(x)) = 1/(√x − 3). Two restrictions apply: the inner function g(x) = √x requires x ≥ 0 (not x > 0, since g(0) = 0 is valid and f(0) = −1/3 is defined). Additionally, the denominator √x − 3 ≠ 0 means x ≠ 9. Together, the domain is x ≥ 0 and x ≠ 9. Excluding x = 0 is a common error since f(g(0)) = 1/(0 − 3) = −1/3 is perfectly defined.

Question 3

Let f(x) = (x + 1)/(x − 2). Which expression is equivalent to f⁻¹(f(x + 1)) for all x in its domain?

Accepted Answer

x + 1

Answer

Since f⁻¹ is the inverse of f, f⁻¹(f(u)) = u for any input u. Here the input to f is (x + 1), so f⁻¹(f(x + 1)) = x + 1. This can be verified algebraically: f(x + 1) = (x + 2)/(x − 1), and applying f⁻¹(y) = (2y + 1)/(y − 1) yields (2(x+2)/(x−1) + 1)/((x+2)/(x−1) − 1) = (3x + 3)/3 = x + 1. The composition f⁻¹ ∘ f returns exactly what was fed into f.

Question 4

The function r(x) = (x² − x − 6)/(x² − 4) has a hole at x = a and a vertical asymptote at x = b. What is the value of a + b²?

Accepted Answer

2

Answer

Factor completely: x² − x − 6 = (x − 3)(x + 2) and x² − 4 = (x − 2)(x + 2). The common factor (x + 2) cancels, creating a removable discontinuity (hole) at x = −2, so a = −2. The remaining denominator factor (x − 2) = 0 gives a vertical asymptote at x = 2, so b = 2. Therefore a + b² = −2 + (2)² = −2 + 4 = 2.

Question 5

The graph of y = f(x) contains the point (3, −1), where f has a local maximum. Which ordered pair represents the corresponding point on the graph of y = −2f(x − 4) + 5?

Accepted Answer

(7, 7)

Answer

Apply transformations to (3, −1) step by step. The (x − 4) inside shifts the graph right 4, so the x-coordinate becomes 3 + 4 = 7. The y-coordinate transforms as: y_new = −2(−1) + 5 = 2 + 5 = 7. The answer is (7, 7). Note also that multiplying by −2 (negative) turns the local maximum into a local minimum on the transformed graph — a subtlety the question deliberately highlights.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Math: Advanced Functions 4