Bluebook SAT Test Math: Advanced Functions 5

Question 1

Let f(x) = (x² − 4) / (x − 2) for x ≠ 2, and f(2) = k. For what value of k is f continuous at x = 2?

Accepted Answer

4

Answer

Factor the numerator: x² − 4 = (x − 2)(x + 2), so (x² − 4)/(x − 2) simplifies to x + 2 for x ≠ 2. As x approaches 2, this limit equals 4. Defining f(2) = 4 fills the removable discontinuity, making f continuous. Any other value of k would create a jump at x = 2.

Question 2

If f(x) = 2x + 1 and g(f(x)) = 4x² + 4x − 3, which of the following defines g(x)?

Accepted Answer

x² − 4

Answer

Let u = f(x) = 2x + 1, so x = (u − 1)/2. Substitute into 4x² + 4x − 3: 4·((u−1)/2)² + 4·((u−1)/2) − 3 = (u−1)² + 2(u−1) − 3 = u² − 2u + 1 + 2u − 2 − 3 = u² − 4. So g(x) = x² − 4. Verify: g(f(1)) = g(3) = 9 − 4 = 5, and 4(1) + 4 − 3 = 5. ✓

Question 3

The function h is defined by h(x) = f(g(x)), where f(x) = √(x − 3) and g(x) = x² + 2x. What is the smallest integer value of x for which h(x) is defined?

Accepted Answer

−3

Answer

h(x) = √(g(x) − 3) = √(x² + 2x − 3). For the square root to be defined, x² + 2x − 3 ≥ 0. Factoring: (x + 3)(x − 1) ≥ 0, which holds when x ≤ −3 or x ≥ 1. Among the answer choices, −3 satisfies this (it's a boundary of the left solution interval), and it is the smallest listed valid integer.

Question 4

A function f satisfies f(f(x)) = x for all real x, and f(3) = 7. If f(x) = ax + b for constants a and b, what is the value of a?

Accepted Answer

−1

Answer

Since f(f(x)) = x, f is its own inverse. Computing f(f(x)) = a(ax + b) + b = a²x + b(a + 1). Setting this equal to x requires a² = 1 (so a = ±1) and b(a + 1) = 0. If a = 1, then f(x) = x, but f(3) = 3 ≠ 7. So a = −1, which satisfies b(0) = 0 for any b. Then f(3) = −3 + b = 7 gives b = 10, consistent with a = −1.

Question 5

The graph of y = f(x) passes through (2, 5). The graph of y = f(−x + 6) − 3 must pass through which point?

Accepted Answer

(4, 2)

Answer

We need the input to f to equal 2, so set −x + 6 = 2, giving x = 4. At x = 4: y = f(−4 + 6) − 3 = f(2) − 3 = 5 − 3 = 2. The transformed graph passes through (4, 2). The replacement x → −x + 6 reflects over x = 3, and the −3 shifts the output down by 3.

Bluebook SAT Practice Test

Bluebook SAT Practice Test

Bluebook SAT Test Math: Advanced Functions 5