Math: Advanced Functions Flashcards

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?

Answer: 4

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.

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

Answer: x² − 4

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. ✓

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?

Answer: −3

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.

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?

Answer: −1

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.

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

Answer: (4, 2)

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.

Let f(x) = (3x − 2) / (x + 1) for x ≠ −1. If f⁻¹ denotes the inverse of f, what is f⁻¹(f⁻¹(1))?

Answer: 7/3

Find f⁻¹: set y = (3x − 2)/(x + 1), then y(x + 1) = 3x − 2 → xy + y = 3x − 2 → x(y − 3) = −(y + 2) → x = (y + 2)/(3 − y). So f⁻¹(x) = (x + 2)/(3 − x). Step 1: f⁻¹(1) = (1 + 2)/(3 − 1) = 3/2. Step 2: f⁻¹(3/2) = (3/2 + 2)/(3 − 3/2) = (7/2)/(3/2) = 7/3.