Math: Advanced Functions Flashcards

Let f(x) = (2x − 3)/(x + 1). What is the value of f⁻¹(f⁻¹(1))?

Answer: -7/2

First, find f⁻¹(x). Setting y = (2x−3)/(x+1) and solving for x: y(x+1) = 2x−3 → x(y−2) = −3−y → x = (3+y)/(2−y). So f⁻¹(x) = (3+x)/(2−x). Now apply it twice: f⁻¹(1) = (3+1)/(2−1) = 4. Then f⁻¹(4) = (3+4)/(2−4) = 7/(−2) = −7/2.

Consider r(x) = (x² − 4)/(x² − x − 2). The graph of r has a hole (removable discontinuity) at x = 2. What is the y-coordinate of that hole?

Answer: 4/3

Factor both the numerator and denominator: x²−4 = (x−2)(x+2) and x²−x−2 = (x−2)(x+1). The (x−2) factor cancels, creating a hole at x=2. The simplified form is (x+2)/(x+1). Substituting x=2 gives (2+2)/(2+1) = 4/3. Note that x=−1 remains a vertical asymptote (no cancellation there).

The function f has a maximum value of 5. If g(x) = −2f(x + 3) − 4, what is the minimum value of g(x)?

Answer: -14

Because the coefficient of f in g(x) is −2 (negative), every output of f is reflected and scaled: where f reaches its maximum of 5, g reaches its minimum. The minimum of g = −2(5) − 4 = −10 − 4 = −14. The horizontal shift (+3 inside) moves where that extreme occurs on the x-axis but does not affect the extreme value itself.

If 9ˣ − 4 · 3ˣ⁺¹ + 27 = 0, what is the product of all real solutions?

Answer: 2

Rewrite: 9ˣ = (3ˣ)² and 4·3ˣ⁺¹ = 12·3ˣ. Let u = 3ˣ, giving u² − 12u + 27 = 0. Factoring: (u−3)(u−9) = 0, so u = 3 or u = 9. Back-substituting: 3ˣ = 3 → x = 1, and 3ˣ = 9 → x = 2. The product of all real solutions is 1 × 2 = 2.

How many distinct real solutions does the equation |x² − 4| = |2x − 1| have?

Answer: 4

Split into two cases. Case 1: x²−4 = 2x−1 → x²−2x−3 = 0 → (x−3)(x+1) = 0 → x = 3, x = −1. Case 2: x²−4 = −(2x−1) → x²+2x−5 = 0 → x = (−2 ± √24)/2 = −1 ± √6. This gives x ≈ 1.449 and x ≈ −3.449. All four values are distinct, so there are 4 real solutions.

Let f(x) = √(4 − x) and g(x) = x² − 2x. What is the domain of the composite function f(g(x))?

Answer: 1 − √5 ≤ x ≤ 1 + √5

f(g(x)) = √(4 − (x²−2x)) = √(−x²+2x+4). For the square root to be defined, we need −x²+2x+4 ≥ 0, or equivalently x²−2x−4 ≤ 0. Using the quadratic formula: x = (2 ± √20)/2 = 1 ± √5. The parabola opens upward, so the inequality holds between the roots: 1−√5 ≤ x ≤ 1+√5.