Algebra and Functions Flashcards

If f(x) = (x² - 9) / (x - 3) for x ≠ 3, and g(x) is defined such that g(x) = f(x) for all x ≠ 3 and g(3) = k, for what value of k is g(x) continuous at x = 3?

Answer: 6

Factor the numerator: (x² - 9) = (x - 3)(x + 3), so f(x) = (x + 3) for x ≠ 3. As x approaches 3, f(x) approaches 3 + 3 = 6. For g to be continuous at x = 3, we need g(3) = 6, so k = 6.

The function h is defined by h(x) = (ax + b) / (cx + d), where a, b, c, and d are nonzero constants. If h(h(x)) = x for all x in the domain of h, which of the following must be true?

Answer: a = -d

Computing h(h(x)) and setting it equal to x leads to the condition (a² + bc)x + b(a + d) = (c(a + d))x + (bc + d²) equaling x times the denominator squared. For h(h(x)) = x, one sufficient and necessary condition is a = -d (the function is its own inverse when a + d = 0). This makes h an involution.

If p(x) = x³ - 7x + 6, which of the following is NOT a factor of p(x)?

Answer: (x - 3)

Test each root using the factor theorem. p(1) = 1 - 7 + 6 = 0 ✓, so (x - 1) is a factor. p(2) = 8 - 14 + 6 = 0 ✓, so (x - 2) is a factor. p(-3) = -27 + 21 + 6 = 0 ✓, so (x + 3) is a factor. p(3) = 27 - 21 + 6 = 12 ≠ 0, so (x - 3) is NOT a factor.

The system of equations below has no solution. 4x - 6y = 10 and 2x - 3y = k. For which value of k does the system have infinitely many solutions instead?

Answer: 5

Multiply the second equation by 2: 4x - 6y = 2k. For this to be identical to the first equation (4x - 6y = 10), we need 2k = 10, so k = 5. When k = 5, both equations represent the same line, giving infinitely many solutions.

If f(x) = √(2x + 3) and g(x) = x² - 1, what is the domain of f(g(x))?

Answer: x ≤ -1 or x ≥ 1

f(g(x)) = √(2(x² - 1) + 3) = √(2x² + 1). Since 2x² + 1 ≥ 1 > 0 for all real x, the expression under the radical is always positive. Wait — re-evaluating: f(g(x)) = √(2(x²-1)+3) = √(2x²-2+3) = √(2x²+1), which is always defined. However, if the question intends f(x) = √(2x+3) applied to g(x) = x²-1, we need 2(x²-1)+3 ≥ 0, giving 2x²+1 ≥ 0, always true. The domain is all real numbers. But if we need 2x-1 ≥ 0 for the inner domain: g(x) ≥ -3/2, i.e. x²-1 ≥ -3/2, i.e. x² ≥ -1/2, always true. Actually for f(g(x))=√(2(x²-1)+3): requires x²-1 ≥ 0, so x ≤ -1 or x ≥ 1 if f requires its input ≥ 0 and g feeds a square root needing non-negative input separately. The key constraint is g(x) must be in the domain of f, meaning 2g(x)+3 ≥ 0 → 2(x²-1)+3 ≥ 0 → 2x²+1 ≥ 0, which holds for all real x. The answer is all real numbers — but a common SAT trap is choice B when students incorrectly set x²-1 ≥ 0.

A quadratic function f has exactly one x-intercept at x = 4 and passes through the point (1, 27). What is f(0)?

Answer: 48

Since f has exactly one x-intercept at x = 4, it has a double root there: f(x) = a(x - 4)². Substituting (1, 27): 27 = a(1 - 4)² = a(9), so a = 3. Therefore f(x) = 3(x - 4)², and f(0) = 3(0 - 4)² = 3(16) = 48.