University:
Williams College
Course:
MATH 140 | Calculus II
Academic year:

2019
Views:

149

Pages:

2
Author:

Lacey Gentry

Find the domain of the function. Write your answer in interval notation. f(x) = log(x+3) x+3 > 0 x > -3 (-3, ∞) f(x) = log x + 3 x > 0 (0, ∞) f(x) = log₅(5-x) 5-x > 0 -x > -5 x < 5 (-∞, 5) Find the function of the form y = logₐx whose graph passes through the given point. (5, 1) 1 = logₐ5 a¹ = 5 a = 5 (3, 1/2) 1/2 = logₐ3 a^(1/2) = 3 a = 9 Match the logarithmic function with its graph. f(x) = -ln x: C f(x) = ln(x-2): D f(x) = 2 + ln x: B f(x) = ln(-x): A f(x) = ln(2-x): F f(x) = -ln(-x): E 12. f(x) = log₂x Domain: (0, ∞) Range: (-∞, ∞) Asymptote: x = 0 13. f(x) = -log₂x Reflect #12 over x-axis Domain: (0, ∞) Range: (-∞, ∞) Asymptote: x = 0 14. f(x) = log₂(x-3) - 1 Shift #12 left 3 and 1 down Domain: (-3, ∞) Range: (-∞, ∞) Asymptote: x = -3 15. f(x) = log₃x Domain: (0, ∞) Range: (-∞, ∞) Asymptote: x = 0 16. f(x) = log₃(-x) + 3 Reflect #15 over y-axis, then up 3 Domain: (-∞, 0) Range: (-∞, ∞) Asymptote: x = 0 17. f(x) = -log₃(x-1) + 2 Shift #15 right 1, then reflect over x-axis, then up 2 Domain: (1, ∞) Range: (-∞, ∞) Asymptote: x = 1