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01:640:111 Lecture Notes - Lecture 6: Piecewise

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School
Rutgers University
Department
Mathematics
Course
01:640:111
Professor
Threlfall

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Document Summary

Example: height is a function of age: (x) f, (x) f. Definition: a function is a role that assigns to each element x in set d (domain). Do not simplify & only work on denominator. The set of all numbers that belong in the domain. Example: find the domain of the following functions: F irst we have to set to an inequality econd we have to factor. : = 2 2 olve for x. P ick points (make table) and find where he product is larger than zero t. T he denominator cannot equal zero, so to ind the domain we have to find when x equals 0 f x x = , x = } 1 x: (x)k irst set as an inequality. M ultiply top and bottom and put in numerator ultiply inner and put in denominator & don t expand a 1 a h+1 h(a+h 1)(a 1) (a+h 1)(a 1)

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    5. (18 pts) Circle True if the statement is ALWAYS true; circle False otherwise. No explanations are required. Let f be a one-to-one function and f its inverse. (a) If the point (2, 5) lies on the graph off, then the point (5,2) lies on the graph of f. True False f (2) = 5 f-1 (5) = 2 (b) sin-1 (T) = 0 True False Since >1, ris not in the domain of sin-?; | or sin-1 (7) #0, since sin (0) = 0 # Iff and g are two functions defined on (-1, 1), and if (c) limg (x) = 0, then it must be true that lim[ f(x) g (x)] = 0. True False Let f (x) = 1 / x, if x #0, and f (0) = 4 and g(x) = x. Then lim g (x) = lim (x) = 0, but Lim ( f (x) g(x)] = lim [() x] = 18 0. Iffis continuous on (-1, 1), and if f (0) = 10 and lim g (x) = 2, (a) True False then lin (= 5. Because lim : (*) - Limx-of (*) .f(0) = 20 = 5 If f is continuous on [1, 3], and if f(1) = 0 and f(3) = 4, then the equation f(x) = (e) has a solution in (1, 3). True False Since f continuous on [1, 3), f(1) = 0 < < 4 = f (3), by the Intermidiate Value Theorem there exists a cin (1, 3) such that f (c) = . (f) Let f be a positive function with vertical asymptote x = 5. Then lim f (x) = +0. True False Let f (x) = 1, if x < 5, and f (x) ==, ifx>5. Then lim f (x) = +00, and therefore, X = 5 is a vertical asymptote. 5 - X So, f is a positive function with vertical asymptote x = 5 but lim f (x) + +co, since lim f (x) = + 0 + lim f (x) = 1, which means that lim f (x) does not exist.