MATH116 Lecture Notes - Lecture 17: Hypot, Even And Odd Functions, Unit Circle

42 views8 pages

cyanimpala400

9 Dec 2015

School

University of Waterloo

Department

Mathematics

Course

MATH116

Professor

Fiona Dunbar

For unlimited access to Class Notes, a Class+ subscription is required.

Document Summary

Practice solutions 2: for the given functions, decide whether the function is even, odd, or neither. (a) f (x) = 5x6 2x4 + x2 + 1 (b) g(x) = 3x5 + x3 x + 1. This is neither g(x) nor g(x), so g is neither even nor odd: we de ne the even and odd parts of any function f , respectively, as fe(x) = fo(x) = 2 (f (x) + f ( x)); (f (x) f ( x)). Now suppose that g is an arbitrary odd function. Prove that the odd part of g is g itself, while the even part of g is the constant zero function. We are given that g is odd, so we know that g( x) = g(x). Now, the even part of g is: ge(x) = 2 (g(x) + g( x)) = (g(x) g(x)) = 2: let h(x) be the heaviside function, de ned (as in the textbook) as: (cid:26) 0.

Related Questions

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.

Please I need quick help these questions they are due so soon. Thanks

Show transcribed image text

pinkcrocodile662

Problem 4 Circle your answer. a) (6 points) If f(x) = 2x + 3x2 - 361 + 5, then the tangent line to the graph y= f(x) is horizontal when i) 1 = -2,3 ii) I = 2,-3 iii) r = 0 iv) not listed b) [6 points] Suppose Given h(x) = f(g(T)), where g(2) = 3, g'(2) = 7, f'(2) = -1 and 1'(3) = -2. Then h'(2) is equal to i) - 1 ii) -2 iii) - 14 iv) not listed (x+3)(1 - 1)3 c) (6 points) The function f(1) == (x2 - 2x + 1)(x + 3) wa has vertical asymptotes at i) x = -3,0 ii) = -3,0,1 iii) x = 0,1 iv) not listed d) (6 points] The solution to the initial value problem v'(x) = 473 – cos(x) +e", y(0) = 3 is i) y(x) = x + sin(x) + e* + 2 ii) y(I) = x4 – sin(x) +et+2 iii) y(x) = 2* – sin(1) + et + 3 iv) not listed

MATH116 Lecture Notes - Lecture 17: Hypot, Even And Odd Functions, Unit Circle

Document Summary

Get access

Related textbook solutions

Calculus

Single Variable Calculus: Early Transcendentals

CALCULUS:EARLY TRANSCENDENTALS

Related Questions