MATH 1P66 Lecture Notes - Lecture 6: Counterexample

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periwinklewhale81

13 Oct 2018

School

Brock University

Department

Mathematics and Statistics

Course

MATH 1P66

Professor

Basil Nanayakkara

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MATH 1P66 verified notes

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MATH 1P66 Lecture 5: Math Lecture #5

MATH 1P66 Lecture Notes - Lecture 6: Counterexample

MATH 1P66 Lecture 7: NESTED QUANTIFIERS - Continued

Document Summary

Xp(x) = x p(x) demorgan"s law for quantifiers: domain = r, x(-20 v x<0) F(x): x has a ferret domain = {students in class: a student in your class has a cat, a dog, and a ferret. X[c(x)^d(x)^f(x): some student in your class has a cat and a ferret but not a dog. X[(c(x)^f(x)^ d(x): for each of the three animals, cats, dogs, and ferrets, there is a student in your class who has this animal as their pet. Xc(x)^ xf(x)^ xd(x) no he same as x[c(x)^f(x)^d(x)] says one student has all. Commutative law for addition of real numbers x+y=y+x whenever x,y are real numbers. For every x, there exists y, that x+y = z . Given any x you can choose y to be 0. Given any x you can choose y to be negative of x.

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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.

4. (21 pts) Let g be the function given by ( 6 - X 2 + x if x < 2, x + -2 g(x) = x2 – x-1 if x>2. (a) (6 pts) Use the definition of continuity to determine whether the function g is continuous at 2. Show your work. (b) (3 pts) Evaluate the limit. Justify your answer. Show your work. Write "does not exist" only if the limit does not exist and is not + or -0. lim g(x) = lim -2+ = +00 X 2 + X since the form of the limit is NUM: POS, DENOM: POS (c) (2 pts) Write the equation(s) of all vertical asymptotes of g. Write "none" if appropriate. The result in part (b) implies that the line x= -2 is a vertical asymptote. This is the only vertical asymptote, since the function g is continuous on its domain (d) (2 pts) Determine the largest) intervals of continuity of g. Use interval notation to write your answer. Let g be the function given by 12+ if x < 2, x# -2 2 + x g(x)= x2-x-1 x-1 if x>2. (e) (6 pts) Evaluate each limit. Do not use L'Hôpital's Rule. Write "does not exist" only if the limit does not exist and is not too or -00. Show your work. 6-1 i lim g(x) = lim -00 2 + X = lim -002 + x 0-1 + 1 x -00 x x x -+- 2 +1 0 0 + 11 x2 - x-1 1 ii. lim g(x) = lim x++ V x2 – x-1 X-1 x2 - x-1 X - 1 = lim x +oc x +0 - lim Vam M93001 (f) (2 pts) Write the equation(s) of all horizontal asymptotes of g. Write "none" if appropriate.

Please answer numbers 2c, 3a, and 3b. Please and thank you.

MATH 1P66 Lecture Notes - Lecture 6: Counterexample

MATH 1P66 Lecture 5: Math Lecture #5

MATH 1P66 Lecture Notes - Lecture 6: Counterexample

MATH 1P66 Lecture 7: NESTED QUANTIFIERS - Continued

Document Summary

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