MATH 140 Lecture Notes - Lecture 5: Intermediate Value Theorem, Trigonometric Functions

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cyanwallaby953

5 Feb 2018

School

University of Maryland

Department

Mathematics

Course

MATH 140

Professor

Gulick

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Document Summary

Math140 lecture 5 continuity and the intermediate value theorem. Examples: the function is not continuous at x = -3 because the function is not defined at that at all x in its domain. it is not defined at some numbers. number. Is the function continuous at x = -3: ex 3: let (cid:4666)(cid:1876)(cid:4667)=(cid:4666)+(cid:2870)(cid:4667)(cid:4666)+(cid:2869)(cid:4667) +(cid:2869: ex 4: let (cid:1863)(cid:4666)(cid:1876)(cid:4667)={(cid:883),(cid:1876) (cid:1871) (cid:1870)(cid:1872)(cid:1867)(cid:1866)(cid:1864) (cid:4666)(cid:1876)=(cid:3017)(cid:3018),(cid:1875) (cid:1857)(cid:1870)(cid:1857) (cid:1842) (cid:1866)(cid:1856) (cid:1843) (cid:1870)(cid:1857) (cid:1866)(cid:1872)(cid:1857)(cid:1859)(cid:1857)(cid:1870)(cid:1871) (cid:1866)(cid:1856) (cid:1843) (cid:882, ex 5: let (cid:1868)(cid:4666)(cid:1876)(cid:4667)={(cid:1876),(cid:1876) (cid:884) Is it continuous: the functions is not continuous because, even though lim (cid:2870)(cid:1868)(cid:4666)(cid:1876)(cid:4667) exists and is 2, it (cid:1876), (cid:1876)(cid:883). Is it does not equal p(2), which is . Continuity on intervals: definition, a function f is continuous on [a, b] if f is continuous on (a, b) and continuous from the left at b and continuous from the right at a, ex 7: (cid:1859)(cid:4666)(cid:1876)(cid:4667)=(cid:1876),(cid:882) (cid:1876) (cid:883), graph:

Related Questions

solve all parts asap

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 are due so soon. Thanks