CAS MA 123 Lecture Notes - Lecture 19: Maxima And Minima, If And Only If

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cerisegiraffe538

20 Oct 2015

School

Boston University

Department

Mathematics & Statistics

Course

CAS MA 123

Professor

John Bergdall

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Document Summary

Ma123 lecture 19 extreme value theorem. C is true because that is the definition of an absolute minimum. If f(x) is continuous on [a,b] then absolute extreme values exist (both the minimum and maximum) Absolute extreme values on some small open interval. If d = [a,b] then c=a or c=b can be x-values/absolute extremes f(c) but they cannot be x-values of local extrema because no i exists. Suppose f(x) has a local extreme value f(x) and f"(c) exists then f"(c) = 0. Why is the theorem true: f(c) has a local maximum, f"(c) exists, f"(c) = lim x c f (c) f (x) c x, f(c) is a local maximum f(c) f(x) for x near c f (c) f (x) { xc near c f (c) f (x) o. So 0 lim x c- f (c) f (x) c x. = f"(c) = lim x c+ f (c) f (x) c x.

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.

CAS MA 123 Lecture Notes - Lecture 19: Maxima And Minima, If And Only If

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