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MA100 Chapter Notes - Chapter 7: Differentiable Function, Edison Electric Institute, Nipple

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coppergnu94
13 Dec 2015
School
WLU
Department
Mathematics
Course
MA100
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All

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If f is a di erentiable function and b > 0, b (cid:54)= 1, then: (cid:48) , and by the chain rule, [logb f (x)] F (cid:48) (x: [logb x] (cid:48) (cid:48, [ln x] , and by the chain rule, [ln f (x)] (cid:0)x2 + 1(cid:1) , then: (cid:0)x2 + 1(cid:1) +(cid:0)1 + x2(cid:1)(cid:18) If y = (1 + x2) log2 y(cid:48) = 2x log2. Maximum and minimum function values: absolute max/min: Note: absolute max/min may only occur at endpoints of i (if closed interval) or critical values of f. If f is continuous on a closed interval [a, b], < a < b < , then f has an absolute maximum and an absolute minimum on [a, b: relative max/min: We de ne relative [or local ] minimum value of f similarly. If f has a local extremum at an interior point c and f(cid:48)(c) exists, then f(cid:48)(c) = 0.

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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.