MATH 140 Lecture 6: Bisection Method and Derivatives

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cerisegiraffe538

23 Sep 2015

School

University of Maryland

Department

Mathematics

Course

MATH 140

Professor

Gulick

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If f(c1) > 0, then let a point c2 be the midpoint of a and c1: ex 1: let f ( x)=cos x x . Approximate the zero of f to within 1/8. o o o o f (0)=cos0 0=1 0=1>0 f (1)=cos1 1<0 f (1. 4) . 168<0: etc, ex 2: approximate 7 using the bisection method, let g (x)=x2 7 g( 7)=0, find a zero of g. If lim x a f ( x) f (a) x a exists, then it is the derivative of f at a: derivatives may be written as f " (a) (say: f prime of a ) or as df dx (say: the derivative of f with respect to x. , we say that f is differentiable at a, ex 3: let f ( x)= 1 x2. Find f"(2): limit: f ( x) f (2) x 2. =lim x 2 lim x 2: simplify: lim x 2.

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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.

Let b > 0 and f : [0, b] â R be continuous and f(b) = 0. Assume that f is differentiable on (0, b). Show: For every n â N, there exists some c â (0, b) such that cf â²(c) = ânf(c).

Hint: Apply Rolleâs theorem to the function g(x) = xⁿ f(x).

11. Let In(2+1)-In 2 s() = { f(x)= a when x > 0 when r = 0 when : < 0 (eto If f is continuous at x = 0, then (a) a = In 2 and b=0 (b) a= 1 and b= 1 (c) a = 1 and 6 = 2 (d) a= In 2 and b== (e) a = 5 and 6= In 2

MATH 140 Lecture 6: Bisection Method and Derivatives

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