MATH 141 Lecture Notes - Fall 2018 Lecture 11 - Inverse function

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silverworm80

20 Sep 2018

School

University of Maryland

Department

Mathematics

Course

MATH 141

Professor

Patrick Brosnan

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MATH 141 Lecture Notes - Lecture 10: Inverse Function

MATH 141 Lecture Notes - Fall 2018 Lecture 11 - Inverse function

MATH 141 Lecture 12: MATH 141 - Lecture 12 - Sept 21

Document Summary

Suppose f i -> r is a difference function on an upon interval i containing a point a. Assume that f has inverse f-1 = g. assume: f"(a) not equal to 0, c = f(a) Since dy/dx is > 0 for all x, f is inclusive so f is one-one. So f has an inverse with range(f) = r. If y = -3 , then x= -1. Therefore by the theorem, g"(-3) = dx/dy which is 1/(dy/dx) = 1/7 (ans) Lets see why it is positive when f(x) = mx + b. F"(x) = m. so since we assume f"(a) not equal to 0, then m is not equal to 0. Inverse function of y = f(x) = mx + b is given by solving for x in terms of y (cid:2207)=(cid:2195)(cid:2206)+(cid:2184) (cid:2207) (cid:2184)=(cid:2195)(cid:2206),(cid:2190)(cid:2187)(cid:2196)(cid:2185)(cid:2187) (cid:2206)=((cid:2778)(cid:2195))((cid:2207) (cid:2184)(cid:2195)) (cid:3052)1 (cid:3051)(cid:3052)1. Also recall, for x, y > 0 n(cid:1876)(cid:1877) n(cid:1877)= = (cid:3051)(cid:3052)(cid:3052) (cid:1872)(cid:1853)(cid:1863)(cid:1857) (cid:1873)=(cid:1876)(cid:1877) , (cid:1857)(cid:1866)(cid:1855)(cid:1857) (cid:1857)(cid:1869)(cid:1873)(cid:1853)(cid:1872)(cid:1867)(cid:1866) (cid:1854)(cid:1857)(cid:1855)(cid:1867)(cid:1865)(cid:1857)(cid:1871) (cid:1856)(cid:1873)(cid:1873) =n(cid:1876)

Related Questions

Let Ï : G â Gâ² be an isomorphism from the group G to the groupGâ². In particular, Ï is bijective so that its inverse Ïâ1 : Gâ² â Gexists (see 1.9.1). Prove that Ïâ1 is anisomorphism.

1.9.1 Theorem. A function f : X â Y has an inverse if and only ifit is bijective.
Proof. Letf:XâY beafunction.
(â) Assume that f has an inverse f^â1. (f injective?) Letx,xâ² â X and assume that f(x) = f(xâ²). Then x =f^â1(f(x)) = f^â1(f(xâ²)) = xâ².

Therefore, f is injective. (f surjective?) Let y â Y . Put x =f^â1(y). Then x â X and
f(x) = f(f^â1(y)) = y.

Therefore, f is surjective. Since f is both injective andsurjective, it is
bijective.
(â) Assume that f is bijective. Define f^â1 : Y â X byletting f^â1(y) be the unique x in X for which f(x) = y.(Since f is surjective, there is at least one such x and since f isinjective, there is at most one such x.) For x â X, we havef^â1(f(x)) = x (since f^â1(f(x)) is defined tobe the element that f sends to f(x)). Similarly, for y â Y ,f(f^â1(y)) = y (since fâ1(y) is defined to be the elementthat f sends to y). Therefore, f^â1 is an inverse off.

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.

Very clear and step by step proof would be very appreciated. Thanks in advance.