MATH127 Lecture 6: Sept 19th- Inverse Functions and Logarithms

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cyanhorse768

29 Sep 2016

School

University of Waterloo

Department

Mathematics

Course

MATH127

Professor

David Deepwell

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Document Summary

F^-1(f(x)) = log (subscript 2) (2^x) = xlog (subscript 2) 2 = x [because log (subscript2) and 2 cancel out] This is true for all f(x) = a^x for a>o, a cannot equal 1. Because if a is 1, its inverse becomes a vertical line (a vertical line is not a function because you can"t get the same output for multiple inputs, and log base 1 is undefined) The inverse of f(x) is f^-1 = log(subscript a) (x) Log(subscript a) triangle = a^ = triangle. Recall that the domain of exponentials is (-infinity, infinity) and the range of exponentials is (0, infinity) By knowing these, you can find the logs because they are the inverse of each other. Sketch y = (1/2)^x and log (subscript )x. V for volume, 2 for doubling, t for time, 20 for 20minutes. 10^39 = 2^t/20 log(subscript 2) 10^39 = log(subscript 2) 2^t/20. 39 log(subscript 2) 10 = t/20 log(subscript 2) 2.

Related Questions

1. the folowing information has been gathered about the rationalfunction f(x)

domain: all reals except -1

discontinuous at -1 only

f(-5)=3, f(-3)=4, f(1)=0,f(3)=1

f has vertical asymptotes x = -1

f has a horizontal asymptote of y = 2

f'(-3) = f'(1) = 0

f'(x) >0 on (-infinity,-3) u (1, +infinity)

f'(x)<0 on (-3,-1) u (-1,1)

f''(-5) = f''(-3) = 0

f''(x) >0 on (-infinity,-5) u (-1, 3)

f''(x)<0 on (-5,-1) u (3,+infinity)

a) indicate intervals on which f is increasing anddecreasing

b) indicate intervals ov which f is concave up and down

c) graph f on the above axes.

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.

MATH127 Lecture 6: Sept 19th- Inverse Functions and Logarithms

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