MATH-1036EL Chapter 1: September 21, 2015 - Lecture and textbook lesson

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bronzecrab202

27 Sep 2015

School

Laurentian

Department

Mathematics / MathÈmatiques

Course

MATH-1036EL

Professor

Markovich Tom

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Document Summary

A rational function is continuous everywhere on its domain, as long as the denominator does not equal 0. This function is rational because the two functions (numerator and denominator) are both polynomials. Therefore, g x is continuous on its domain. Its domain is: solution to find inequality: x. 4(2)( 1: therefore, the domain is then x x. This function is composed of two continuous functions on all values of x . Therefore, the denominator cannot be equal to 0, and the domain is x. Since this radical is an even numbered radical, then the function is continuous on the interval . 0, , which represents that the limit from the right side exists, but not from the left. f g x. 0 f g x is continuous at that point. Use continuity to find x lim sin( x. Solution: x lim sin( x x sin ) sin ) (0) sin( ) sin( sin.

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beigecrocodile802

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.

What are the answers for all of these questions? I am confused on how to solve for these.