MAD 4301 Midterm: Test 1 Fall 1990

16 views1 pages

turquoisegnu128

30 Jan 2019

School

Department

Course

Professor

For unlimited access to Study Guides, a Grade+ subscription is required.

Document Summary

Nd the domain and range. (1) (15 pts. ) (a) sketch the graph of x = py + 1 = 0. (b) if this equation de nes y as a function of x, (2) (15 pts. ) If |f (x) 4| < 3, nd a bound for 1/f . (3) (20 pts. ) Evaluate the limits. (a) (b) (c) (d) (a) (b) For f (x) = x 1 evaluate the following (5) (15 pts. ) Suppose > 0 and nd a positive number such that (cid:12)(cid:12)x2 9(cid:12)(cid:12) < whenever |x 3| < . (6) (10 pts. ) Find a value of a so that f is continuous at 0 or indicate that this is impossible. Assume |x 3| 1. f (x) = x a if x < 0 if x = 0 sin x if x > 0 (7) (15 pts. )

Related Questions

1. (25 pts) The graph of a function f is given in the figure below. The domain of fis (-, +.). Use the graph of fto answer the questions below. yax) y=f(x) -3 - 2 3 4 (I) (1 pt) Find the range of f. (II) Find the following values. (Note: Possible answers include + , -, or "does not exist".) (a) (1 pt) (x) = 5 (b) (1 pt) lim f (x) = 5 (0) (1 pt) Tim f (x) = 5 (d) (1 pt) f(3) = 2 (e) (2 pts) lim f (x) = + 0 x-2- (8) (2 pts) 11m = (x) = 0 (g) (2 pts) lim f (x) = DOES NOT EXIST x+-2 (h) (2 pts) Lim f (x) = 0 (i) (2 pts) lim f (x) = + X +00 (III) (2 pts) Find all vertical asymptotes. x = -2 (IV) (2 pts) Find all horizontal asymptotes. y = 0 (v) (2 pts) Find all slant asymptotes. | y = x + 2 (VI) (4 pts) Determine the intervals of continuity for f. (-00, -2), [-2, 3), and (3, +c)

4. (21 pts) Let g be the function given by ( 6 - X 2 + x if x < 2, x + -2 g(x) = x2 – x-1 if x>2. (a) (6 pts) Use the definition of continuity to determine whether the function g is continuous at 2. Show your work. (b) (3 pts) Evaluate the limit. Justify your answer. Show your work. Write "does not exist" only if the limit does not exist and is not + or -0. lim g(x) = lim -2+ = +00 X 2 + X since the form of the limit is NUM: POS, DENOM: POS (c) (2 pts) Write the equation(s) of all vertical asymptotes of g. Write "none" if appropriate. The result in part (b) implies that the line x= -2 is a vertical asymptote. This is the only vertical asymptote, since the function g is continuous on its domain (d) (2 pts) Determine the largest) intervals of continuity of g. Use interval notation to write your answer. Let g be the function given by 12+ if x < 2, x# -2 2 + x g(x)= x2-x-1 x-1 if x>2. (e) (6 pts) Evaluate each limit. Do not use L'Hôpital's Rule. Write "does not exist" only if the limit does not exist and is not too or -00. Show your work. 6-1 i lim g(x) = lim -00 2 + X = lim -002 + x 0-1 + 1 x -00 x x x -+- 2 +1 0 0 + 11 x2 - x-1 1 ii. lim g(x) = lim x++ V x2 – x-1 X-1 x2 - x-1 X - 1 = lim x +oc x +0 - lim Vam M93001 (f) (2 pts) Write the equation(s) of all horizontal asymptotes of g. Write "none" if appropriate.

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.

MAD 4301 Midterm: Test 1 Fall 1990

Document Summary

Get access

Related Documents

MAD 4301 Midterm: Test 5 Fall 1990

MAD 4301 Midterm: Test 3 Fall 1990

MAD 4301 Midterm: Test 2 Fall 1990

Related Questions