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Mathematics
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26 Jan 2018

1. (20 pts) The graph of a function f is given in the figure below. Use the graph of f to answer the questions below. y = f(x) -64 - - - - - - - - - - - - (I) (1 pt) Pind the domain of f. Domain of f = (-0, 3) U (3, +co) (II) (1 pt) Find the range of f. Range of f = (-, +O) (a) (1 pt) lim f (x) = 0 (b) (1 pt) f(3) = DNE (does not exist) (c) (1 pt) f (5) = 2 (a) (1 pt) lim f (x) = 0 (0) (1 pt) lim f (x) = -- (f) (1 pt) lim f (x) = DNE 0 1 (g) (1 pt) lim f (x) = + (h) (1 pt) lim f (x) = -6 X-++00 (IV) (2 pts) Find all vertical asymptotes. (V) (2 pts) Find all horizontal asymptotes. (VI) (6 pts) Determine the intervals of continuity for f. (-0, 0], (0, 3), (3, 5), (5, +00)

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Tod Thiel
Tod ThielLv2
29 Jan 2018
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1. (32 pts) The graph of a function f is given in the figure below. The domain of fis (-0, +). Use the graph of fto answer the questions below. y = f(x) y=f(x) (I) (2 pts) Find the range of f. Range of f = (II) Find the following values. (Note: Possible answers include + co, -, or "does not exist".) (a) (2 pts) lim f (x) = X+1 + (b) (2 pts) lim f (x) = (c) (2 pts) lim f (x) = x+1 (a) (2 pts) f(1) = (e) (2 pts) lim f (x) = (f) (2 pts) lim f (x) = (g) (2 pts) lim f (x) = (h) (2 pts) lim f (x) = (i) (2 pts) lim f (x) = (j) (2 pts) f (0) = (III) (2 pts) Find all vertical asymptotes. (IV) (2 pts) Find all horizontal asymptotes. (V) (2 pts) Find all slant asymptotes. (VI) (4 pts) Determine the intervals of continuity for f.

1. (18 pts) The graph of a function f, defined on (-0, +o), is given in the figure below. - - y = f(x) - - - - - - N - - 4 A -2-1 144 -1 - - - - - - - -- - - -- Use the graph of f to answer questions (a)-(f) below. (a) Determine the value or write "does not exist". i. f(-3) = -2 ii. f(0) = 1 (b) Determine the limit or write "does not exist" (DNE). Write "does not exist" only if a limit does not exist and is not too or -0. i. lim f(x) = 0 iv. lim f(x) = +00 X- 3 x2 ii. lim f(x) = -4 v. lim f(x) = -0 X- 0- x -00 iii. lim f(x) = DNE vi. lim f(x) = -2 x0 x++00 (C) MULTIPLE CHOICE! Circle ONLY one answer! Find a number a for which the following statement is true (the function f is shown above): STATEMENT: The function f is continuous at a, and lim f(x) = 0. i. a=-3 iii. a=0 V. a=3 ii. a = -2 iv. a = 2 vi. Such a number does not exist. (d) Determine the largest intervals of continuity of f. Use interval notation to write your answer. (-0, -3), (-3,0), (0, 2), (2, +00) (e) Write the equation(s) of all vertical asymptotes of f. Write "none" if appropriate. x= 2 (f) Write the equation(s) of all horizontal asymptotes of f. Write "none" if appropriate. y = -2

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.

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