MATH 1225 Study Guide - Midterm Guide: Removable Singularity, Intermediate Value Theorem, Asymptote

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5 Feb 2020

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A math 1225 practice test 1 (32 pts) C (c) lim x 0 (d) lim x 0 sin(ex) = sin(cid:16) lim ex(cid:17) x (cid:0) x x(cid:1) =(cid:16) lim. X(cid:17) (cid:16) lim x x x(cid:17: let a be a constant. What value of a will make f continuous at x = 3 if lim x 3+ f (x) = 2a + 3, lim x 3 f (x) = 5 2a, and f (3) = 5? (a) 0. 1 (b) lim (c) lim (d) lim h h 0" 2h 2 h 0(cid:20) h2 + 4h + 2 + h h (cid:21) Give the limit if it exists, and where no nite limit exists be as precise as possible among + , , or dne. Cite theorems when appropriate. (a) (3 pts) lim x 1 x 5 ln(x) ln(x) . Also, as x 1 , (x 5) 4. Thus, lim x 1 x 5 ln(x) (b) (6 pts)

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4. (16 pts) MULTIPLE CHOICE! CIRCLE THE CORRECT ANSWER IN EACH PART. (I) ( 4 pts) Complete the statement of the Intermediate Value Theorem: Suppose f is continuous on the interval [a, b] and L is a number between f (a) and f (b). Then there is at least one number c in (a, b) such that (a) C = L; (b) f(c) = 0; (c) f(c) = L; (a) f' (c) = L; (e) C = 0; (f) f (a) < c < f (b); (g) NONE OF THE PREVIOUS ANSWERS. (II) (4 pts) The figure shows the graph of a function f. At what point (points) c does the conclusion of the Intermediate Value Theorem hold for f (x) on the interval [1, 5] and L = 2. --- - - - -1 - - - 2 - - I- - - (a) c = 1; (b) C = 0; (c) C = 2; (a) c = 3; (e) C = 4; (f) C = 5; (g) c= -1. (III) (4 pts) Given that cos x s f (x) s et, for all x > 0 evaluate lim f (x) x+0+ (a) lim f (x) = 0; (b) lim f (x) = 1; (c) lim f (x) = e; x0+ X0+ (d) lim f(x) DOES NOT EXIST. WE DON'T HAVE ENOUGH INFORMATION TO ANSWER THE QUESTION *40+ (IV) (4 pts) Given the function 1 f (x) = , find its inverse, f+(x) x -2 (a) f (x) = x -2; (b) f«+ (x) = x+2 ; (c) fº+ (x) = -2; (f) f-+ (x) -2; (e) f (x) DOES NOT EXIST.

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.

MATH 1225 Study Guide - Midterm Guide: Removable Singularity, Intermediate Value Theorem, Asymptote

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