MATH 1225 Study Guide - Midterm Guide: Power Rule, Scilab

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rosecrocodile591

5 Feb 2020

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Circle only one answer choice. f (x) f (0) x (a) lim x 0 (b) 1. D (c) lim h 0 eh 1 h (d) answer choices (a), (b), and (c) are all correct: the velocity of a particle at time t is given by the graph below, y = v(t). Circle only one answer choice. y = v(t) 3 (a) the particle is speeding up on (2, 3). (c) the particle is at rest on (8, ). (b) the particle is speeding up on (6, 8). (d) the particle is slowing down on (0, 1). 12 (a) (4 pts) let h(x) = g(x) + 2x3 + x. Note h (x) = g (x) + 6x2 + . Note h (2) = g (2) + 12(2) = 12 + 24 = 36. (b) (4 pts) let k(x) = g(cid:18)f (x) + 1(cid:19). Note k (x) = g (cid:18)f (x) + 1(cid:19)f (x)

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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

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.

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: Power Rule, Scilab

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