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27 Feb 2018

5. Let (2) be the function of the previous problem. If f(0) = 1, then what is (5)? (A) 13 (B) 14 (C) 15 (D) 16 ANSWER:

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Patrina Schowalter
Patrina SchowalterLv2
1 Mar 2018
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3. (20 pts) Let f be a differentiable one-to-one function on the interval (0,4) with the particular values of f and f' given in the table below, and let f-1 be the inverse function of f. Let g be the function given by the graph below. y y=g(x) 12 2 3 5 1 2 3 4 5 6 34-2 For (a), (b), and (c), circle the correct choice. lim 8(3+h) – g(3) h iii. 5 V. -2 vi. does not exist vii. none of the previous choices iv. O (b) dip [6^x +2)/(x) at x = 3 iii. 1 V. -1 vii. none of the previous choices iv. 0 vi. does not exist (0) [r_w) at x = 2 V. vii. none of the previous choices iv. I vi. does not exist (d) Consider the curve given by xey + y = f(x) - 1. Use implicit differentiation to computed at the point (2.0). Show your work.

4. (16 pts) Let f be a differentiable one-to-one function with domain (0,6) with the particular values of f and f' given in the table below, and let f-1 be the inverse function of f. Let g be the function given by the graph below. x f(x) f'(x) --- 23 = g(x) Find the value or write "not defined". Show your work! (a) lim 5(x) – 8(5) x+5 X-5 (b) [6'«x+2)(x) at x = 3 (c) [p-(x)) at x = 2 (0( vecm) at x = 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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