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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

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Elin Hessel
Elin HesselLv2
11 Apr 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.

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.

1. (16 pts) Let f be the function with domain [0, 6] given by the graph below. y=f(x) 3 4 5 6 Use the graph off to answer the following questions about f. (a) List the critical point(s) of f on the interval (0,6). Write "none" if appropriate. (b) List the inflection point(s) of f on the interval (0,6). Write "none" if appropriate. Use the graph off to answer the following questions about the derivative f' of f. (c) List the intervals(s) on which f' is negative. Write "none" if appropriate. (d) List the intervals(s) on which f' is decreasing. Write "none" if appropriate. (e) List the x-coordinates(s) at which f' has a local minimum. Write "none" if appropriate.

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