- if x > 0 6. f(x)= is continuous on the intervals if x<0 A) (-0,040,) B) (-4,-5)(-5,-3)(-3,00) C) (--,-5)(-5,0)(0.) D --,-3)(-3,0)(Q.) E) none of these

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(c) Sketch a graph of a continuous function y = f(1) that satisfies the following: • f(-5) = f(0) = 0; • f'(x) < 0 for -5 < <-3 and 1 < x < 5; • f'(x) > 0 for -3 < x < 1; • f'(x) < 0 for -2 < x < 3; • F"(x) > 0 for -5 < x < -2 and 3 < x < 5; Clearly label and local extrema and points of inflection.

11. Let In(2+1)-In 2 s() = { f(x)= a when x > 0 when r = 0 when : < 0 (eto If f is continuous at x = 0, then (a) a = In 2 and b=0 (b) a= 1 and b= 1 (c) a = 1 and 6 = 2 (d) a= In 2 and b== (e) a = 5 and 6= In 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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- if x > 0 6. f(x)= is continuous on the intervals if x<0 A) (-0,040,) B) (-4,-5)(-5,-3)(-3,00) C) (--,-5)(-5,0)(0.) D --,-3)(-3,0)(Q.) E) none of these

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