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Suppose f(x) is a function such that if p < q, f(p) < f(q). Which statement best describes f(x)?

A.) f(x) can be odd or even.
B.) f(x) can be odd but cannot be even.
C.) f(x) can be even but cannot be odd.
D.) f(x) cannot be odd or even.

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Keith Leannon
Keith LeannonLv2
8 Apr 2020

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if f(x) = (xm + 9)^2, which statement about f(x) is true?

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48 f(x) = 12 + (x + 1)2 4. (22 pts) The function f, its derivative f', and its second derivative f" are given by the expressions below. f'(x) = -96(x+1) (12+(x + 1)2) (12+(x + 1)2) (a) Circle the correct statement. f"(x)288(x2 + 2x - 3) i. f is even ii. f is odd. iii. f is neither even nor odd. (b) Give an equation for each horizontal asymptote of f. Write "none" if appropriate. Show your work. Show work for parts (c), (d), (e), and (f) in the space below. (c) List the interval(s) on which f is increasing. Write "none" if appropriate. (d) List the interval(s) on which f is decreasing. Write "none" if appropriate (e) List the x-value(s) at which the local minima of f occur. Write "none" if appropriate (f) List the x-value(s) at which the local maxima of f occur. Write "none" if appropriate. (8) List the interval(s) on which f is concave up. Write "none" if appropriate. (h) List the interval(s) on which f is concave down. Write "none" if appropriate. (i) List the x-coordinate(s) of the inflection points of f. Write "none" if appropriate. () Sketch the graph of y = f(x). Mark the locations of any local extrema and/or inflection points with filled-in circles. Your sketch should be consistent with your answers to (a)-0). -10 -9 -8 -7 -6 -5 -4 4 5 6 7 8 9 10

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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