MATH 1225 Study Guide - Midterm Guide: Intermediate Value Theorem, Negative Number

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5 Feb 2020

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Multiple choice (1 pt each) no partial credit will be given: clearly circle your answer, no calculator, the graph of y = f (x) is given below. y = f (x) Statement 1 if 0 < |x 3| < 1, then |f (x) 1. 5| < 0. 5. C: the function y = f (x) is de ned below. Which of the following statements must be true? f (x) = if x 1 if 1 < x < 1. Hint: it may be helpful to graph f . (a) f is continuous on ( , ) (b) f is discontinuous on ( , 0) If lim x 5 f (x) = 2 = lim x 5 u(x) does not exist u(x) = 1 (a) lim x 1 (b) lim x 5. B (c) u(5) = 100 (d) lim x 5 u(x) 6= u(5)

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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.

3. (a) Assume that a function f(x) is concave down on its whole domain, and f(0) = -2 and f'(0) = -1. Circle each statement that must be true: • The function f(2) has no inflection points. • The derivative f'(2) is always negative. • The derivative satisfies f'(3) < -1. • The derivative satisfies f'(-3) < -1. • The function f(x) <-2 for all 1 >0. (b) Sketch a graph of a function f() satisfying all of the following conditions: • The derivative function f'(x) is decreasing when x < 1 and when r > 3. • The derivative function f'(2) is increasing when 1

Out of these, which is the false statement?

a) f(x) is concave-up on at least two disjoint open intervals

b) f'''(x) is defined everywhere

c) f''(x) is defined everywhere

d) f'(x) is defined everywhere

e) f''(x) < 0

f) f(x) is concave-up on at least two disjoint closed intervals

g) f(VI) = 0

h) There are at least 2 critical numbers

i) f'(0) < 0

j) There is at least 1 inflection point

MATH 1225 Study Guide - Midterm Guide: Intermediate Value Theorem, Negative Number

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