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MAT135Y5 Lecture Notes - Lecture 3: Rational Number, Piecewise, Negative Number

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UTM
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Mathematics
Course
MAT135Y5
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Burazin, A.

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MAT135Y5 Lecture 2: MAT135Y Lecture 2.5 - Appendix A cont’d, Appendix B, and Section 1.1

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MAT135Y5 Lecture Notes - Lecture 3: Rational Number, Piecewise, Negative Number

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MAT135Y5 Lecture Notes - Lecture 4: If And Only If

Document Summary

Mat135y lecture 3 - section 1. 1 cont"d and section 1. 2. Q: what"s the domain of each function? a) b) c) = 1 (x)g (x)h (x)m = x 2 1 x 3 a) g (x) So, domain = , ) x x. 2 0 ( 2. So, domain = , 3) (x)m = x 2 1 c) x 2 1 0 ** x 2 = x = x| x| | a a x a x| ( 1 [ x+2. A function takes all input values, which a part of the domain of the function and maps it to only one output value. Draw a vertical line through the graph. For x value of the graph the vertical line touches the graph only once! arabola (facing down) x. = 1 2 = 1 so, ( , 1) 1 x 2 + 1 (x) f f (1) 2 + 1 = 0 pretend" f(1) 1) F ind f( ), f(1), f(2), and f(0)

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3. (18 pts) Show your work. Let f be a function defined by f(x) = x12 (a) Write the form of the limit and evaluate the limit. Do not use L'Hôpital's Rule. Show your work. 12 12 - 40=(0) - -= lim - 2 = * + 2) = Lima (68–1672 - FORM: (b) Find the natural) domain off. Write your answer in interval notation. (-0, -2) (-2, +00) (c) Let f-1 be an inverse of f. Find an expression for f-1(x), for x in the domain of f-1. = y(x + 2) = 12 yx + 2y = 12 = xy = 12 - 2y = 12 - 2y x= -- x= 12 - 2, and by switching x and y, we get y= 12 - 2 f(x) = 12 (d) Find the value or say "does not exist" (DNE). 1.84(f(n)) = 1 ii. f(p-10)) = DNE, because 0 is not in the domain of f-1

4. (21 pts) Let g be the function given by ( 6 - X 2 + x if x < 2, x + -2 g(x) = x2 – x-1 if x>2. (a) (6 pts) Use the definition of continuity to determine whether the function g is continuous at 2. Show your work. (b) (3 pts) Evaluate the limit. Justify your answer. Show your work. Write "does not exist" only if the limit does not exist and is not + or -0. lim g(x) = lim -2+ = +00 X 2 + X since the form of the limit is NUM: POS, DENOM: POS (c) (2 pts) Write the equation(s) of all vertical asymptotes of g. Write "none" if appropriate. The result in part (b) implies that the line x= -2 is a vertical asymptote. This is the only vertical asymptote, since the function g is continuous on its domain (d) (2 pts) Determine the largest) intervals of continuity of g. Use interval notation to write your answer. Let g be the function given by 12+ if x < 2, x# -2 2 + x g(x)= x2-x-1 x-1 if x>2. (e) (6 pts) Evaluate each limit. Do not use L'Hôpital's Rule. Write "does not exist" only if the limit does not exist and is not too or -00. Show your work. 6-1 i lim g(x) = lim -00 2 + X = lim -002 + x 0-1 + 1 x -00 x x x -+- 2 +1 0 0 + 11 x2 - x-1 1 ii. lim g(x) = lim x++ V x2 – x-1 X-1 x2 - x-1 X - 1 = lim x +oc x +0 - lim Vam M93001 (f) (2 pts) Write the equation(s) of all horizontal asymptotes of g. Write "none" if appropriate.

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.