St. Augustine, De Genesi ad Litteram, Book II, xviii, 37. (1) Note, however, that mathematici was most likely used to refer to astrologers.

Transcription

1 Quote: [...] Beware of mathematicians, and all those who make empty prophecies. The danger already exists that the mathematicians (1) have made a covenant with the devil to darken the spirit and to confine man in the bonds of Hell. (1) Note, however, that mathematici was most likely used to refer to astrologers. (Quapropter bono christiano, sive mathematici (1), sive quilibet impie divinantium, maxime dicentes vera, cavendi sunt, ne consortio daemoniorum irretiant.) St. Augustine, De Genesi ad Litteram, Book II, xviii, 37.

2 Midterm 1 - Data Overall (all sections): Average Median Std dev Section 80: Average Median Std dev 14.70

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4 Real Grades VS Expected Grades

5 So... the grading of the first midterm exam was... (A) You were way too harsh. Take it easy on us! (B) Tough! (C) Fair. (D) Kind of easy grading. (E) So easy... can t believe I got away with these mistakes. Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 5 / 40

6 Read This First! Please read each question carefully. In order to receive full credit on a problem, solution methods must be complete, logical and understandable.

7 First Midterm Exam, Problem 1: If the statement is always true, circle the printed capital T. If the statement is sometimes false, circle the printed capital F. In each case, write a careful and clear justification or a counterexample. (a) If log 3 a = x and 3 b = y, then y x = a b. T F Justification: (b) If f (x) = log 5 (x), then f 1 (x) = ln(5)e x for all x > 0. T F Justification:

8 First Midterm Exam, Problem 1: f (x) (c) If lim f (x) = 0 and lim g(x) = 0 then lim is not defined. x 2 x 2 x 2 g(x) T F Justification: (d) If f (x) = 7 for all x, then f (x) = 1 2 for all x. T F 7 Justification: (e) If f (5) exists, then lim f (x) = f (5). x 5 T F Justification:

9 First Midterm Exam, Problem 2: (a) Simplify cos(tan 1 (2x)). (b) Find the inverse of the function f (x) = e 7x + 1.

10 First Midterm Exam, Problem 3: Determine the following limits, using algebraic methods to simplify the expression before finding the limit. Evaluate each of the following limits. If the limit does not exist but goes to or, indicate so. If the limit does not exist for any other reason, write DNE with a justification. (a) lim x 2 x 2 + x 6 x 2 (b) lim x 1 x 1 x 1

11 First Midterm Exam, Problem 3: Determine the following limits, using algebraic methods to simplify the expression before finding the limit. Evaluate each of the following limits. If the limit does not exist but goes to or, indicate so. If the limit does not exist for any other reason, write DNE with a justification. (c) lim t 0 e 2t 1 e t 1.

12 First Midterm Exam, Problem 4: For the continuous function f (x) graphed below, f (2) = 3 and f ( 2.5) = 3. (a) Estimate a value of δ 1 > 0 such that 0 < x 2 < δ 1 f (x) 3 < 1. Your answer should be supported by what you draw in the figure. y x 1

13 4 First Midterm Exam, Problem 4: (b) Estimate a value of δ 2 > 0 such that 0 < x ( 2.5) < δ 2 f (x) ( 3) < 1/2. y x 2 3

14 First Midterm Exam, Problem 5: Find the derivative of f (x) = 3 x for x = 0, using the limit definition of the derivative. (No credit for using any other method.)

15 First Midterm Exam, Problem 6: Let x 2 + x if x < 1, g(x) = a if x = 1, 3x + 5 if x > 1. (a) Determine the value of a for which g would be continuous from the left at 1. (b) Determine the value of a for which g would be continuous from the right at 1. (c) Is there a value of a for which g would be continuous at 1? Explain.

16 First Midterm Exam, Problem 7: The graph of y = f (x) is pictured below on the left. Compute the following derivatives. If it does not exist, write DNE. Briefly justify your answers. f (0) = f (1) = f (2) = f (2) = y = f (x) y 3 2 y = f (x) y x x

17 First Midterm Exam, Problem 8: Find the derivatives of the following functions using the rules of differentiation. It is not necessary to simplify after finding the derivative. (a) f (x) = 3x (b) f (x) = 3e x 3 x sin(x)

18 First Midterm Exam, Problem 8: Find the derivatives of the following functions using the rules of differentiation. It is not necessary to simplify after finding the derivative. (c) f (x) = (4x 2 + 2)e x (d) f (x) = 5ex 1 2x 2

19 First Midterm Exam, Problem 8: Find the derivatives of the following functions using the rules of differentiation. It is not necessary to simplify after finding the derivative. (e) f (x) = x sin(x) x 3 + 1

20 First Midterm Exam, Problem 9: Find the points on the curve f (x) = 1 3 x 3 x where the tangent line is parallel to the line y = 3x.

21 First Midterm Exam, Problem 9: Find the points on the curve f (x) = 1 3 x 3 x where the tangent line is parallel to the line y = 3x

22 First Midterm Exam, Problem 9: Find the points on the curve f (x) = 1 3 x 3 x where the tangent line is parallel to the line y = 3x

23 First Midterm Exam, Problem 9: Find the points on the curve f (x) = 1 3 x 3 x where the tangent line is parallel to the line y = 3x

24 MATH 1131Q - Calculus 1. Álvaro Lozano-Robledo Department of Mathematics University of Connecticut Day 13 Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 23 / 40

25 Derivatives (Rates of Change)

26 Let f (x) and g(x) be differentiable functions (i.e., f (x) and g (x) exist at every point x). Derivatives f (x + h) f (x) f (x) = lim. (c) = 0, where c is a constant. (x n ) = nx n 1, where n is real. (a x ) = ln(a) a x, where a > 0. (e x ) = e x. (sin x) = cos x. (cos x) = sin x. Rules (cf ) = cf, where c is a constant. (f + g) = f + g. Product rule: (fg) = f g + fg. Quotient rule: f f g = g fg. g 2 Chain rule: (f (g(x))) = f (g(x))g (x). Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 25 / 40

27 Example (Chain Rule) (f (g(x))) = f (g(x)) g (x) Find the derivative of F(x) = sin(cos(tan(x))).

28 (f (g(x))) = f (g(x)) g (x) Example (Implicit Differentiation) Find an expression for dy dx where 1 + x sin y = exy.

29 (f (g(x))) = f (g(x)) g (x) Example (Implicit Differentiation) sin 1 (x) 1 = (arcsin(x)) = 1 x 2 cos 1 (x) 1 = (arccos(x)) = 1 x 2 tan 1 (x) = (arctan(x)) = 1 loga (x) = 1 (ln(x)) = 1 x, x ln(a) 1 + x 2

30 (f (g(x))) = f (g(x)) g (x) Example (Implicit Differentiation) sin 1 (x) 1 = (arcsin(x)) = 1 x 2 cos 1 (x) 1 = (arccos(x)) = 1 x 2 tan 1 (x) = (arctan(x)) = 1 loga (x) = 1 x ln(a) 1 + x 2 (ln(x)) = 1 x, and (ln( x )) = 1 as well. x

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32 The number e Recall that we defined e as the unique real number a such that a h 1 lim = 1. This is not a very useful definition. Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 37 / 40

33 The number e Recall that we defined e as the unique real number a such that a h 1 lim = 1. This is not a very useful definition. Instead, we have also seen that f (x) = ln(x) has derivative f (x) = 1/x. Thus, f (1) = 1/1 = 1, and by definition Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 37 / 40

34 The number e Recall that we defined e as the unique real number a such that a h 1 lim = 1. This is not a very useful definition. Instead, we have also seen that f (x) = ln(x) has derivative f (x) = 1/x. Thus, f (1) = 1/1 = 1, and by definition 1 = f (1) Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 37 / 40

35 The number e Recall that we defined e as the unique real number a such that a h 1 lim = 1. This is not a very useful definition. Instead, we have also seen that f (x) = ln(x) has derivative f (x) = 1/x. Thus, f (1) = 1/1 = 1, and by definition 1 = f (1) = lim h 0 f (1 + h) f (1) h Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 37 / 40

36 The number e Recall that we defined e as the unique real number a such that a h 1 lim = 1. This is not a very useful definition. Instead, we have also seen that f (x) = ln(x) has derivative f (x) = 1/x. Thus, f (1) = 1/1 = 1, and by definition f (1 + h) f (1) ln(1 + h) ln(1) 1 = f (1) = lim = lim Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 37 / 40

37 The number e Recall that we defined e as the unique real number a such that a h 1 lim = 1. This is not a very useful definition. Instead, we have also seen that f (x) = ln(x) has derivative f (x) = 1/x. Thus, f (1) = 1/1 = 1, and by definition f (1 + h) f (1) ln(1 + h) ln(1) 1 = f (1) = lim = lim ln (1 + h) = lim Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 37 / 40

38 The number e Recall that we defined e as the unique real number a such that a h 1 lim = 1. This is not a very useful definition. Instead, we have also seen that f (x) = ln(x) has derivative f (x) = 1/x. Thus, f (1) = 1/1 = 1, and by definition f (1 + h) f (1) ln(1 + h) ln(1) 1 = f (1) = lim = lim ln (1 + h) 1 = lim = lim ln (1 + h) Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 37 / 40

39 The number e Recall that we defined e as the unique real number a such that a h 1 lim = 1. This is not a very useful definition. Instead, we have also seen that f (x) = ln(x) has derivative f (x) = 1/x. Thus, f (1) = 1/1 = 1, and by definition f (1 + h) f (1) ln(1 + h) ln(1) 1 = f (1) = lim = lim ln (1 + h) 1 = lim = lim ln (1 + h) = lim ln (1 + h) 1/h h 0 Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 37 / 40

40 The number e Recall that we defined e as the unique real number a such that a h 1 lim = 1. This is not a very useful definition. Instead, we have also seen that f (x) = ln(x) has derivative f (x) = 1/x. Thus, f (1) = 1/1 = 1, and by definition f (1 + h) f (1) ln(1 + h) ln(1) 1 = f (1) = lim = lim ln (1 + h) 1 = lim = lim ln (1 + h) = lim ln (1 + h) 1/h h 0 Since e x is continuous, we can exponentiate both sides and obtain: Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 37 / 40

41 The number e Recall that we defined e as the unique real number a such that a h 1 lim = 1. This is not a very useful definition. Instead, we have also seen that f (x) = ln(x) has derivative f (x) = 1/x. Thus, f (1) = 1/1 = 1, and by definition f (1 + h) f (1) ln(1 + h) ln(1) 1 = f (1) = lim = lim ln (1 + h) 1 = lim = lim ln (1 + h) = lim ln (1 + h) 1/h h 0 Since e x is continuous, we can exponentiate both sides and obtain: e = e 1 = e lim h 0 ln((1+h) 1/h ) = lim h 0 e ln((1+h)1/h ) = lim h 0 (1 + h) 1/h. Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 37 / 40

42 The number e Hence: e = lim h 0 (1 + h) 1/h. Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 38 / 40

43 The number e Hence: e = lim h 0 (1 + h) 1/h. Or changing h by 1/n: e = lim n n n. Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 38 / 40

44 The number e e = lim n = n n n n n Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 39 / 40

45 This slide left intentionally blank Álvaro Lozano-Robledo (UConn) MATH 1131Q - Calculus 1 40 / 40