...__________PREVIEW________________ Name and student number (print clearly) MATH1013A 3.0 Time: 45 minutes Value:100 Test no. 2 Preview F2013 No calculators or written material allowed. Attempt as many questions as you please. Maximum is 100. Part marks will be awarded for relevant and correct intermediate work. While the focus on this test is Ch.3, you must know material in Ch.1 and Ch.2. It is not sensible to expect to do well with Ch.3 while being unfamiliar with earlier material. It is the nature of mathematics where one topic builds on an earlier one. §3.10 & §3.11 are not examinable. [Note: While it is true that doing well in assignments should help you in this test, you must recognize that the format of an electronic assignment does not enable certain types of questions to be asked. Therefore, you need to prepare for different types of questions in addition to those you encounter in assignments. This is especially true in view of the fact that, for instance, it is difficult in an e-assignment to ask for intermediate work in obtaining a derivative starting with its definition – namely using a limit. In the following I will attempt to give a flavour of how some questions may be worded.] There will be 4 questions of equal weight, drawn from various topics. It is entirely possible that a topic found in the text but not appearing below may appear in your test, i.e. what appears below does not constitute an exhaustive list of examinable topics. 5 possible questions are presented...
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...My name is Sun Liying. I am majoring in software engineering in Shandong University. Four-year college study awards me excellent mathematics knowledge and organized thought. Apart from my major courses, I have been crazy about attending lectures of Economy Department in my college. The effect of market and economy attracted me a lot. Therefore, I completed macroeconomics, microeconomics and accounting in my sophomore year. In addition, my father is taking managing position in an entrepreneur. Accordingly, I have been interested in business courses and I decide to take financial related courses, especially Actuarial Studies as the direction of my master’s study. I have been interested in numbers since I was a high school student. I felt satisfied even though I had to contribute more than one hour to solve a mathematic problem. I often spent time to think about other methods to solve mathematic problems that my teacher had provided answers. My enthusiasm about mathematics was inspired again when I began my college study. I took some basic mathematics concepts, such as limit, series, calculus and differential coefficient. I also learned some basic theories and the application of related concepts, such as differential coefficient of function of one variable, calculus, partial derivative of function of many variables, differential equation, and Taylor's formula, intermediate value theorem and infinite series which help me to know the nature of function, and the independent vector...
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...Managerial Mathematics(QQM 1023) Tutorial 2 – Introduction to Function 1. Which of the following equations define y as a function of x? a) y = 3x + 1 b) y = 2x2 c) y = 5 d) y = 2x e) x = 3 f) y2 = x g) y = x3 h) y = [pic] i) y = x j) y = [pic] 2. Determine types of function for the following equations: a) f(x) = 2 b) g(x) = [pic] c) f(x) = 4 – x d) f(x) = 2x e) g(x) = x2 + 3x f) h(x) = 2x g) h(x) = ex h) g(x) = x2 i) h(x) = [pic] j) h(x) = [pic] 3. Find the values for each function based on the inputs given: a) f(x) = 3x – 5, f(-1) and f(0) b) g(x) = x2 – 3x, g(4) and g(-2) + g(0) c) f(t) = [pic], f(2) and f(0) d) g(t) = 2t , g(3) and g(0) – g(1) e) h(x) = [pic] , h(-1) and h(3) f) h(x) =[pic] , h(-3) and h(1) + h(0) g) g(x) = [pic] , g(1) and g(-1) h) f(x) = [pic] , f(1) and f(9) – f(-1) i) g(x) = [pic] , g(2) and g(0) + g(-2) j) f(x) = 5 , f(0) and f(-1) – f(5) 4. Find the domain of each function: a) f(x) = [pic] b) h(x) = [pic] c) g(x) = [pic] d) g(x) = [pic] e) h(x) = [pic] f) f(x) = [pic] g) f(x) =[pic] h) h(x) = [pic] i) f(t) = 4t2 – 5 j) f(x) = 4 k) g(x) = [pic] l) f(x) =[pic] m) f(x) = [pic] n) h(x) = [pic] 5. Based on the graphs, determine the domain and range of the given functions. a) y = x...
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...Analysis II Solutions: June 2010 April 2013 Question 1. a) (i) (ii) The sequence an converges to a if for any 1 1 > 0 there exists an N such that |xn − a| < for n ≥ N (3n + 22n ) n = (3n + 4n ) n → max {3, 4} = 4 by a standard result. 4n2 2n +2n3 4n 8n4 2n +n3 4n = 1 4 ( n )( 1 )n +2 2 8n( 1 )n +1 2 1 4 → 2 by the product, sum and quotient rule. 1 1 4 1 1 1 1 1 1 5 5 (iii) (5n3 +6n4 ) n = 6 n n n ( 6n +1) n and limn→∞ 6 n n n = limn→∞ 6 n (n n ×n n ×n n ×n n ) = 1 and limn→∞ ( 6n +1) n = 1 1 4 1 7 and hence limn→∞ 5 n n n ( 5n + 1) n = 1 by the product rule. b) A sequence an eventually has some property if ∃N such that aN +n has this property ∀n > 0. b Assume bn → b and choose = 2 then ∃N such that for n > N , |bn − b| < b 2 then 3b b < bn < 2 2 2 b2 b so eventually bn > 2 . If b > 0 then eventually > 1 bbn so 2 |bn − b| 1 1 |bn − b| > = | − | > 0. b2 |bbn | bn b Hence by the sandwich theorem an a bn → b . c) 1 bn → 1 b and then if we let an → a then by the product rule an × 1 bn →a× 1 b and hence n+1 Ratio Lemma: If | aan | ≤ l where 0 ≤ l < 1 then an → 0. n+1 Want to prove that if | aan | ≤ l where 0 ≤ l < 1 then 0 < an+1 < ln a1 We will use proof by induction. First for n = 1, |a2 | ≤ l|a1 | which is true by the assumption on the sequence. Now assume that 0 ≤ |an+1 | ≤ ln |a1 |. Then for the induction step since |an+2 | < l|an+1 | then |an+2 | ≤ ln+1 |an+1 |. So since 0 ≤ l < 1, ln → 0 and then...
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...Name: PAULIN T. Assignment 1 1. Let[pic], where [pic] is the set of integers and[pic]. F(x) is a one-to-one function if we can show that: For [pic] and[pic], [pic]=[pic]===> [pic]=[pic] Let’s find out: [pic]==> [pic]=[pic], and [pic]=[pic] So, [pic]=[pic]==> [pic]=[pic]. Subtracting 101 on both sides gives [pic]==> [pic]=[pic]. Since we’re able to show that [pic]=[pic] ==> [pic]=[pic], we then conclude that [pic] is a one-to-one function. 2. Let’s prove that [pic] is a one-to-one function. To prove that, we have to show that For [pic] and[pic], [pic]=[pic]===> [pic]=[pic] So, [pic]==> [pic] and[pic]. [pic]=[pic]==> [pic]=[pic] Subtracting -3 on both sides gives [pic]=[pic] By simplifying 4, we easily have [pic]=[pic], hence the result!!! Since we proved that [pic]=[pic]===> [pic]=[pic], we easily conclude that [pic] is a one-to-one function. 3. The ceiling function maps every real number to the smallest integer greater than or equal to that number,[pic], where [pic] is the smallest integer greater than or equal to [pic]. a) Is the ceiling function a one-to-one function? Let’s find out: The ceiling function is a one-to-one function if and only if for [pic] and[pic], [pic]=[pic]===> [pic]=[pic] By considering these real numbers, [pic]=5.60 and [pic]=5.88, if we apply the ceiling function to both real numbers; we’ll have the following output: [pic]=6 and [pic]=6. We notice here that [pic]=[pic]=6, but[pic][pic][pic]. We see that using the...
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...modify the numerator using similar approach as in the x−1 lecture example) (d) Let f (x) = 4x2 . Show that f (5 + h) − f (5) = 40h + 4h2 . Hence, f (5 + h) − f (5) = 40 + 4h h Using this result, find f (5). Problem 3. Use differentiation rules (not the formal definition) to do the following: (a) Differentiate y = √ 3 − 6x2 + 49x − 54 2x 2 (b) Differentiate y = x − x 2 (c) Differentiate y = (x + 3x − 5)3 (d) Find the equation of the tangent line to the graph of f (x) = x/(x2 + 2) at x0 = 3. Tip: Given a point on a line and the slope, one can construct an equation of the line. Problem 4. Give an example of a function (if possible), which is defined for all real numbers and (a) is not continuous and not differentiable at x = 0 (b) is continuous, but not differentiable at x = 0 (c) is differentiable, but not continuous at x = 0 Problem 5. (Optional bonus question) (a) Use the formal definition of derivative, i.e. f (x) = limh→0 f...
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...GNU Scientific Library Reference Manual Edition 1.14, for GSL Version 1.14 4 March 2010 Mark Galassi Los Alamos National Laboratory Jim Davies Department of Computer Science, Georgia Institute of Technology James Theiler Astrophysics and Radiation Measurements Group, Los Alamos National Laboratory Brian Gough Network Theory Limited Gerard Jungman Theoretical Astrophysics Group, Los Alamos National Laboratory Patrick Alken Department of Physics, University of Colorado at Boulder Michael Booth Department of Physics and Astronomy, The Johns Hopkins University Fabrice Rossi University of Paris-Dauphine Copyright c 1996, 1997, 1998, 1999, 2000, 2001, 2002, 2003, 2004, 2005, 2006, 2007, 2008, 2009, 2010 The GSL Team. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.3 or any later version published by the Free Software Foundation; with the Invariant Sections being “GNU General Public License” and “Free Software Needs Free Documentation”, the Front-Cover text being “A GNU Manual”, and with the Back-Cover Text being (a) (see below). A copy of the license is included in the section entitled “GNU Free Documentation License”. (a) The Back-Cover Text is: “You have the freedom to copy and modify this GNU Manual.” Printed copies of this manual can be purchased from Network Theory Ltd at http://www.network-theory.co.uk/gsl/manual/. The money raised from sales of the manual...
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...HEALTH, SAFETY AND ENVIRONMENT TITLE: FRACTIONAL DISTILLATION PROCESS CONTENTS 1. Title………………………………………………………………………….1 2. Summary…………………………………………………………………….3 3. Introduction of Case Study…………….…………………………………….4 4. Risk Scenario Development…….…………………………………………...5 5. Justification of Fault Tree Analysis…………………………………………6 6. Procedures of Fault Tree Analysis…………………………………………..7 7. Fault Tree Analysis…………………………………………………………8 8. Possible Risk Associated with Hazards….………………………………...11 9. Accident Consequences…………………………………………………….13 10. Method to Control the Risk………………………………………………...15 11. Solution to Minimize the Risk……………………………………………..17 12. Conclusion………………………………………………………………….18 13. References………………………………………………………………….19 SUMMARY Figure 1: Fractioning Column (Copyright of science-resources.co.uk, 2009) Figure 1: Fractioning Column (Copyright of science-resources.co.uk, 2009) Crude oil is one the most important non-renewable sources on Earth. Demand for this black viscous liquid is growing every day in this era if modern technology. Electricity, vehicles and synthetics are among the major consumers of petroleum fluids or crude oil. Crude oil could be referred to as the ‘black gold’ due to its expensive price and complicated production process. Unlike gold, crude oil naturally is useless in its primary form. A process called fractional distillation or petroleum refining need to be carried out onto the crude oil to separate...
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...Composition and Inverse MAT 222 Instructor Amy Glidewell 19 January 2015 Composition and Inverse When you think of math there are several types of problems that are ment to help prepare for the different skills that you may have. I am going to solve three different problems that were assigned to me. Within he paper I will define what composition and inverse functions are. As stated in the reading , “the notation uses for composition is f composed with g of x or f of g of x” (Dugopolski, 2012). Something that is noticed is that the letters stay in he same order in each expression for the composition. These problems will be graphed and solved further in the paper. The following information was given so I can solve the following expressions. In step one I was given the following finctions: f(x) = 2x + 5, g(x) = x^2 – 3, and h(x) = 7 – x/ 3. To start I am going to compute (f-h) (4) f(4) = 2*4 + 5 = 13 h(4) = (7-4)/3 = 1 (f-h)(4) = f(4) -h(4) = 13 - 1 = 12 Answer: (f-h)(4) = 12 Evaluate the following: (f o g)(x) and (h o g)(x) (f o g)(x) = f(g(x)) = f(x^2-3) = 2(x^2-3)+5 = 2x^2 - 6 + 5 = 2x^2 - 1 (h o g)(x)=h(g(x)) = h(x^2-3) = (7-(x^2-3))/3 = (10-x^2)/3 Transform the g(x) function so that the graph is moved 6 units to the right and 7 units down. Function will be G(x) = (x-6)^2 - 10 Find the inverse function: f^-1(x) and h^-1(x) For finding f^-1(x), replace x by y and f(x) by x in f(x)=2x+5 x = 2y + 5 2y = x-5 y = (x-5)/2 Thus f^-1(x) = (x-5)/2 For...
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...Composition and Inverse Tara Fuentes Mat 222 Week 5 Assignment Kristopher Childs 1/19/15 In this week's assignment we need to complete composite and inverse functions. The x and y interchange when a function is inversed, otherwise the points are identical. This is the following function. f(x)=2x+5 g(x)=x2-3 h(x)= 7-x/3 First we need to compute (f-h)(4) (f*h)(4)=f(4)-h(4), each function can be done separately f(4)=2(4)+5 f(4)=8+5 f(4)=13 H h(4)=(7-4)/3 same process as above h(4)=3/3=h(4)=1 (f-h)(4)=13-1 (f-h)(4)=12 this is the solution after substituting and subtracting The next part we need to replace the x in the f function with the g (f*g)(x)=f(g(x)) (f*g)(x)=f(x2-3) (f*g)(x)=2x2-1 is the result Now we need to do the h function (h*g)(x)=h(g(x)) (h*g)(x)=h(x2-3) (h*g)(x)=7-(x2-3) (h*g)(x)=10-x2 end result The inverse function-- f-1(x)=x-5h-1(x)=-(3-7) By doing problems this way it can save a person and a business a lot of time. A lot of people think they don't need math everyday throughout their life, but in all reality people use math almost everyday in life. The more you know the better off your life will...
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...4-1 Exponential Functions 1. What is the definition of an exponential function? Page 412 An exponential function f with base b is defined by f(x) = bx or y = bx, where b is a positive constant other than 1 (b > 0 and b is not equal to 1) and x is any real number. Example: g(x) = 10^x 2. What is the inverse of an exponential function? Page No horizontal line can be drawn that intersects the graph of an exponential function at more than one point. This means that the exponential function is one-to-one and has an inverse. Example: fx = b(x) Steps for solving problem: Replace: fx with y: y = b(x) Interchange x and y: x=b(y) Solve for y 3. What are the characteristics of an exponential function? Page 415 * The domain of f(x) = b^x consists of all real numbers. The range of F(x) = b^x consists of all positive real numbers (0, to infinity). * The graphs of all exponential functions of the form f(x) = b^x pass Through the point (0, 1) because f(0) = b^0 =1 (b not equal to 0) the y intercept is 1. There is no x intercept. * If b > 1, f(x) – b^x has a graph that goes up to the right and is an increasing function. The greater the value of b, the steeper the increase. * If 0 < b < 1, f (x) = b^x has a graph that goes down to the right and is a decreasing function. The smaller the value of b, the steeper the decrease. * F(x) = b^x is one-to-one...
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...series of sine and cosine functions: 1 f x a0 n 1 a n cos nx a1 cos x b1 sin x bn sin nx a3 cos 3x b3 sin 3x a0 a2 cos 2x b2 sin 2x Earlier, Daniel Bernoulli and Leonard Euler had used such series while investigating problems concerning vibrating strings and astronomy. The series in Equation 1 is called a trigonometric series or Fourier series and it turns out that expressing a function as a Fourier series is sometimes more advantageous than expanding it as a power series. In particular, astronomical phenomena are usually periodic, as are heartbeats, tides, and vibrating strings, so it makes sense to express them in terms of periodic functions. We start by assuming that the trigonometric series converges and has a continuous function f x as its sum on the interval , , that is, 2 f x a0 n 1 a n cos nx bn sin nx x Our aim is to find formulas for the coefficients a n and bn in terms of f . Recall that for a power series f x cn x a n we found a formula for the coefficients in terms of derivn atives: cn f a n!. Here we use integrals. If we integrate both sides of Equation 2 and assume that it’s permissible to integrate the series term-by-term, we get y f x dx y a 0 dx y n 1 a n cos nx bn sin nx dx 2 a0 n 1 an y cos nx dx n 1 bn y sin nx dx But y cos nx dx 1 sin nx n sin nx dx f x dx 1 sin n n 0. So 2 a0 sin n 0 because n is an integer. Similarly, x y 1 2 ❙❙❙❙ FOURIER...
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...definition as the limit except it requires x 0 and sgn ( a ) = -1 if a < 0 . 1. lim e x = ¥ & x®¥ x ®¥ x®- ¥ lim e x = 0 x ®0 - 5. n even : lim x n = ¥ x ®± ¥ 2. lim ln ( x ) = ¥ 3. If r > 0 then lim x ®¥ & lim ln ( x ) = - ¥ 6. n odd : lim x n = ¥ & lim x n = -¥ x ®¥ x ®- ¥ b =0 xr 4. If r > 0 and x r is real for negative x b then lim r = 0 x ®-¥ x 7. n even : lim a x + L + b x + c = sgn ( a ) ¥ n x ®± ¥ 8. n odd : lim a x n + L + b x + c = sgn ( a ) ¥ 9. n odd : lim a x n + L + c x + d = - sgn ( a ) ¥ x ®-¥ x ®¥ Visit http://tutorial.math.lamar.edu for a complete set of Calculus notes. © 2005 Paul Dawkins Calculus Cheat Sheet Evaluation Techniques Continuous Functions L’Hospital’s Rule f ( x) 0 f ( x) ± ¥ If f ( x ) is continuous at a then lim f ( x )...
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... 5 5 1 3 1 6 2 2 4 6 7 4 8 3 3. The modulo function (a mod n or a modulo n) maps every positive integer number to the remainder of the division of a/n. For example, the expression 22 mod 5 would evaluate to 2 since 22 divided by 5 is 4 with a remainder of 2. The expression 10 mod 5 would resolve to 0 since 10 is divisible by 5 and there is not a remainder. a. If n is fixed as 5, is this function one-to-one? b. List five numbers that have the exact same image. 4. Find limn→∞n2-1n3+n 5. Find limx→∞x1001-xx1000+x 6. Find limx→∞8x6+4x4-3x2x6+x5-7 7. Find lim2x99+x+1,000,000x100+2,000,000 as x→0 8. Prove that fx=x2+7 is a continuous function. 9. Find derivatives of the following functions using differentiation rules. (Do not use the definition of a derivative!) c. fx=2x+5 d. fx=2x5+x3-6x+100 e. fx=ex+2(4x-1) f. fx=7ex g. fx=x2(x-1) h. fx=3x2+7 i. fx=ex(2x+5) j. fx=ex+2+4x-1 10. Find the derivative of the following function at : fx=2x3+x2+9 11. Find the derivative of the following function fx=2x2-x using the definition of a derivative. (Hint: though you can use the rules of differentiation to check your answer, you must use the definition of a derivative to solve this problem in order to receive any credit...
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...Recall our main theorem about vector fields. Theorem. Let R be an open region in E2 and let F be a C1 vector field on R. The following statements about F are equivalent: (1) There is a differentiable function f : R → R such that ∇f = F. (2) If C is a piecewise C1 path in R, then C F · dx depends only on the endpoints of C. (3) C F · dx = 0 for every piecewise C1 simple, closed curve in R. Furthermore, statements (1)–(3) imply (4) curl F = 0, and (4) implies (1)–(3) when R is simply connected (so all four are equivalent when R is simply connected). The purpose of this handout is to explore what happens when R is not simply connected, that is, when there exist non-gradient vector fields with zero curl. An important example Consider the vector field F = −y x2 + y2 i + x x2 + y2 j defined on R = R2\{(0, 0)}, that is, on all of R2 except the origin. Letting P = −y/(x2+y2) and Q = x/(x2 + y2), it is a simple matter to show that ∂Q ∂x = ∂P ∂y , and so curlF = 0. We can show that (1)–(3) are false for F by finding a simple, closed curve C for which C F · dx = 0. Let C be the unit circle parameterized counterclockwise by x = cost, y = sin t, 0 ≤ t ≤ 2π. We have C F · dx = C −y x2 + y2 dx + x x2 + y2 dy = 2π 0 − sin t 1 (− sin t) dt + cost 1 cost dt = 2π 0 dt = 2π. It follows from the theorem that F is not the gradient of any...
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