MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)


 Carol Payne
 3 years ago
 Views:
Transcription
1 MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give n exmple of function with infinitely mny verticl symptotes. Solution. The function y = (x )(x 2) (x n), with domin of definition R \ {, 2,..., n}, hs exctly n verticl symptotes, nmely t x =,..., n. In generl, given n distinct numbers,..., n, the function y = (x )(x 2) (x n), with domin of definition R \ {,..., n }, hs exctly n verticl symptotes, nmely t x =,..., n. The function y = tn x, with domin of definition R \ {..., 3π 2, π 2, π 2, 3π 2,... }, hs infinitely mny verticl symptotes, exctly t x {..., 3π 2, π 2, π 2, 3π 2,... }, which is where cos x = 0. Exercise 2. Let f be function which is differentible everywhere. Suppose tht f (x) > for ll x. Show tht x f(x) =. Solution. Let x < x 2 be rel numbers. By the Men Vlue Theorem, there exists some c (x, x 2 ) such tht f (c) = f(x2) f(x) x 2 x, nd since f (c) > we hve f(x 2 ) f(x ) > x 2 x. In prticulr, substituting x 2 = x nd x = 0, we hve f(x) > x + f(0) for ll x > 0. Thus, since x (x + f(0)) =, we hve x f(x) =. An lternte solution is to use the Fundmentl Theorem of Clculus. We hve f(x) f(0) ( ) = x where we used FTC in the equlity mrked ( ) bove. 0 f (t) dt > x 0 dt = x Be creful bout using indefinite integrls. It is not necessrily true tht f(x) > x for ll x; it is only true for ll x greter thn some number. For exmple, consider the cse f(x) = 2x; then f (x) = 2 so f (x) > for ll x. But f(x) > x only if x > 0. Exercise 3. Grph the function f(x) = x 3 + 6x 2 + 9x. Indicte domin, criticl points, inflection points, regions where the grph is incresing/decresing, x intercepts nd yintercepts, regions of concvity (up or down), locl mxim nd minim, ny symptotes nd behvior t infinity. Solution. The domin is R. A point (x, f(x)) on the grph of the function is criticl point if f (x) = 0 or is not defined; f(x) is differentible everywhere, so the only wy point cn be criticl point is if f (x) = 0. We hve f (x) = 3x 2 + 2x + 9 = 3(x + )(x + 3)
2 2 FRI, OCT 25, 203 so the only criticl points of f re (, f( )) = (, 4) nd ( 3, f( 3)) = ( 3, 0). A point (x, f(x)) on the grph of the function is n inflection point if f (x) = 0 or is not defined; gin, the only wy point cn be n inflection point is if f (x) = 0. We hve f (x) = 6x + 2 so the only criticl point of f is ( 2, f( 2)) = ( 2). We cn fctor f(x) = x(x + 3) 2, so the xintercepts re x = 0, 3. The yintercept is f(0) = 0. The function f is incresing (resp. decresing) in the intervl I if f(x ) < f(x 2 ) (resp. f(x ) > f(x 2 )) for ll x, x 2 I such tht x < x 2 (this is the definition on pge 9). This is equivlent to sying tht f (x) > 0 (or f (x) < 0) for ll x in the intervl. Since f (x) > 0 on (, 3) nd (, ) nd f (x) < 0 on ( 3, ), f is incresing on (, 3), decresing on ( 3, ), nd incresing on (, ). A function f is concve up (resp. down) on the intervl I if f (x) > 0 (resp. f (x) < 0) for ll x I. Since f (x) = 6(x + 2), f (x) > 0 on the intervl ( 2, ) nd f (x) < 0 on the intervl (, 2). So f is concve up on ( 2, ) nd concve down on (, 2). A function which is differentible on ll of R hs locl mximum (resp. minimum) t x = if nd only if f () = 0 nd f () < 0 (resp. f () > 0). We hve f ( ) = 0 nd f ( ) = 6 > 0, so f hs locl minimum t x = ; we hve f ( 3) = 0 nd f ( 3) = 6 < 0, so f hs locl mximum t x = 3. There re no verticl symptotes becuse the function is continuous on the entire rel line. Suppose there is slnt or horizontl symptote; then there exist rel numbers m, b such tht either (f(x) (mx + b)) = 0 or (f(x) (mx + b)) = 0. x x But f(x) (mx+b) is polynomil of degree 3, so we hve contrdiction (recll the exercise bout endpoint behvior of nonconstnt polynomils). So there re no slnt or horizontl symptotes. y ( 3, 0) x ( 2, 2) (, 4) Figure. Grph of f(x) = x 3 + 6x 2 + 9x Notice tht the function f(x) = x defined on R\{0} hs the property tht f (x) < 0 for ll x R\{0}, but it is not decresing on R \ {0} (since f( ) < f()). This hs to do with the fct tht f is not defined t x = 0.
3 MORE FUNCTION GRAPHING; OPTIMIZATION 3 Exercise 4. Find t 6 in three wys: (i) using methods lerned up to nd including the first midterm; (ii) by relizing the it s f (c) for some function f(t) nd some vlue c; (iii) using L Hospitl s Rule. Solution. (i) We hve (ii) Set f(t) = t. Then (iii) We hve t 6 = ( )( t + 4) = t + 4 = 8. t 6 = f(t) f(6) = f (6) = /2 = t /2 t 6 = t = 8. Exercise 5 (Section 4.7, #9). Find the point on the line y = 2x + 3 tht is closest to the origin. Solution. At the point (x, 2x + 3), the distnce to the origin is d(x) = (x 0) 2 + (2x + 3 0) 2 = x 2 + (2x + 3) 2 = 5x 2 + 2x + 9 by the Pythgoren theorem. The function d(x) is defined for ll x nd is differentible everywhere. We wnt to find globl minimum of d(x). To ese computtion, I mke the following (perhps unconventionl) rgument. Let p(x) = 5x 2 + 2x + 9. For ny two nonnegtive rel numbers, the condition x x 2 is equivlent to x x 2. Thus, for ny two rel numbers x, x 2, the condition d(x ) d(x 2 ) is equivlent to p(x ) p(x 2 ) (since d(x) = p(x)). Thus d() is globl minimum vlue of the function d(x) if nd only if p() is globl minimum vlue of the function p(x). Let s find the globl minimum vlue of p(x). For qudrtic polynomils x 2 + bx + c with > 0, the minimum vlue tkes plce t x = b 2 since tht s the vertex of the prbol. 2 So in our cse the minimum of p(x) occurs t x = = 6 5. Thus the desired point is ( 6 5, 2( 6 5 ) + 3) = ( 6 5, 3 5 ).3 Exercise 6 (Section 4.7, #2). Find the points on the ellipse 4x 2 + y 2 = 4 tht re frthest wy from the point (, 0). Solution. I m going to find the point on the upper hlf of the ellipse which is frthest from the point (, 0), then note tht the ellipse is symmetric with respect to the xxis, so the reflection will be frthest, too. The eqution of the upper prt is given by y = 4 4x 2, with domin of definition [, ]. It is continuous on [, ] nd differentible on (, ). The distnce from the point (x, 4 4x 2 ) to (, 0) is given by d(x) = (x ) 2 + ( 4 4x 2 0) 2 = 3x 2 2x + 5 by the Pythgoren theorem. It is continuous on [, ] nd differentible on (, ). Set p(x) = 3x 2 2x + 5. By the rgument given in the solution to Exercise 5, the problem reduces to finding the mximum 2 Exercise: Prove this. 3 Note tht the line joining ( 6 5, 3 ) to the origin is perpendiculr to the line y = 2x + 3. This will lwys be the cse for 5 problems of the form find the point on the line l tht is closest to the point P.
4 4 FRI, OCT 25, 203 of the qudrtic p(x) on the intervl [, ]. If the qudrtic were defined over ll of R, then its globl mximum would occur t x = = 3, which is inside the intervl [, ], so this globl mximum of the qudrtic on the intervl [, ]. Thus the point ( 3 4, 4( 3 )2 ) = ( 3, ) is the point on the upper hlf of the ellipse which is frthest wy from (, 0). On the bottom hlf, the point ( 3, ) is frthest wy. So the desired points re ( 3, ) nd ( 3, ). Exercise 7 (Section 4.7, #24). Find the re of the lrgest rectngle tht cn be inscribed in the ellipse x 2 / 2 + y 2 /b 2 =. Solution. Assume without loss of generlity tht, b re positive. If b =, then the ellipse is ctully circle nd ny rectngle inscribed in circle of rdius is squre of side length 2, so it hs re 2 2. Now ssume b <. I m going to ssume without proof tht, for ny rectngle inscribed in n ellipse which is not circle, its sides re prllel to the xes of the ellipse. 4 Since the sides of our rectngles re prllel to the x nd y xes, its vertices re (x, y), ( x, y), ( x, y), (x, y) for some x [0, ] nd y = b x2. 2 Thus the re of the rectngle is A(x) = 2x 2y = 4bx x2 2 where A(x) is function on the domin [0, ]. It is continuous on [0, ] nd differentible on (0, ). We mximize A(x). Let p(x) = x 2 2 x 4. Then A(x) = 4b p(x). By the rgument given in the solution to Exercise 5, the problem reduces to finding the mximum of p(x) on the intervl [0, ]. Considering p(x) to be qudrtic polynomil in the vrible x 2, we hve tht it tkes mximum when x 2 = = , or when x = 2, which is inside the intervl [0, ]. For this vlue of x, we hve y = b (/ 2) 2 = b 2 2, so the rectngle hs re b 2 = 2b. 5 Exercise 8 (Section 4.7, #54). At which points on the curve y = + 40x 3 3x 5 does the tngent line hve the lrgest slope? Solution. Let f(x) = + 40x 3 3x 5. At (x, f(x)), the tngent line to the curve y = f(x) hs slope f (x) = 20x 2 5x 4. So our problem reduces to mximizing f (x). We hve f (x) = 240x 60x 3 = 60x(x 2)(x + 2), which is 0 only if x = 0, ±2. We hve f (x) = x 2 nd f (0) = 240 > 0, f (2) = 480 < 0, f ( 2) = 480 < 0, so 0 is locl minimum while ±2 re locl mxim. Since f (x) > 0 on the intervls (, 2) nd (0, 2) nd f (x) < 0 on the intervls ( 2, 0) nd (2, ), we hve tht f (x) is incresing on the intervls (, 2) nd (0, 2) nd f (x) is decresing on the intervls ( 2, 0) nd (2, ). Thus ( 2, f( 2)) = ( 2, 223) nd (2, f(2)) = (2, 225) re the globl mxim of f (x). Alterntively, you cn complete the squre in f (x). We hve f (x) = 5(x 4 8x 2 ). Completing the squre gives f (x) = 5(x 4 8x 2 + 6) = 5(x 2 4) , so f (x) tkes its mximum whenever 4 This is minor technicl point which is less relevnt to Mth A but is still necessry in complete solution: if rectngle is inscribed in n ellipse which is not circle, how do I know tht its sides re going to be prllel to the x nd y xes? (For exmple, if the ellipse is circle, then inscribed rectngles re squres, whose sides don t hve to be prllel to the xes.) I hope this following rgument will stisfy you. Assume without loss of generlity tht b < nd rectngle is inscribed in the ellipse such tht its sides ren t prllel to the xes of the ellipse. Apply the liner trnsformtion which shrinks everything long the xxis by the fctor k = b. So the ellipse now becomes circle of rdius b. Under this trnsformtion, the rectngle tht ws inscribed in the ellipse becomes prllelogrm. The four sides of the originl rectngle hve slope m, m, m, m for some nonzero m, so the prllelogrm hs four sides whose slopes re mk, k m, mk, k. Since k >, djcent sides of the m prllelogrm ren t perpendiculr, nd the prllelogrm is not rectngle. Thus the prllelogrm hs n cute ngle, nd such prllelogrms cnnot be inscribed in circles, contrdiction. 5 Note tht this gives the sme nswer s bove for the specil cse b =.
5 MORE FUNCTION GRAPHING; OPTIMIZATION 5 (x 2 4) 2 is minimized, i.e. when x 2 4 = 0, or x = ±2. Thus our points re ( 2, f( 2)) = ( 2, 223) nd (2, f(2)) = (2, 225).
Main topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationPolynomials and Division Theory
Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More informationMath 1102: Calculus I (Math/Sci majors) MWF 3pm, Fulton Hall 230 Homework 2 solutions
Mth 1102: Clculus I (Mth/Sci mjors) MWF 3pm, Fulton Hll 230 Homework 2 solutions Plese write netly, nd show ll work. Cution: An nswer with no work is wrong! Do the following problems from Chpter III: 6,
More informationCalculus III Review Sheet
Clculus III Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationMath 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
More informationThe Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.
Clculus Li Vs The Fundmentl Theorem of Clculus. The Totl Chnge Theorem nd the Are Under Curve. Recll the following fct from Clculus course. If continuous function f(x) represents the rte of chnge of F
More informationIntegral points on the rational curve
Integrl points on the rtionl curve y x bx c x ;, b, c integers. Konstntine Zeltor Mthemtics University of Wisconsin  Mrinette 750 W. Byshore Street Mrinette, WI 5443453 Also: Konstntine Zeltor P.O. Box
More informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More informationMathematics 19A; Fall 2001; V. Ginzburg Practice Final Solutions
Mthemtics 9A; Fll 200; V Ginzburg Prctice Finl Solutions For ech of the ten questions below, stte whether the ssertion is true or flse ) Let fx) be continuous t x Then x fx) f) Answer: T b) Let f be differentible
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationProblem Set 3
14.102 Problem Set 3 Due Tuesdy, October 18, in clss 1. Lecture Notes Exercise 208: Find R b log(t)dt,where0
More informationAP Calculus Multiple Choice: BC Edition Solutions
AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this
More informationMath& 152 Section Integration by Parts
Mth& 5 Section 7.  Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More informationSOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014
SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 014 Mrk Scheme: Ech prt of Question 1 is worth four mrks which re wrded solely for the correct nswer.
More informationWe know that if f is a continuous nonnegative function on the interval [a, b], then b
1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going
More informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
More informationThe graphs of Rational Functions
Lecture 4 5A: The its of Rtionl Functions s x nd s x + The grphs of Rtionl Functions The grphs of rtionl functions hve severl differences compred to power functions. One of the differences is the behvior
More informationMATH SS124 Sec 39 Concepts summary with examples
This note is mde for students in MTH124 Section 39 to review most(not ll) topics I think we covered in this semester, nd there s exmples fter these concepts, go over this note nd try to solve those exmples
More informationIntegration Techniques
Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u
More informationFirst midterm topics Second midterm topics End of quarter topics. Math 3B Review. Steve. 18 March 2009
Mth 3B Review Steve 18 Mrch 2009 About the finl Fridy Mrch 20, 3pm6pm, Lkretz 110 No notes, no book, no clcultor Ten questions Five review questions (Chpters 6,7,8) Five new questions (Chpters 9,10) No
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More information4.4 Areas, Integrals and Antiderivatives
. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationMAT137 Calculus! Lecture 20
officil website http://uoft.me/mat137 MAT137 Clculus! Lecture 20 Tody: 4.6 Concvity 4.7 Asypmtotes Net: 4.8 Curve Sketching 4.5 More Optimiztion Problems MVT Applictions Emple 1 Let f () = 3 27 20. 1 Find
More information( ) Same as above but m = f x = f x  symmetric to yaxis. find where f ( x) Relative: Find where f ( x) x a + lim exists ( lim f exists.
AP Clculus Finl Review Sheet solutions When you see the words This is wht you think of doing Find the zeros Set function =, fctor or use qudrtic eqution if qudrtic, grph to find zeros on clcultor Find
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationAB Calculus Review Sheet
AB Clculus Review Sheet Legend: A Preclculus, B Limits, C Differentil Clculus, D Applictions of Differentil Clculus, E Integrl Clculus, F Applictions of Integrl Clculus, G Prticle Motion nd Rtes This is
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationx 2 1 dx x 3 dx = ln(x) + 2e u du = 2e u + C = 2e x + C 2x dx = arcsin x + 1 x 1 x du = 2 u + C (t + 2) 50 dt x 2 4 dx
. Compute the following indefinite integrls: ) sin(5 + )d b) c) d e d d) + d Solutions: ) After substituting u 5 +, we get: sin(5 + )d sin(u)du cos(u) + C cos(5 + ) + C b) We hve: d d ln() + + C c) Substitute
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More information( ) as a fraction. Determine location of the highest
AB Clculus Exm Review Sheet  Solutions A. Preclculus Type prolems A1 A2 A3 A4 A5 A6 A7 This is wht you think of doing Find the zeros of f ( x). Set function equl to 0. Fctor or use qudrtic eqution if
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Tody we provide the connection
More informationSummary Information and Formulae MTH109 College Algebra
Generl Formuls Summry Informtion nd Formule MTH109 College Algebr Temperture: F = 9 5 C + 32 nd C = 5 ( 9 F 32 ) F = degrees Fhrenheit C = degrees Celsius Simple Interest: I = Pr t I = Interest erned (chrged)
More information( ) where f ( x ) is a. AB Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x).
AB Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 A3 Find the intersection of f ( x) nd g( x). Show tht f ( x) is even. A4 Show tht f
More informationFirst Semester Review Calculus BC
First Semester Review lculus. Wht is the coordinte of the point of inflection on the grph of Multiple hoice: No lcultor y 3 3 5 4? 5 0 0 3 5 0. The grph of piecewiseliner function f, for 4, is shown below.
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationAPPLICATIONS OF THE DEFINITE INTEGRAL
APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its crosssection in plne pssing through
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationSection 7.1 Integration by Substitution
Section 7. Integrtion by Substitution Evlute ech of the following integrls. Keep in mind tht using substitution my not work on some problems. For one of the definite integrls, it is not possible to find
More information0.1 Chapters 1: Limits and continuity
1 REVIEW SHEET FOR CALCULUS 140 Some of the topics hve smple problems from previous finls indicted next to the hedings. 0.1 Chpters 1: Limits nd continuity Theorem 0.1.1 Sndwich Theorem(F 96 # 20, F 97
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More informationB Veitch. Calculus I Study Guide
Clculus I Stuy Guie This stuy guie is in no wy exhustive. As stte in clss, ny type of question from clss, quizzes, exms, n homeworks re fir gme. There s no informtion here bout the wor problems. 1. Some
More informationP 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)
1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this
More informationMain topics for the Second Midterm
Min topics for the Second Midterm The Midterm will cover Sections 5.45.9, Sections 6.16.3, nd Sections 7.17.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationObjectives. Materials
Techer Notes Activity 17 Fundmentl Theorem of Clculus Objectives Explore the connections between n ccumultion function, one defined by definite integrl, nd the integrnd Discover tht the derivtive of the
More information38 Riemann sums and existence of the definite integral.
38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the xxis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
More informationMATH 115 FINAL EXAM. April 25, 2005
MATH 115 FINAL EXAM April 25, 2005 NAME: Solution Key INSTRUCTOR: SECTION NO: 1. Do not open this exm until you re told to begin. 2. This exm hs 9 pges including this cover. There re 9 questions. 3. Do
More informationHow can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?
Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those
More information( dg. ) 2 dt. + dt. dt j + dh. + dt. r(t) dt. Comparing this equation with the one listed above for the length of see that
Arc Length of Curves in Three Dimensionl Spce If the vector function r(t) f(t) i + g(t) j + h(t) k trces out the curve C s t vries, we cn mesure distnces long C using formul nerly identicl to one tht we
More informationapproaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below
. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.
More informationMATH , Calculus 2, Fall 2018
MATH 362, 363 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly
More informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More information( ) where f ( x ) is a. AB/BC Calculus Exam Review Sheet. A. Precalculus Type problems. Find the zeros of f ( x).
AB/ Clculus Exm Review Sheet A. Preclculus Type prolems A1 Find the zeros of f ( x). This is wht you think of doing A2 Find the intersection of f ( x) nd g( x). A3 Show tht f ( x) is even. A4 Show tht
More informationLecture 3. Limits of Functions and Continuity
Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live
More informationContinuous Random Variables
STAT/MATH 395 A  PROBABILITY II UW Winter Qurter 217 Néhémy Lim Continuous Rndom Vribles Nottion. The indictor function of set S is relvlued function defined by : { 1 if x S 1 S (x) if x S Suppose tht
More informationTopics Covered AP Calculus AB
Topics Covered AP Clculus AB ) Elementry Functions ) Properties of Functions i) A function f is defined s set of ll ordered pirs (, y), such tht for ech element, there corresponds ectly one element y.
More information13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS
33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in
More informationMASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK 11 WRITTEN EXAMINATION 2 SOLUTIONS SECTION 1 MULTIPLE CHOICE QUESTIONS
MASTER CLASS PROGRAM UNIT 4 SPECIALIST MATHEMATICS WEEK WRITTEN EXAMINATION SOLUTIONS FOR ERRORS AND UPDATES, PLEASE VISIT WWW.TSFX.COM.AU/MCUPDATES SECTION MULTIPLE CHOICE QUESTIONS QUESTION QUESTION
More informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More information5.4, 6.1, 6.2 Handout. As we ve discussed, the integral is in some way the opposite of taking a derivative. The exact relationship
5.4, 6.1, 6.2 Hnout As we ve iscusse, the integrl is in some wy the opposite of tking erivtive. The exct reltionship is given by the Funmentl Theorem of Clculus: The Funmentl Theorem of Clculus: If f is
More informationQuadratic Forms. Quadratic Forms
Qudrtic Forms Recll the Simon & Blume excerpt from n erlier lecture which sid tht the min tsk of clculus is to pproximte nonliner functions with liner functions. It s ctully more ccurte to sy tht we pproximte
More information20 MATHEMATICS POLYNOMIALS
0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of
More informationPrerna Tower, Road No 2, Contractors Area, Bistupur, Jamshedpur , Tel (0657) ,
R rern Tower, Rod No, Contrctors Are, Bistupur, Jmshedpur 800, Tel 065789, www.prernclsses.com IIT JEE 0 Mthemtics per I ART III SECTION I Single Correct Answer Type This section contins 0 multiple choice
More informationMath 231E, Lecture 33. Parametric Calculus
Mth 31E, Lecture 33. Prmetric Clculus 1 Derivtives 1.1 First derivtive Now, let us sy tht we wnt the slope t point on prmetric curve. Recll the chin rule: which exists s long s /. = / / Exmple 1.1. Reconsider
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Theorem Suppose f is continuous
More informationLecture 14: Quadrature
Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be relvlues nd smooth The pproximtion of n integrl by numericl
More informationUSA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year
1/1/21. Fill in the circles in the picture t right with the digits 18, one digit in ech circle with no digit repeted, so tht no two circles tht re connected by line segment contin consecutive digits.
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More information2008 Mathematical Methods (CAS) GA 3: Examination 2
Mthemticl Methods (CAS) GA : Exmintion GENERAL COMMENTS There were 406 students who st the Mthemticl Methods (CAS) exmintion in. Mrks rnged from to 79 out of possible score of 80. Student responses showed
More information6.5 Numerical Approximations of Definite Integrals
Arknss Tech University MATH 94: Clculus II Dr. Mrcel B. Finn 6.5 Numericl Approximtions of Definite Integrls Sometimes the integrl of function cnnot be expressed with elementry functions, i.e., polynomil,
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationa < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1
Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the
More informationf a L Most reasonable functions are continuous, as seen in the following theorem:
Limits Suppose f : R R. To sy lim f(x) = L x mens tht s x gets closer n closer to, then f(x) gets closer n closer to L. This suggests tht the grph of f looks like one of the following three pictures: f
More informationRecitation 3: More Applications of the Derivative
Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech
More information. Doubleangle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =
Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos(  1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin(  1 ) = π 2 6 2 6 Cn you do similr problems?
More informationChapter 1: Fundamentals
Chpter 1: Fundmentls 1.1 Rel Numbers Types of Rel Numbers: Nturl Numbers: {1, 2, 3,...}; These re the counting numbers. Integers: {... 3, 2, 1, 0, 1, 2, 3,...}; These re ll the nturl numbers, their negtives,
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationMathematics for economists
Mthemtics for economists Peter Jochumzen September 26, 2016 Contents 1 Logic 3 2 Set theory 4 3 Rel number system: Axioms 4 4 Rel number system: Definitions 5 5 Rel numbers: Results 5 6 Rel numbers: Powers
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationFINALTERM EXAMINATION 2011 Calculus &. Analytical GeometryI
FINALTERM EXAMINATION 011 Clculus &. Anlyticl GeometryI Question No: 1 { Mrks: 1 )  Plese choose one If f is twice differentible function t sttionry point x 0 x 0 nd f ''(x 0 ) > 0 then f hs reltive...
More informationI. Equations of a Circle a. At the origin center= r= b. Standard from: center= r=
11.: Circle & Ellipse I cn Write the eqution of circle given specific informtion Grph circle in coordinte plne. Grph n ellipse nd determine ll criticl informtion. Write the eqution of n ellipse from rel
More informationUnit 1 Exponentials and Logarithms
HARTFIELD PRECALCULUS UNIT 1 NOTES PAGE 1 Unit 1 Eponentils nd Logrithms (2) Eponentil Functions (3) The number e (4) Logrithms (5) Specil Logrithms (7) Chnge of Bse Formul (8) Logrithmic Functions (10)
More informationTHE KENNESAW STATE UNIVERSITY HIGH SCHOOL MATHEMATICS COMPETITION PART I MULTIPLE CHOICE NO CALCULATORS 90 MINUTES
THE 08 09 KENNESW STTE UNIVERSITY HIGH SHOOL MTHEMTIS OMPETITION PRT I MULTIPLE HOIE For ech of the following questions, crefully blcken the pproprite box on the nswer sheet with # pencil. o not fold,
More informationMath 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8
Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite
More information