Study Guide for the Advanced Placement Calculus AB …

1 Copyright 1996,1997 Elaine CheongAll Rights ReservedStudy Guide for theAdvanced PlacementCalculus ABExaminationBy Elaine Cheong1 Table of ContentsINTRODUCTION2 TOPICS TO STUDY3 Elementary Functions3 Limits5 Differential Calculus7 Integral Calculus12 SOME USEFUL FORMULAS16 CALCULATOR TIPS AND PROGRAMS17 BOOK REVIEW OF AVAILABLE Study GUIDES19 ACKNOWLEDGEMENTS192 IntroductionAdvanced Placement1 is a program of college-level courses and examinations that giveshigh school students the opportunity to receive Advanced Placement and/or credit in college. TheAdvanced Placement Calculus AB Exam tests students on introductory differential and integralcalculus, covering a full-year college mathematics are three sections on the AP Calculus AB Examination:1.

2 Multiple Choice: Part A (25 questions in 45 minutes) - calculators are not allowed2. Multiple Choice: Part B (15 questions in 45 minutes) - graphing calculators are required forsome questions3. Free response (6 questions in 45 minutes) - graphing calculators are required for somequestionsScoringBoth sections (multiple choice and free response) are given equal are reported on a 1 to 5 scale:Examination GradeExtremely well qualified5 Well qualified4 Qualified3 Possibly qualified2No recommendation1To obtain a grade of 3 or higher, you need to answer about 50 percent of the multiple-choicequestions correctly and do acceptable work on the free-response section. In both Parts A and Bof the multiple choice section, 1/4 of the number of questions answered incorrectly will besubtracted from the number of questions answered correctly.

3 1 Advanced Placement Program and AP are trademarks of the College Entrance Examination to StudyElementary FunctionsProperties of FunctionsA function is defined as a set of all ordered pairs (x, y), such that for each element x, therecorresponds exactly one element domain of is the set range of is the set of FunctionsIf (x) = 3x + 1 and g(x) = x2 - 1a) the sum (x) + g(x) = (3x + 1) + (x2 - 1) = x2 + 3xb) the difference (x) - g(x) = (3x + 1) - (x2 - 1) = -x2 + 3x + 2c) the product (x)g(x) = (3x + 1)(x2 - 1) = 3x3 + x2 - 3x - 1d) the quotient (x)/g(x) = (3x + 1)/(x2 - 1)e) the composite ( g)(x) = (g(x)) = 3(x2 - 1) + 1 = 3x2 - 2 Inverse FunctionsFunctions and g are inverses of each other if (g(x)) = x for each x in the domain of gg( (x)) = x for each x in the domain of The inverse of the function is denoted find -1, switch x and y in the original equation and solve the equation for y in terms of :If (x) = 3x + 2, then -1(x) =(A) 132x+(B) x3 - 2(C) 3x - 2(D) 12x + 3(E) x 23 The answer is = 3y + 23y = x - 2y = x 23 Even and Odd FunctionsThe function y = (x) is even if (-x) = (x).

4 Even functions are symmetric about the y-axis ( y = x2)The function y = (x) is odd if (-x) = - (x).Odd functions are symmetric about the origin ( y = x3)4 Exercise:If the graph of y = 3x + 1 is reflected about the y-axis,then an equation of the reflection is y =(A) 3x - 1(B) log3 (x - 1)(C) log3 (x + 1)(D) 3-x + 1(E) 1 - 3xThe answer is reflection of y = (x) in the y-axis is y = (-x)Periodic FunctionsYou should be familiar with the definitions and graphs of these trigonometric functions:sine, cosine, tangent, cotangent, secant, and cosecantExercise:If (x) = sin(tan-1 x), what is the range of ?(A) (- /2, /2)(B) [- /2, /2](C) (0, 1](D) (-1, 1)(E) [-1, 1]The answer is range of sin x is (E), but the points at which sin x = 1 ( /2 + k ),tan-1 x is undefined.)

5 Therefore, the endpoints are not : The range is expressed using interval notation:(,)abaxb <<[,]abaxb Zeros of a FunctionThese occur where the function (x) crosses the x-axis. These points are also called theroots of a :The zeros of (x) = x3 - 2x2 + x is(A) 0, -1(B) 0, 1(C) -1(D) 1(E) -1, 1 The answer is B. (x) = x(x2 - 2x + 1) = x(x -1)25 Properties of GraphsYou should review the following topics:a) Interceptsb) Symmetryc) Asymptotesd) Relationships between the graph ofy = (x) andy = k (x)y = (kx)y - k = (x - h)y = | (x)|y = (|x|)LimitsProperties of LimitsIf b and c are real numbers, n is a positive integer, and the functions and g have limits as xc ,then the following properties are Scalar multiple:limxc [b( (x))] = b[limxc (x)]2.

6 Sum or difference:limxc [ (x) g(x)] = limxc (x) limxc g(x)3. Product:limxc [ (x)g(x)] = [limxc (x)][limxc g(x)]4. Quotient:limxc [ (x)/g(x)] = [limxc (x)]/[limxc g(x)], if limxc g(x) 0 One-Sided Limitslimxa + (x)x approaches c from the rightlimxa (x)x approaches c from the leftLimits at Infinitylimx + (x) = Lorlimx (x) = LThe value of (x) approaches L as x increases/decreases without = L is the horizontal asymptote of the graph of .Some Nonexistent Limitslimx 012x limx 0||xx limx 0 sin1xSome Infinite Limitslimx 012x = limx +0ln x = 6 Exercise:What is limx 0 sinxx ?(A) 1(B) 0(C) (D) 2(E) The limit does not answer is should memorize this function is continuous at c if:1. (c) is defined2. limxc (x) exists3.

7 Limxc (x) = (c)Graphically, the function is continuous at c if a pencil can be moved along the graph of (x)through (c, (c)) without lifting it off the :If fxxxxfk()()=+= 3202 ,for x 0and if is continuous at x = 0, then k =(A) -3/2(B) -1(C) 0(D) 1(E) 3/2 The answer is 0 (x) = 3/2 Intermediate Value TheoremIf is continuous on [a, b] and k is any number between (a) and (b), then there is at least onenumber c between a and b such that (c) = CalculusDefinition '(x) = lim x 0fxxfxx()()+ andif this limit exists '(c) = limxc fxfcxc()() If is differentiable at x = c, then is continuous at x = RulesGeneral and Logarithmic Differentiation Rules1. ddx[cu] = cu'2. ddx[u v] = u' v'sum rule3. ddx[uv] = uv' + vu'product rule4.

8 Ddx[uv] = vuuvv'' 2quotient rule5. ddx[c] = 06. ddx[un] = nun-1u'power rule7. ddx[x] = 18. ddx[ln u] = uu'9. ddx[eu] = euu' [ (g(x))] = ' (g(x)) g' (x)chain ruleDerivatives of the Trigonometric Functions1. ddx[sin u] = (cos u)u'2. ddx[csc u] = -(csc u cot u)u'3. ddx[cos u] = -(sin u)u'4. ddx[sec u] = (sec u tan u)u'5. ddx[tan u] = (sec2 u)u'6. ddx[cot u] = -(csc2 u)u'Derivatives of the Inverse Trigonometric Functions1. ddx[arcsin u] = uu'12 2. ddx[arccsc u] = uuu'||213. ddx[arccos u] = uu'124. ddx[arcsec u] = uuu'||21 5. ddx[arctan u] = uu'12+6. ddx[arccot u] = +uu'12 Implicit DifferentiationImplicit differentiation is useful in cases in which you cannot easily solve for y as a function of :Find dydx for y3 + xy - 2y - x2 = -2dydx [y3 + xy - 2y - x2] = dydx [-2]3y2dydx + (xdydx + y) - 2dydx - 2x = 0dydx(3y2 + x - 2) = 2x - ydydx = 2322xyyx + Higher Order DerivativesThese are successive derivatives of (x).

9 Using prime notation, the second derivative of (x), ''(x), is the derivative of '(x). The numerical notation for higher order derivatives is representedby: (n)(x) = y(n)The second derivative is also indicated by :Find the third derivative of y = ' = 5x4y'' = 20x3y''' = 60x2 Derivatives of Inverse FunctionsIf y = (x) and x = -1(y) are differentiable inverse functions, then their derivatives are reciprocals:dxdydydx=1 Logarithmic DifferentiationIt is often advantageous to use logarithms to differentiate certain Take ln of both sides2. Differentiate3. Solve for y'4. Substitute for y5. SimplifyExercise:Find dydx for y = xx221311+ /ln y = 13[ln(x2 + 1) - ln(x2 - 1)]yy' = 13111122xx+ 9y' = + + 21111222213()()/xxxxy' = + 211243223()()//xxMean Value TheoremIf is continuous on [a, b] and differentiable on (a, b), then there exists a number c in (a, b) suchthat '(c) = fbfaba()() L'H pital's RuleIf lim (x)/g(x) is an indeterminate of the form 0/0 or /, and if lim '(x)/g'(x) exists, thenlim fxgx()() = lim fxgx'()'()The indeterminate form 0 can be reduced to 0/0 or /so that L'H pital's Rule can : L'H pital's Rule can be applied to the four different indeterminate forms of /: /, ()/ , /(), and ()/() Exercise:What is limsinxxx +01 ?

10 (A) 2(B) 1(C) 0(D) (E) The limit does not answer is 01 = 1 Tangent and Normal LinesThe derivative of a function at a point is the slope of the tangent line. The normal line is the linethat is perpendicular to the tangent line at the point of :The slope of the normal line to the curve y = 2x2 + 1 at (1, 3) is(A) -1/12(B) -1/4(C) 1/12(D) 1/4(E) 410 The answer is ' = 4xy = 4(1) = 4slope of normal = -1/4 Extreme Value TheoremIf a function (x) is continuous on a closed interval, then (x) has both a maximum and minimumvalue in the SketchingSituationIndicates '(c) > 0 increasing at c '(c) < 0 decreasing at c '(c) = 0horizontal tangent at c '(c) = 0, '(c-) < 0, '(c+) > 0relative minimum at c '(c) = 0, '(c-) > 0, '(c+) < 0relative maximum at c '(c) = 0, ''(c) > 0relative minimum at c '(c) = 0, ''(c) < 0relative maximum at c '(c) = 0, ''(c) = 0further investigation required ''(c) > 0concave upward ''(c) < 0concave downward ''(c) = 0further investigation required ''(c)