April_Week_1_review

April_Week_1_review - Vat, UH “WW/‘7. 4P Calculus AB t...

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Unformatted text preview: Vat, UH “WW/‘7. 4P Calculus AB t Review: Week 1: April 7’" 2003 —-April 11'” 2003. Sample Review Questions: 1. Multiple Choice: 2. Multiple Choice: Free Response: Explorations: . Non Calculator: V Approidmately 2 minutes per question: Functions; Graphs & Limits ' Derivatives: Concepts and Computation: #15, 21,2, 4, 7,10,12,14, 17,19. A calculator may. be used: ApproXimately 2.6 minutes. per question. Derivatives: Concepts and Computation # 76, 79, 80, 86, ' 3 Questions: Approximately 15 minutes pcr'questionn Definition of the Defivative [NO Calculator] . 'Analysis off given the graph off ' [NO'Calculator] g ' 'Implicitly defined function. [NO Calculator] ‘ Family of-Parabolas - Relating f and f ’ Inverse functions. ‘IVUIV LALLULAI UK REVIEW: WEEK 1 : MULTIPLE CHOICE: 15. The graph ofthe function f is shown in the figure above. Wthh ofthe following statements about f is‘true? ' - o A. nmflx): 1imf(x) _x—)a x—I)b OBS griflxlfl' oo- ggmka ~ OD. Egg,ka ' O E. does not exist. 0 E. nonexistent 7, image): ' ‘ dx 0A 595ian (x3) on.“6x"cos[x3) Oc. 331128) c D. we sin (x?) mm OE. —2$inv('x3)'cos(x3) . 10. An equation ofthe lino tangent to the géph of y: cos(2x) a_t X: % is OK IIy—1=—(xf-£) OB y—1=—2[x—£) " ,5. OC WW?) OD, y=— X‘ZE) 4 v__ _E OE y” 2(x 4) 12. At fiat-point on the graph of y5 —Jr2 -is the‘tangent fine parallel to the line 2x— 4y= 3‘? l l 0* (3'7) - .1 1 GB [5%) 00 (1'4) 4 :14. Let fbe a differentiable function such that. = 2 and f'(3) = 5. If the 'line to the graph of f at x =3 is used to find—an approximation to a zero of j: that . approximation is CA. 0.4' “ QB. .05 O‘c. _2.6 Ob. 3.4 0 EL 3 3 REVIEW’ WEEK 1: MULTIPLE CHOICE: WITH A CA LCULATOR 2x 76. Iff(x)=e , then f'm: 2X OA‘ 1 en 1"2X) OB} 2 2 V 70 OC‘ €21 'ej"(2x+1) OD‘ X2 : ' " ‘ . — 2 ' . v . 79. Let fbeafimction suchthat %f(2+hg f( ) =5. Whlph of the followmg mustbe true? ' 1. fis continuous at is 2. 11. ' fis differentiable at x: 2. ' . ' III. The derivative of 'f is confinuousiat x’= OIL 1013157 013. ‘IIonly' Oc. I_andIIonly _O_D. IandIIlonly V O E. 11 and 111 'only 80. Let f be the function given by f[x) = 29‘":. For what value of x is the slope of the line tangent to the graph of f at (x,f(x)) equal t0.3 ? v OA. 0.168 QB. 0276' 0C. 0.318 OD. 0.342 05,0551 5 t i E i K E i z 86. Let f(x)= Ifthe rate of change off at x=cv is twice its rate 0.. :.,-hangé at x: 1, thenc= 0. P! [H '. -1 ’ D.‘——' t o 2. E____, o. 22 /. Wat/4 we 7775M V/iw .' _ _ For all real numbers x and )7, let f be a function such that f (x mfm=_ * I . h—+0 )7 (3) Find f(-). Justify your answer. (b) Use the definition ofthe derivative to find f‘(x), (C) Find f(x). + y): for) + fly) + .andisuch that (“1% Note: This is thegraph ofthc derivative 1 _ , off. not Ihc‘ graph off. The figure above shows the graph of f’, the denvafive of a function f. The domain ofthe functionj' is the setpfall x such that—3 S XS 3:, (a) For what values of X,- 3 <x < 3, does f have-a relative maximurh? A relafive minimum? : _ 7 _ Justify your answer. _ . (b) For what values of xis the graph off concave up? Justify youranswer. (c) Use the infatuation found in parts (a) and (b) and the fact that f (-3) = U to sketch a possible graph 0f f on the axes grovided. ‘ - ' - - E i i i i I z i g ‘ § E E é Consider the curve defined by the equation y+ cosy= x+l for U S 315 2ft. (8) Find g— interms of'y (b) Writé an equation for each vertical tangent to the curve. 3 , (c) Find “1—33 interms 0fy. ; dx? L . l % . 1 € € . 1 l REVIEW: WEEK 1: Mub‘fiV’éé c Ira/as .' SOLUTIONS: Adi/“Ho u” (/mc a; arc-z, From the figure, lim f(x)= lirn f(x)= lim f(x)= 2 x—9a+ 3—m— X—m Because of the "jump" discontinuity of f at x: b, 1i_>rr}’f(x) does not exist, and Choices A, X C, and D cannot be true. Choice E results horn assuming that li_r>n f(x) X a does not exist when f(a) is not defined. The correct answer (B) was chosen by 59% of test takers. limit is evaluated using the property lim f (x) - £99.- Ha Sign) ‘ grx) As 2: approaches 1 from the right, the limit of the numerator is 1 and the denominator approaches 0 with positive values. Thus, lirn ——x— = +00. x—)1+ As x approaches 1 from the left, the limit of the numerator is 1 and the denominator approaches 0 . . . x With negative values. Thus, 11 -—-= —oo. x—yl- In x Therefore, by the definition of a limit, lim 3- 1—91 In x does not exist. The correct answer (E) was chosen by 34% of test takers. , V I _ . g “wwwfiwmcnfinm The Product Rule and the Chain Rule are both used in computing this derivative. The resulting expression can be simplified by combining two fractions over a common denominator. f’(x) = g—(xrizx‘i 3) d = E(x)-m+x-— =1-sz— 3+ x-[é.(2x—3)'i-2] =xl2x—3+ x J2x—3 = 2x—3 \f2x— 3 =(2x—3)+x «bx—3 3x—3 \l2x— 3 Computation of this derivative requires the use ofthe PowerRule for eachterm . 4’ d 1 ".9 f'(x)=Zx-(-x3+x+-£) - ' d d d -1 =.gx_(—x3)+a(x)+a;[x ) = 432:2 + 1+ ("X-2) 1 Therefore, f'(—l)-—-—3-(--l)2 +1—( i)? =‘3. The correct answer (D) was chosen by 78% of test takers. d dx L. / Choices B and C result from common Product Rule errors. Choice E results from omitting the factor of 2 required by the Chain Rule. (m) The correct answer (A) was chosen by 57% of test takers. X + \l2x— 3 The Chain Rule is applied twice to compute the requested derivative. 7. éccsglfihétosmlz v ’ dissent-goose?» =2wslx3ll-sinlfllélfl =2wslx3l-(-Sin(x3J)-(3x’) =—6x2‘sin[x3)-cos[x3) Choices B and E have Chain Rule errors, and Choice A has a Sign error in the derivative of the cosine function. Choice C, in addition to having a Chain Rule error and a‘ sign error, does not acknowledge that the given function 13 a power ofa cosine function. The correct answer (D) Was chosen by 70% of test takers. The slope ofthe tangent line at 7:: % is equal to the derivative of y= cos(2 x) evaluated at the point of tangency. ,0 , / £[cos(2X)) dx = (—sin(2x) -2)|X 3:5 4 - if =-2 -— 8111(2) =5 4 =4 By substitution, the y-coordinate of the point of tangency is y = 003(2- = 0. The equation ofthe line can be written in point—slope form as (y — 0) = —2 [x— Choices C and D result from sign and Chain Rule errors, respectively, in computing the derivative. Choice B uses an incorrect y-coordinate, and Choice A results from a combination of these errors. The correct answer (E) was chosen by 56% of test takers. The slope of the tangent line to the curve y: f(x)= é—xz when x=ar is far). a. . .'1 / The slope ofthe line 2x-4y=3 is —2-. 1 -the tangentline andthe line 2x—4y=3 areparallel, rep; By the Power Rule, f'(x)= r, so a: f'(a)i=a,vand the point is (a, f(a)) = The graph of f has slope f'(3) at the point (3, f(3))_ hf. - . / In point-slope form, the equation of the tangent line is (y—f(3))= f'(3)-(X—3) equafiori’ and Chaices D and. E both “fleet 6mm in Alter substituting the information given in the question, this calculating the slope of the line. equation becomes For Choice A, x=l2 was substituted into the wrong The correct answer (B) was chosen by 51% of test takers. (y ‘ 2) = 5 ' (X ‘ 3) The x-coordinate of the point Where the tangent line crosses the x-axis approximates the value of a zero of f This is found by ; substituting y= U in_the equation of the tangent line: m-n=sa—n —2=5x—15 n =—=u x 5 Choice B results from interchanging the roles of f (3) and f '(3) ' The correct answer (C) was chosen by 34% of test takers. i The first derivative of dy y, a}, can be found using implicit differentiation. gwgm gas) Finding the second derivative Substituting 3: = 4 and 2 = 3, 0f y, requires the use y - a? 4 ofthe uotient Rule. — =—_ Q “(4,3) 3 a”)? _ 0’ dy 3, 1 _ 4 9 "Elfii 9’1 =_ l) 4 ‘3; 2 =i[_£] m (4,3) 32 dx y 16 d 3+? __}’ “(U1'50) -- 9 y2 =_g§ ‘ fl 27 2 . y Chmces 3: C, and E allresult from various sign errors. The correct answer (A) was chosen by 22% of test takers. Finding this derivative requires use of the When u < I], 1n|u [ =ln(—u) and “1 ChanRuk. ‘/ fix); 1 ghm) g—Gn(—u))=-_3;%(-u) {—1&' __1qu = __1_ . (2 x) It 031 ‘1 =1 Q! = 2x :4: dx' ‘ x2 ’1 Choices A, B, and C result from errors in Note that the absolute value inside the logarithm does not affect the derivative, as demonstrated by comparing the two possible cases. When u > U, ln|q|=ln u and fianwkg-‘g. handling the absolute value, While Choice E ignores the need for the Chain Rule. The correct answer (D) was chosen by 38% of test takers. \a.. 4/543“ /‘ Me, AJ/‘flr (4/.au44roIC. REWEQ’ Determining this derivative requires use of the Quotient Rule and the Chain Rule. 72,, / Choice B reverses the terms in the numerator of the Quotient Rule formula, and Choice D treats the numerator of the Quotient Rule as a sum instead of as a difference. The correct answer (E) was chosen by'75% of test takers. Stateinent II is TRUEbecause the givenlimit is the definition of f'(x) at x= 2. ’ Statement I is TRUE because any function differentiable at x= 2 is alSo continuous at x = 2. Statement III is false. For any Emotion, the fact that the derivative exists at apoint does not guarantee that the derivative is continuous at that point. 7. fl (For example, the fianction x2 sin me 0 gm = X 0, x= 0 ‘ is differentiable at x= [1, but g' is not continuous at if =' 0.) The correct answer (C) was chosen by 40% of test takers. ’ g E l t s t l The slope of the line tangent to the gaph of a function at a given point is the value of the derivative of the function evaluated at that point. The computation of this derivative requires use of the Chain Rule. . a, 5’0. X2 f(x)=2e‘ 5(4):“) / =294‘2-8x 2 =16xe” , 2 . . 2 The equation 16x2” = 3 can be solved usmg the graphical or g the solver capabilities of a graphing calculator. The only solution is x = 0.168. ? Choices B and D both reflect Chain Rule errors, while Choice C results from neglecting the Chain Rule altogether. Choice E ; incorrectly uses the Power Rule to differentiate an exponential function. | The correct answer (A) was chosen by 50% of test takers. The rate of change of f at a given So, / point is the value of the derivative f,(c)_ 2 f, (1) f '(x) evaluated at that point.‘ fungi?) ' "275 dx - = J; l 4 1 l ‘2 - 2X 5- Choice D results from a common - ~ 1 algebra error in solvmg J— = —2-. 1 _ #24; It is given that the value of f ‘(c) A h en is twice the value of flu). The correct answer ( ) was 0 os by 35% oftest takers. WW_,_i._.TMWM,.,iwwufiw.“www.mwmwww.mm. “WWW. WA A, V.ii,ci.,i......,i._.. (a) f has a relative maximum at r = —2 because 0 f ' changes from positive to negative at x = —2 OR o f changes from increasing to decreasing at x = —— 2 OR - f‘(-2)= 0 and f”(—2) <0 f has a relative minimum at x = 0 because 0 f ' changes from negative to positive at x = U OR o f changes from decreasing to increasing at x = 0 OR - f‘(0)=-0 and f"(0) >0 (h) f is concaveup on(-1,l) and(2, 3) because 0 f' is increasing on those intervals OR ' o f "(x) > 0 on those intervals ' (c) 4 points were awarded in part (a), as follows: 1 point was awarded for indicating that f has a relative maximum at x = —2, 1 point was awarded for justification of the correct maximum (note that only one of the three bulleted reasons was needed to receive this point), 1 point was awarded for indicating that f has a relative minimum atx= U, and 1 point was awarded for justification of the correct minimum (note that only one ‘ ofthe three bulleted reasons was needed to receive this point). 2 points were awarded in part (b), as follows: 1 point was awarded for-identifying both intervals (open or closed), and 1 point was awarded for ju$tification of the correct answer (note one of the two bulleted reasons was needed to receive this point). ' 3 points were awarded in part (c), as follows: r 1 point was. awarded for extrema consistent with those found in part (a), 1 point was awarded for concavity consistent with that found in part (b), and 1 point 'was awarded for drawing a smooth curve on the domain [—3, 3]with fC—3) = U 3 l l i i s 5 REV/ea: Maize /. NWT/5“" ‘I' (a) Let x=y=0. Then f(0+0)=f(0)+f(0)+2-0~0 f(0) = MOD 1%) = o (‘9 f ’(x)= 33% [Sig—.419 = lim f(x)+f(h)+ 23*“ fix) h—m h = lim [foo + 2x] h—zoo h =7+ 2x (0) f’(x)=7+2x I j"(x)=’7x+7:2 +0 Usex=01 ' o=f(o_)=o+ c therefore, 0 = E] therefore, f(x) = 7x + x? 2 points were awarded in part (a), as follows: 1 point was awarded for the correct- answer of 0, and '1 point was awarded for a correct Justification. 5 points were awarded in part (b), as follows: 1 point was awarded for the definition of the derivative, 1 point was awarded for expanding for + h) using the functional equation 1 point was awarded for reducing to Ni) 7+ 22:, and 2 points were awarded for correctly evaluating the limit. 2 points were awarded in part (c), as follows: ' 1 point was awarded for correctly taking the antiderivative Of f‘(x) found in part (b) and including the constant, and 1 point was awarded for finding the constant and substimting it into f (x). 3 points were awarded in part (c), as follows: 2 points were awarded for implicit differentiation, but these points were not awarded in either of the following cases: - if dy ——— was not a quotient or product in y, dz 0 if there was any calculus error, and . . . . dy I paint was awarded for substituting the Ex— obtained in part (a) after differentiation, and d 2 y solving for wem.~m_mr_fimw"WWW"..MW..._..»M«Whmmfiwwwwmyem l 3(a) Q _ sinyfl = 3 points were awarded in part (a), as follows: ‘ dx dx 2 points were awarded for implicit dy. . differentiation, but this amount was reducedio Ea _ 3m 3’) =1 1 or 0 points for any of the following errors} at}, l o 1 point was deducted if the differentiat. a; = 1_ sin y was not perforrned with respect to x, o 2 points were deducted for either a Chain Rule error or an error in differentiating the right—hand side of the equation, - 1 point was deducted for a sign error resulting in ~31- + any-3‘:— =1, - and a5! 1 point was awarded for solving for a (however, this point was not awarded if I factoring out 6%:— was not required). ‘ (b) 2;:— mdefinedwhen smy=l pomtswere assigned forpart(b), as follows: ; I point was awarded for emblishing an a east; ' 1 point was awarded forsolvin "at equation 1 x g £— 1 for y (note that the equafionr?_' _ _ intake a 2 trigonometric fiinetion tash‘efifigihl‘ag this FEM-‘3‘. and h l l 1 point was awarded for usingflla't solution for 5 l a l l a l I y = .2: ‘ . equation that detennines where doe-s not 8 y to give an equation of a linefilote that a solution was not efigib‘felvfor point if it came from a value where is defined). 0)) Sir; undefined when siny=1 (Note thatamagmmn ofl point was awarded if an incorrect was computed that exists at ; ' . x . ’ . - i y = j ‘ all paints and the solution stated that there are ' o no vertical tangents. A maximum of 2 out of = x+ l the 3 points were awarded for a sclution which a51_1 5 x=£—1 used —_ 2 d7: l~sinx') ...
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April_Week_1_review - Vat, UH “WW/‘7. 4P Calculus AB t...

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