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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 ofParabolas 
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—£) "
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OC WW?)
OD, y=— X‘ZE)
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v__ _E
OE y” 2(x 4)
12. At ﬁatpoint on the graph of y5 —Jr2 is the‘tangent ﬁne parallel to the line 2x— 4y= 3‘?
l l
0* (3'7)
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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
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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!
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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 havea 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
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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 .
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l REVIEW: WEEK 1: Mub‘ﬁV’éé c Ira/as .' SOLUTIONS:
Adi/“Ho u” (/mc a; arcz, 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—
=1sz— 3+ x[é.(2x—3)'i2] =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+ ("X2)
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 ’
dissentgoose?»
=2wslx3llsinlfllélﬂ
=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 ycoordinate 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 ycoordinate, 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 2x4y=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 pointslope 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 xcoordinate of the point Where the tangent line crosses the
xaxis approximates the value of a zero of f This is found by ; substituting y= U in_the equation of the tangent line: mn=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], 1nu [ =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, lnq=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
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t
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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‘28x 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 foridentifying 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
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s
5 REV/ea: Maize /. NWT/5“" ‘I' (a) Let x=y=0.
Then f(0+0)=f(0)+f(0)+20~0 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 + any3‘:— =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 does 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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 Spring '10
 John
 Calculus

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