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Unformatted text preview: n No‘reL logo x =
.. . (Iog4 42xlogi7>GZ§z§>(qui99ﬁe). a "1 +Iogzzza+ 3x —(lnez)(a) ' x: e:>logze<l°‘935:> (Doc#011p.16.02‘l) I I Practice for Prelim 1 — Fall 2011 
MaTh1011 l * ! Learning Sfra‘regies Qenfer "Simplify: loglzé + log12 8+ %laglzl81 — l09123 loglzb +log12 8 + log12 M40912 3 =7 ' 7 L H [0912(6 * 8 * JET/ 3) = 10914144) :> 10914122) ; 21091212 =2
Simblifw ('09416)(093 7><Iog79xlogg 3) long 7 r r
log,J a v== (2) (Io937>(::::3><%)_ , = (2)  1  (I09332)(%)=(2)1 (2) . e) =2
Simplify:logz(2  4a) +1092 8" — (l093(ez))(ln3)(lag3(3“)
l0922+logz4a+logZZ3X—(%)(ln3)(a) . , ' 1 +20
.1 + 2a + 3X' —(2ne)(a)
'+ 3x ~20 ‘ =1+3X TRUE or' FALSE: for' X > O, lbgzx < log3X
FALSE by CounTerexample. Need ‘ro show only 933 counterexample
x =1:>l0921<0931: O < O which'is false ' x = 2 :> loQZZ < {0932 ;> 1 < 10932 .2; 31.<'3"°932 :> 3 < 2 which is false ‘
x = 3 : l0923 <logs3 :> l0923 <1; 2'°923 <21:> 3 <2 which is false V
I715 < "+3 :> ln2'> ln3 which is false me: in;
n2 < MS 2) . . 2,
MathIOll .:_'L' 3 ~ Determine Whether the followmg statements are true or false. Please
give a brief explanation (in complete sentences!)  a reason why it's ' trUe, or an example where it fails. r
(a) If f(x) is an odd function, then so is f(x)+ 3 FALSE: f(x.) is odd indicates f(—X) = —f(x). 7 If'we consider g(yx)=‘ f(X)+3, then g(x) is odd if
' g(—X) .—. —_q(X). We have 9(4) = rem+3 = '—f(x’) 4" 3 and '— g(x) = f (X) — 3. v HoWever, g(X) #2 g(x) as
' """¥’t(x)'+3 does not e‘a’uaIItZRFBTEB‘tUFSTs not odEf
(b) 4 Suppose t(x’) is a function defined for all real numbers,
and —x2 3 ﬁx) 3 x2 for every x. Then f(0) = O and f(x) is '
continuous at x = 0. ~ 4
TRUE: f(x) is continOus at x =0 if f (X) = f(0).
As ~x2 svf(x) s x2, at x = O we have 0 3 f(x) 3 O so.f(0)=0.
The sandwich theorem says that as —x2 3 f(x) 3 X2 and “mlf): lingxaz 0, them mm) =0. As Wm = f(0) = o, ‘ f(X) = 0 and f(x) is continuous at O. I
(c) The expressions cos’1(cos(37z=)) and cos"(cos(57z))  are equal.
' TRUE. cos(37z) = —1 and cos(57z) = 1, so our equation boils down to cos‘1(—1) = cos‘1(—1),.I
which is true. cos'1(cos(37z)) = cos1(cos(57z)) = 7r. (Doc #Cllp.26.021) .3) MaTh1011 ' "y‘ir'.w"r’.}'ﬁ i '
1. ’1‘ . An amoun‘r of money _x is invested a‘r 10°/o confinusiJsly
cbmoundin infehes’r rate. Find x as a func’rion Of Time, a , ..... If ‘The initial inves‘rmen’r is $500,000, show ‘rha’r The amoun’r of mom will double inless Than 10 cars. V
. X = 50000020“ ‘ {Se‘r x = 1,000,000 and solve for ’r 1,000,000 = 500,00060'”
2 = 80.11 ,
n2 % Me“? = 0.1flne = 0.17‘ f:10'n2 V ‘
’As 2<e:_>n2<ne :'>In2<1:>101n2<10 The double ﬁrm is less Than 10 years. (Doc #1011p.15.01) Ma’rh1011 ,_ Consider The ‘fUhC'l‘iOl’l f(x) = Wha’r are The domain and range of. 1‘he func‘rion f(><). ' Domain: ' ’
lln(x) is defined for all posi’rive numbers, so x > O.
’ ﬁx can Take on all values, except when lnx=0, so x #1 * 0(f).=(o,1)u(1,oo> w lunexgainq‘ake—onollmealignumbersTsoM/hen weilakew ' The recipribal, we ge‘r all numbers excep’r for 0. SW”) = (—00.0) U (0,00) , Wri’re an equa’rion for The inverse func’rion f'1(x).
1 ' . o ' (solve forx in Terms of y) ' (cross mul’riply) ‘<
l
fl interchange x and y No'fe :
D(f”1)= (—oo,0) u (0,00) which = W)
93(f’1)=(0,1)‘u(1,oo) which = w)  (Doc #011p.16.011) ' — 5.
Ma’rh1011 l Shdw Tha’rg’rhe following func‘rion is confinuous at x = O. r sin(2x) 7
' f(x)= x 2ex farx >0 for X :0 , Defini‘rion ofﬂCon‘rinuilyf. 03c {+5 domain ’
: cuff lim'f()()E"f(c')""“" ~ 7 ~ f(x) is continuous of on inlenionpoin’r 7X7 x
Le’r's figure oul fhe limi‘r from The left and The limi’r from The righ’r X—90— x>O+ infl‘fﬁ ~=le.%'5‘"2‘1“ .
 lim S—‘M—Zﬂ Consl‘qn’r Mulliple Rule X——>O—. 2X 221 Cons’ron‘r Mulliple Rule V Nole: lim 55—? =1
z—>0 Thus,in(\)f(x) = 2 as‘ ligfk) = lilaﬂx) = 2 Also, f(0)=2e° :92 I
As lingﬂx) = f(O) :2, f(x) is conl‘inuous of x = 0. (Doc #1011p.25.02) 6 MaThiOll The following 'sTaTen'ienTs are someTimes True and
r someTimes false. Please give an example ThaT shows V when The sTaTemenT is True, and when if is false.
If a funcTion f(x), defined for all real numbers x, is onTinuous aT x = 1," Then lirnﬂx) does noT exisT. X—)1 f (X) is naT conTinuous of x = 1
f(1)= o :sggﬂx) * =1 = —00
vx—>1—  linli f(x) = 00
for) is noT conTinuous aT x = 1 >= oo, Ligﬂxwm ‘= o
Examle where false
f(X)=x2:3li_:rgf(x)=O ' If Lin; 7‘ (X) 4— O and Egg“
Examle where True . ‘
'f(x)=x3§li_ngf(x);0 I g(X) =4; :>'im g(x) = 00 X x+0
imf(x)g(x)
x>O ‘
= lil'nX2 ~17: liml =1 x—+O X x—90 g(x) = :37 => m9“) = 00 T (Doc #011p.26.011) 3‘ MaTh1011' Please answer The fOllowing quesTions abouT The funcTion f(x)
’ 7 whose graph is shown in The figure below. ' Is f(x) a oneTo—one funcTion?’ Wh or why noT? A funcTion is one—To—one if f(a)¢f(b) for aib.
' ﬁx) is NOT oneTo=one by counTer example, as f(0) = f(4) = —1. ,
We could also say f(x) fails The horizonTal line TesT for y = 2. ' On whaT inTervals ‘beTween —5 and 5 is f(x) increasing?
f(x) is increasin on [—5,—1] and [2, 4). . . For which values of is lim f(x).= —1?y ' X—>c c=O. NoTe: lin}t f(X) = 00 so 4 isNOT a correcT answer. I
x—> ‘ For which values of c is lim ' f(X) :4 lim‘ f(X)? X—)C+‘ X—>C ue c = 4 as well. c=—1,‘c = 2 and one could or f(x) is conTinuou (Doc #1011p.25.03) Mafh1011  ' ‘ 'c5 ~63 +c+2=10'
Consider f(c) 2' c5 — c3 + c + 2
f is a polynomial and Thus con‘rinuous forall c. Le’r's fry ’ro find ‘rwo values such that f (a) <10 < f(b)
f(1) = 15 413 +1+2 =3 1‘(2)=‘25 —23 +2+2 =32—8+2+2=28 ‘ E; Intermediate Value Theorem : "As 1‘ is conﬁnuous on [1,2] and «15:10 'g'f'cz'); uneﬁhereﬂlh
exists a number c in (1, 2)ysuch Tha‘r f(c) =10.
_50, There is a posil‘ive number c such Tha’r c5 — c3 + c + 2 =10. l If‘f(x) = x2 + 10 sin x, sh0w ’rha‘r There is a number c such ' '; Tha‘r f(c) 21000. Give comple’rereasons for your answer.
1‘ is continuous on (709,00) as it is ’rheSum of a polynomial and a “(rig function. v ‘ '  Le’r's Try To find TWo values such ’rha’r
f(0) = 02 +105in(0) = 0 +0 = 0 < 1000
f(1007r) =' (1007:)2 + 10 sin(1007'r) = 10,00 Infermedia’re Value Theorem
As 1‘ is conTinuous on [0,100] and f(0) < 1000 < f(1007r), "(hen
lha’r f(c) =1000. f(a) . <1000 <' f(b) ' 0 +0 '.—. 10,000 >1000' (Doc #011p.26.041) Mo’rh1011
l . l Prelim exis’r. You must show all your work; Note Evolua‘re we following limii’s,'if 'l‘hey
full credi‘r. lha‘r simply ploh‘ing poinl‘s is no’r sufficient To receive 10 . :9 DNE 'ismoll number close to O —1.ssin(x)sl :> ﬁg ' lim 5‘25") = O by Sandwich Theorem 1 _" — 3+o—o = 3 (Doc #011p.25.031) Mofthll Factor ou’r (x+3) X2+2X—3 : X X
x—>—3 Xe +35 xa—B (x+3)(x—1) = (x—l) = (—3—1) = _4e3 ex(x+3) X_9_3 e" e Iim. ———“’ 7:1; 2
1—) 3 Multiply by Conjugate  W—z _  $742 \/¢+1+2 _  {+1—4 _  f—3
I'm ¢—3 ‘Im 7—3 'Jﬁhz _,.'_rg(r—3)(JM+2) —Lm(f—3X«/m+2) Cancél ou’r (1—3)
~ ' . 1 __ 1 _ l»
'23; (We) ' J§+’I+z ’ 4 (Doc #011p.25.021) When x a 3+, ‘x—3 is posi’rive, so Kelsi = 955—31
x —9 X —9 = lim 713:}; as 715 is con‘l’inuous a'r x=3. X—>3+ . 3 _ . _ 3 __ . é _ _ __ . . ’ __
_~}[_gg\__,‘,;.9';yg__(————x;2§ols—. )1 Lugs—ii 952% 3... is. cooilnu_oush9i2$—3. The funclion is no‘r defined of x = 3. If possible, give a continuous eXTension of: This funcl‘ion a‘r x = 3.. If if i ' is no’r possible, please explain Why not.
If is NOT possible To give a conﬁnuous. ex‘lension ‘ ‘. limsi‘g; DNE as The one sided—limits diSagree. Thus, There'is no Such number we can define for f(3)
such lha’r ling H3), as “mks; DNE. X)3X ' ' V—1 x+1
hm sun (2X24) X—>°O 2  . 2 2 2 . ‘ 1 x /x +1/X _
lll’T\————X + = lIm—»————— I X900 2x2—1 X_m 2x2/x2—1/x2 PM As sin'lx is coniinous of x = —§—, ‘
c c X2+1 _ . _1 . X2+1 _ . _1 sm (2X2_1)_ snn (lingo—2X24) — sm (
(Doc #1011p.26.01) ' Ma’rh1011 i Find The vertical and horizontal asymp’ro’res of
__ 2x3—3x+4 '
. ﬁx) — rizon‘ral Asympfo‘re; Iim 7‘ (X) i 6' X—)00‘ Ho O 3
I'm 2x —3)g+4
x—x  2—3/x2+4/X3 2—0+o  ' ‘ .
 Ilm ——————— = = ——2. .
_ X600 1/x2—1 0—1 . “ Herizonml Asympfo‘re a’r y =  Verﬁcal Asy‘mp’roTe: Iim f(X) =ioo or Iim {(x) : 4:00
~ ' ‘ *X—)C— . X—)C+ ‘  “m 2x3—3x+4 X—)C X’X3 __ v ' 2x3—3x44 __ (P05) _ I
X “ 1 ' I'm x(1—X)(1frx) ‘ (p05) ’ +00 x—> 1— 'Ver’rical aéymp‘l‘o’res when x = 91, 0,1 (Doc #011p.25.011) /3 Math1011 Determine whether the following statements are true or false. Please
give a brief explanation (in complete sentences!)  a reason why it's true, or an example where it fails. (0’) A one—to—one function can have at most two vertical asymptotes.
TRUE. A one—to—one function must pass the horizontal line 7 test. If a one—toone function has three (or more) vertical
asymptotes, then two must ggto the same infinity, If Vitudoems this, however, the function would have the same y valuefor different x values, andthus fail the horizontal line test. ,
(e) If f(x) is continuous at x = c, then it has a tangent line at x = c. .
. FALSE. f(x) = lxl is continuoUs, but has a corner at x =0. The slope of the secant line coming from the left is —1
~ and the slope of the secant line coming from the right is 1. t The slope of the graph M at x = 0 does not exist.
3 (7‘) Suppose f(x) is a continous function on the interval ' ‘ [—3, 3], and that f(—3) =14 and f(3) = 1. Then 1‘ must , ' have a root (or zero) somewhere in the interval.
TRUE. The Intermediate Value Theorem says that if f(x) is continuous on [a, b] then it takes on all intermediate values
between f(a) and f(b). As f(x) is continuous and“ f(a) s O s f(b),
f(c) ‘= O for some c between a and b. '  (Doc #011p.26.03t) E MaTthll
LeT ﬁx): X+4 ‘ X+4 = —L for x i —4 v x2 + 3X — 4 : (x+=4)(x—1) x—l
\ '(a) SkeTch The graph of f(x) , (b) WhaT is The domain of ﬁx)
T ' ' (°°.4) U (—4.1) U (1.°°) ' (c) WriTe The equaTion(s) of The '
verTical and horizonTal as mToTe(s). ‘ HorizanTal AsympToTes: y = O ’
VerTical AsympToTes: .x = 1 (d) CompuTe limﬁx) if This limiT exisTs._ Show all work. WV“) =“r‘l‘ 2’34 =. imW=lim x1=.‘5=% (6) Is 1‘ conTinuous aT x = 4
 05 lingﬂx) = —% and f(—4) DNE f(x) is NOT cohTinuoUs as ﬁx) 7: {(4) If no: I WhaT Type of disconTinuiT exisTs aT x = —4? Removable DisconTinuiT , How can on modif f To make iT conTinuous?
Define f(4)=—% because Then (X) .= f(’—4) (Doc #1011p.26.02) l ' /5'
Math1011 ' j ‘Simplifyi i v . \/x2+1oo—10 llm a
X—)O X Multiply by'Conjugate:
Jx2+10040 , x/x2+100+10 : “m x24100—100 ________——— x2 Jx2+100+1o , x_,o x2(\/X2+100+10) . \/1+secx ' . , ' lim ' ‘ Evaluate: '
1 __ 1 81:0 __ _ 2 _ J5 __
«lasecx «’45 + _ J— _Tl___J—2_ LID?) cos(7r+tanx) — cos(7z+1an0) — cos(7r+0) cos(7r) lim 'X—)0' ‘ Iim (ln(1+ X»2 sin(—};) X—aO
We knew ~1s sin(1/x)s1for' any x at 0. Multiply by (In(1+x))2 3 '(n(1+x))z s (ln(1+x))2 sin_(1/ x) s (ln(1+><))2 for any x at 0
As yg<(In(1+x»2) = malnamm = o ' ' ‘
Then ym(ln(1+x))2 sin(1/X) = o by the Sandwich Theorem! 2x a: Iim ——————I
i} H'” x/x2 +1 'Divide by Highest Order in Denominator:
' ZX/X ' 2 ' Z ' 2 ’ 2
—————,._— = I ——————r.= I ———r—=II ,.—= I =—2 X2+1IX Klan—1:0 X2+1/(t'\/¢—\’E) Jamar) _. K331 X—E‘jno _ 5.2.4752. xl._)n:\OO _m J27 rather than X = +x/7 ? Divide by x (why?) ‘ Note: Why is X = — (Doc #011p.25.05t)(5p08, spas, 5po7, Faoa) l6
‘ Math1011 Let f(x) be a continuous function on the interval [0, 1] .
and suppose um n0) = —1 and f(1) = 0. Show that there
is at least one number c in [0, 1] such that t(c)2 = c. _
(Hint: Consider the function g(x) = for)2 — 5],;X'j‘)
Consider g(X) = 7‘()()2 ,— X I .
As f(x) is continuous, ﬁx)2 is continuous (product of '1‘: fais cont.) or
and. for)2 — xis'continuous (ditferences are continuous).
9(0) = 76(0)2 — o = (—1)2 —o :1 Intermediate Value Theorem. .5 . As 9 is continuous on [0,1] and 9(1) < 0 < 9(0), then there
. exists a number c in (0, 1) such that g(c) = 0. ' v 4
' . As g(c) = f(c)2 —c = o ‘:.> f(c)2 = c There exist a number c in [0,1] such that 7‘(c)‘2 = c . 6x2cosx
Iim —,———,————
H0 snnxsuan Iim 6cosx X X xao sinx sinZX Iim 6cosx——— X_)0 sinx 2 sinZX Iim 3cosx—£——?i‘——=3(1)11 X_)0 sinX sian 3 ' (Doc #oi1p.25.o4) ...
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This note was uploaded on 03/01/2012 for the course MATH 1110 at Cornell University (Engineering School).
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 MARTIN,C.

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