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ZIRBES MATH 251X FINAL EXAM SUMMER 2016
Your Name End Time
Page Total Points Score
2 22
3 10
4 20
5 16
6 15
7 12
8 22
9 20
10 13
Total 150
• You will have 2.5 hours to complete the test.
• This test is closed notes and closed book and you may not use a calculator.
• In order to receive full credit, you must show your work. Please write out your computations on the
exam paper.
• PLACE A BOX AROUND YOUR FINAL ANSWER to each question where appropriate.
1. (22 pts) Find the following limits. If a limit does not exist, explain why and give ±1 when appropri-
ate.
(a) (3 pts) limx!5
2x2 � 50
x2 � 3x� 10(b) (3 pts) lim
x!0
x
tan�1(4x)
(c) (3 pts) limx!�1
px2 + 4x
4x+ 1(d) (3 pts) lim
x!5+
2� x2
x� 5
(e) (5 pts) limx!2
x�p3x� 2
x2 � 4(f) (5 pts) lim
x!1(ex + x)1/x
2
Elim 4-not )aims ¥55442,
⇐" →= @ ✓
= linn 21×-5*5 )
xss ( × - 5) ( XTZ )
= "nmst¥It=
#=
-@✓= him VIII. as ×→5+
, 2- x2 → 2-25=-23
X→o 4C - x ) +1as ×→gt
,× -5 → a small Pos '
= lim 1M¥ ' Yx # .
XIN -
-4×+1 Yx we have a negative number
Him FIEF # ✓ divided by a small positive ...
xsa
. 11+-3=2( xt -3×-2 )
X'
- (3×-2) y= ( extx )"×
=lim -
xsz LXTZKX - 2) 4+-3×-2 ) my = h(e×tX)"×= k¥+1)
= him112-3×+2*21×+211×-211×+62 ) lxingnlny =lxismn M¥+1)
= , ;mKAI ±lyhl±¥)
×→z 1×+41×-41×+537 ),
* I= lim - = Ixxlxtzxxt* )
thus yygnhy⇒ andlxihnY@ ✓= 1-412+56-2 )
⇒ ✓
2. (6 pts) Sketch the graph of f(x) =
8><
>:
p�x if x < 0
x2 � 3 if 0 x 3
2x� 5 if x > 3
. Then, find the numbers at which f(x)
is discontinuous. Specify whether f is continuous from the right, from the left, or neither at any
discontinuities.
3. (4 pts) Use the definition of the derivative to find f 0(x) if f(x) = 7 + 5x� 3x2.
3
µdiscontinuous at ×=o
and X =3 .
•
Continuous from left at 11=0%•¥°° continuous from right at ×=3
:
f 'C×)= himflXth)-f(×h→oh
= lim 7+5 lxth ) - 3L xth )2 - (7+5×-3×2)
no =
= him 7+5×+54 -3×2 - bxh - 3h27 -5×+3×2hso
-
h
= lim 5h-6X=3h2hso h
= lim ( 5- bx - 3h )
:-5€
4. (10 pts) Find and simplify the derivatives of the following functions.
a) y = t2 arctan(2t) b) f(x) =e1/x
x2
5. (10 pts) Find and simplify the derivatives of the following functions.
a) y = ln
✓ px
x2 + 1
◆b) g(x) = tan2(sin 5x)
4
yt2tarc1anlztftt2.2Hl2tJ.anotaniatna@tkIeIgIYMezxy-ztinx-lnlx4Dgtxt2tanlsinl5xDseElsinl5XD.5vosl5XJYt2xt-xIf-1Otanlsinl5xDseo4sml5xDvosl5x_ytXtt_x.x
2×1×2+1 )
5taYu¥€
6. (9 pts) Find the derivatives of the following functions.
(c) (5 pts) y = (cosx)2x
(d) (4 pts) F (x) =
Z 1
x5
t2p1� t4
dt
7. (7 pts)
(a) (5 pts) Given x2 � xy + y2 = 1 finddy
dx.
(b) (2 pts) Find an equation of the tangent line to x2 � 2xy + y2 = 1 at (1, 0)
5
my =D In tos X )
¥ = 2 in ( no SX ) + 2x . (Tong )
y'
= ( 2 In @SX ) - 2 × Ian X ) y
y'=l2lnloosxh2xtan×)lcosx)2×= - t.IE# at
Ftx ) = Fx t.ME# at )
=IxI÷t*'
- TIE
2 x - × y'
- y t Zyy '= 0
2 y y'
- × y'
= y- 2 ×
y'
( 2y - x ) =
y- 2×
FEIGm =
0 - 24 )
#= 2
y -
y , = m ( x - xD
y - 0 = 2 ( x-D
y=2@
8. (5 pts) A ladder 10 feet long is leaning against the wall of a house. The base of the ladder is pulled
away from the wall at a rate of 2 feet per second. How fast is the top of the ladder moving down the
wall when it’s base is 3 feet from the wall?
9. (4 pts) Find the absolute maximum and absolute minimum of f(x) = x3 � 6x2 + 9x+ 3 on [�1, 2]
10. (6 pts) An open-top box is to be made whose base length is twice its width. If 600 square centimeters
is to be used in its construction, what dimensions maximize the volume of the box?
6
XZT Y2= 100Know d×1d+=z
want dyldt when 11=3Zx # +
Zydatf=O if X=3 ,y=jFy¥, 31 2) +591da¥=0
ftp.t-6/ffHsecf
slept : find critical #S,
make step 2 : Testcnstendptsm
sure they are in domain in fcx ).
TEEIEIEYE'sFIFE's:IEEI30=31×-31×-1 ) fl2)=8 - 24+18+3=5
x⇒c not in domains×Fu
AnbgmmianxtfyY¥@
. V=2w2h,
A=2w2t 2.2Wh +2Wh
h 600 = 2w2t 6Wh
600--2-2- =h
w 6W:o
, v=2w2( 60¥52 )=tzw( 600 - 2w2 )
V= 200W -
}w3 - 10
V '=zoo - 2w2
2w=2o
a600-200--400=200-20To 6
200 =W2
WE no Basekoomxzoomtf ,h=20@
11. (12 pts) Let f(x) = xe�x
(a) (3 pts) Find the intervals of increase/decrease.
(b) (2 point) Find the local extrema. Classify as a maxima or minima.
(c) (3 point) Find the intervals of concavity.
(d) (2 point) Find any inflection points.
(e) (2 pt) Find the horizontal asymptotes, if any.
7
ftxt e- ×
- × e- ×
0 = e-× ( l - × ) =) ×= I
¥-02,594of t '
fincOntN,DtdeconU)N)_fCD=etisa1oca1maXf'
Cx ) = e-×
- × e-×
f' '
( ×) = - e-×
- e- ×+ ×5×0= 5×(11-2) ⇒ ×= 2
¥-0,3 ,Sign of f
"
fisCuonl2.DtoDontARD_ftD-2e-2isiPoI_xiggxtxttigetxEtxinsaetxa@xlismnxtkxhjm.n
them
=wins
.txi
' #
HA@y=Of
12. (6 pts) Given f(x) = x2 + 1, estimate the area under the curve from [0, 8] with n = 4 using:
a) Left endpoints. b) Midpoints.
13. (8 pts) Find the following indefinite integrals.
a)
Z ✓x� 1
x
◆2
dx b)
Zsecx(secx+ tanx)dx
14. (8 pts) Find the following indefinite integrals.
a)
Zepx
pxdx b)
Zx3
1 + x4dx
8
¥nnn¥¥tit
bigiii. to
Ly=21f6)+fw)tfuDtfl6)) My=2lfU)tfC3)tfes)tftD)= 211+5+17+37 ) =
212+10+26+503=2188)= 21603=1200
= @
=/(xtz. x.tt#dX= /lseo2x tsecxtanx )d×
= 11×2-2+ ×' 2) dx
=
,,|.
=/ # ezrxau
duty "2d×- Zfe
"
du
j =aeu@
III'x"y=¥g¥m¥
"
= lqlnlultc=2er×tc |=tlnlHxYtcf
15. (10 pts) Find the following indefinite integrals.
a)
Zx
(x� 3)2dx b)
Zx+ 4
x2 + 1dx
16. (10 pts) Find the following definite integrals.
a)
Z 9
1
pu� 2u2
udu b)
Z ⇡/4
0sec2 x tan2 xdx
9
KEY,a¥f .
= ⇐ " =h¥+x¥a×
- 1H+3i4du auutyltat,f=lx¥a×tf¥ndx=m1ult3uj+c |=1¥f÷t4tantx+c
= tzlnlultltardanxtc=mlx-3l-÷@ |=zW*xyt4awtan×t@
=ftFi- Hau = folurduu=tanx=f{n'
" 2-2u)du
arseaxdx= +343£
arm : III;:F¥:f@=
6-81-(2-1)=6-81-1
=
6-82=-760
17. (5 pts) A particle is moving with the given data. Find the position of the particle. a(t) = 2t+sin t+1,
s(0) = 1, v(0) = �2
18. A particle moves along the x-axis with velocity function v(t) = t2 � 4, where v is measured in meters
per second.
(a) (3 pts) Find the displacement over the time interval [0, 3]
(b) (5 pts) Find the distance traveled over the time interval [0, 3].
10
v A) = t2 - cost + t + C
- ⇐ ooosototc
fI¥I÷⇒EInII¥e*,g-2=-1 + c ⇒ c = - I
vlt )=t2 - cost + t -1
sltktgtsisintttzettb
disp = So }t2- 4) dt = 9- 12
=(tzt3- 4t ) 103= -3€
pbackwards 3 M=
tz(27 ) - 12from start position .
dist = felt -41 at 1¥=- GTE. 4) at + §t2- 4) at
=ftzt3t4t)ly + ( ft'-44123
= -8-3+8 + q - 12 - (813-8)
= -163+16-3
=
-311 + 13 =
-131+313 = 2}@