Math 180 Exam II Solutions (the graphs are provided in the last page)
1) % error =
3
2)
dv
(100)
v
and
v  x 3  dv  3x 2 dx .
Thus
dv
3x 2 dx
dx
(100) 
(100)  3 (100) 
3
v
x
x
0.1
 100  3% .
10
If f(x) is continuous on [a,b] and f(a) and f(b) have opposite signs, then f(x)=0 in some point in (a,b).
f(x) is continuous on [0,2] since it is a quotient of two polynomials and its denominator is not 0 in
[0,2]. f(0) is negative and f(2) is positive. Thus there is a real root between 0 and 2.
3) Let x= half of the width of the box, y=half of the length of the box.
Then the volume of the box is V
 (2 x)(2 x)(2 y)  8x 2 y .
We need to maximize the volume. Since
there are two variables in the volume equation, use the Pythagorean theorem to eliminate x. From
the picture,
x 2  y 2  100  x 2  100  y 2 (Do not solve for y.
Solving for y would result in
V  8x 2 100  x 2
It is much easier to eliminate
x2 .
. It is much harder to compute the derivative of this
equation.)
V  8x 2 y  V  8(100  y 2 ) y  800 y  8 y 3 .
dV
100
10
 800  24 y 2  0  y 2 
y
dy
3
3
10 2 10 2 10


dimensions are
3
3
3
4)
1
Find the critical point by setting
. Now find x:
x  100  y 2 
dV
0.
dy
10 2
.
3
Thus the
n
n
2
2i 2
ba 2
2i
  xi  x0   i i  .  (3x 2  5)dx  lim  f ( xi )xi  lim  f ( )
0
n


n


n n
n
n
n
i 1
i 1
n
n
2i
2
24
10
 lim  [3( ) 2  5]( )  lim  [ 3 i 2  ]
n 
n
n n i 1 n
n
i 1
n
n
24
10
24 n(n  1)( 2n  1) 10
 lim [ 3  i 2  1]  lim [ 3
 n]  8  10  18
n  n
n  n
n i 1
6
n
i 1
x0  0,  i 
5)
It is given that
dV
2.
dt
1
V  r 2 h . Eliminate r using
3
10 h
3
1
1 3
3
  r  h . V  r 2 h  V   ( r ) 2 (h)  h 3 . Differentiating both sides
proportion:
6 r
5
3
3 5
25
dV
9
dh
dV
dh
2
 h 2
 2, h  5, we have

with respect to t,
. Using
in/sec
dt
25
dt
dt
dt 9
You are asked to find
dh
dt
when h = 5. We know that formula
6) First find all the critical points in the domain. Then compare the y-value at the critical points and end
points.
f ' ( x)  
2
3( x  1)
and 2.
min is
is never zero, but is undefined at x = 1. Now check the y-values at -1, 1,
f (1)  1  3 4 , f (1)  1 , f (2)  1  3 4 .
3
1 4
X
-1
The absolute max is 1 attained at x = 1, the absolute
attained at -1 and 2.
F(x)
f (1)  1  3 4
f (1)  1
1
2
f (2)  1  3 4
7)
V  r 2 h .
Treat h as a constant since there is no measurement error for the height.
dr
dV
dv 2rh
dr
 1%,
 2(1%)  2%
 2 dr  2 . Since
r
V
V r h
r
1
1
1
x is odd and cos x is even. f ( x)  x  cos x  f ' ( x)   sin x . To
There is no symmetry since
2
2
2
1
7 11
,
find all the critical points, let f ' ( x)  0  sin x    x 
. To find possible points of
2
6 6
 3
,
inflection, set f " ( x )  0 . f ' ' ( x )  0  cos x  0  x 
. Making a sign graph f ' ( x ) , f (x ) is
2 2
7
11
7 11
][
,2 ] and f (x) is decreasing in [ ,
] . Thus a local minimum is
increasing in [0,
6
6
6 6
dv  2rhdh 
8)
2
1
3
attained at
x
 3
2
,
2
x  0,
11
6
and a local maximum is attained at
, concave down in [0,
X
y
1
0

2
][
7
, 2
6
3
 3
,2 ] , up in [ , ],
2
2 2
. The points of inflection are
.
1
x  cos x
2

7 /12 1/ 2 
3 / 4 
11 /12  1/ 2 

 /2
7 / 6
3 / 2
11 / 6
2
L( x)  f (a)  f ' (a)( x  a) . Use a = 100, x = 102, f ( x) 
1
L( x)  100 
(102  100)  10  0.1  10.1
2 100
10) The domain is [ 6, ) . There is no symmetry.
9)
x.
Then


1
1
12  3x
is undefined at x  6 , but this is
f ' ( x)  (6  x) 2  x( (6  x) 2 )  (6  x) 2 (2(6  x)  x) 
2
2
2 6 x
not a critical point since it is an end point. Set the numerator 0 12  3x  0 and x  4 is a critical
point. By constructing a sign graph, the function is decreasing in (6,4) , increasing in (4, ) .
1
1
Thus a local minimum is attained at
1
2
1
x  4 .
Compute the second derivative and get
1

2
1
3(2(6  x) )  (12  3x)( 2 (6  x) )
24  3x
2
f " ( x) 

3
2
(2 6  x )
4(6  x) 2
. f”(x)=0 when x is -8, which is not in the
domain. Making a sign graph (only one section: pick x = 0), we see that the graph is always concave
up.
3
X
y  x 6 x
-4
 4 2  5.6
0 (x-intercept)
11)
x2  x1 
0
f ( x1 )
8  6 1 5
 2

f ' ( x1 )
9
3
e x is continuous and differentiable on all real numbers, the assumptions of the mean value
theorem are met. Using the mean value theorem with a  0, b  x , there is a c between 0 and x such
12) Since
that
f ( x)  f (0)  f ' (c)( x  0) .
ec x  x .
Thus
e x  1  ec x
Simplifying this expression, we have
becomes
ex 1 x .
Thus we obtain
e x  1  ec x .
ex  x 1
for all
But since
e c  1,
(0, )
13) Let x = the length of the shadow, y = the distance between the boy and pole. Set up a proportion:
x x y
5
dx 5 dy
dy
dx 20

x y

 4,

Since
ft/sec.
5
16
11
dt 11 dt
dt
dt 11
1
3
4
3
2
1
2
8 3 4 3 4 3
x  x  x (2  x) , thus f ' ( x) is
f ( x)  8 x  x
3
3
3
x  2 . Using a sign test, x  2 is a local maximum, x=0 is a local
14) The domain is all real number:
f ' ( x) 
undefined at x  0 , f’(x) is 0 at
min. (observe that at x = 0, the first derivative is undefined, it is in the domain, it is not a local
5
max/min. Thus the graph has a vertical tangent at x = 0).
2
5
16 
4 
4 
f " ( x)   x 3  x 3  x 3 (4  x)
9
9
9
This is 0 at x = 4, undefined at x = 0. Making a sign graph, they are both points of inflection.
Increasing on
X
-2
0
4
(2, ) , decreasing on (,2) , concave up on (4, ), (,0) , concave down on (0,4)
1
3
y  8x  x
8(2)
0

1
3
4
3
 (2)
4
3
4
8(4)

1
3
4
 (4) 3
y '  tan x  1  0  tan x  1  x  
15) First set the first derivative equal 0 or undefined.
x
sign test,

4
is a local minimum.
y"  sec 2 x  0 is impossible.
f ( x) 
4
. Using a
No points of inflection.
1
1
16)

1
1

2( x  2) 2  2 x ( x  2) 2
x
( x  2) 2 [2 x  4  x]
2
 f ' ( x) 


( x  2)
x2
( x  2)2
x4
( x  2)
3
2
. The critical
point is -4, which is not in the domain. Thus evaluate the function at the end points:
f (1)  1, f (1) 
1
3
. Max
1
3
@ x  1 , min  1@ x  1
X
f
f ( x) 
x
x2
1
-1
1
1/ 3
17) Find
dr
r
given
we obtain
dV
 3% . V  r 2 h  dV  2rhdr .
V
dr
 0.015 .
r
Therefore, it is
Thus 0.03 
1.5%
18) Let x = the side of the rectangle parallel to a, y=side parallel to b.
5
dV 2rhdr
dr

2 .
2
V
r
2r h
Solve for
dr
,
r
x b y
b
b

 x  (b  y ) . Area = xy  (by  y 2 ) . Next find the critical point:
b
a
a
a
b
A'  b  2 y  0  y  . Since the optimization function is quadratic, this must be the absolute
2
a
ab a
a
b
 . Thus the dimensions are by .
maximum. x  a  y  a 
b
b2 2
2
2
3 2x
3
19) A) ) Take ln of both sides, compute the limit, then take e: lim ln( 1  ) = lim 2 x ln( 1  )
x 
x


x
x
1
3
( )
3 x2
3
1
ln( 1  )
3
x
x
 2 lim
2 lim
 6 . Taking e, your answer is e 6 B)
 2 lim
x 
x 
x 
1
3
1
 2
1
x
x
x
2
sin x
2 sin x cos x
lim
 lim
 lim 2 cos 2 x  2
x 0 ln(cos x )
x 0
x 0
sin x
cos x
Set up a proportion:
20) A) Do log differentiation:
1
2x
2x
y '  ln( 2 x  1) 
 y '  (2 x  1) x (ln( 2 x  1) 
) B) Apply the change
y
2x  1
2x  1
ln x
1
)  f ' ( x)  cos(log 4 x)
of base formula: f ( x)  sin(log 4 x)  sin(
C)
ln 4
x ln 4
1
1
y  ln(tan 1 x)  y ' 
1
tan x 1  x 2
1
d) Apply log differentiation: ln f ( x)  ln( 2 x  1)  ln( x  1)  ln 3  ln x 
3
(2 x  1)3 x  1
1
2
1 1
1
2
1
1
f ' ( x) 

  f ' ( x) 
(

 )
f ( x)
2x  1 3 x  1 x
3x
2x  1
3( x  1) x
ln y  x ln( 2 x  1) 
dy 2 x 2  1

, y (1)  2 , first compute y by integrating.
dx
x
2x 2  1
2x 2 1
1
y
dx   (
 )dx   (2 x  )dx  x 2  ln x  c . To find the constant, let x = 1, y = 2.
x
x
x
x
2
2
We get 2  1  ln 1  c  c  1 . Therefore y  x  ln x  1 b)
21) A)To solve the IVP
1
1
1
x
x
1
 ( e x  1  x 2 ) dx   (e  1  x 2 )dx  e  tan x  c
u  e 3 x  du  6 xe3 x dx
2
1
xe3 x
1 e
0
22)
6
2
2
3x
2
1
c)
To find the new limits, if x =0 then
1 e3 1
1
dx  
du  ln 1  u
6 0 1 u
6
u e3
u 0
xe3 x
1 e
2
3x2
dx :
Let
0
u  1and if x = 1 then u  e 3
1
 ln( 1  e 3 )
6
dx
dy
 35,
 25 . After four
dt
dt
dz (290)(35)  100(25)
2
2

hours, x = 140, y= 100, thus z  290  100  307 . Therefore,
dt
307
2
1
1


2
23) The domain is all real numbers. f ( x )  2 x  3 x 3  f ' ( x)  2  2 x 3  2  1 . 2  2 x 3  0  x  1
z 2  ( x  150) 2  y 2  2 z
dz
dx
dy
 2( x  150)  2 y .
dt
dt
dt
It is given that
x3
X = 0 is a critical point since the derivative is undefined. x= -1 is another critical point. By
performing the sign test, x=0 is a local minimum and x = -1 is a local maximum. Note that there is a
cusp at x = 0 since it is a local minimum with undefined derivative.
f " ( x)  
2
3x
4
3
is undefined at x
= 0. It is not a point of inflection since the sign of f”(x) does not change. But use this points to make
a sign graph and we see that the second derivative is always negative.
, decreasing on (-1,0), concave down on
X
Increasing on
(,1), (0, )
( , , )
2
y  2 x  3x 3
-1
2
2(1)  (1) 3  1
0
0
24) Let
f ( x)  1  x , a  0.b  x .
Using the mean value theorem, there is a c between 0 and x such that
f (b)  f (a)  f ' (c)(b  a)  1  x  1 
1
( x  0) .
But since c is between 0 and x,
1
2 1 c
2 1 c
1
1
( x  0)  x  1  x  1  x
. Thus 1  x  1 
2
2
2 1 c
g
(
x
)

f
(
x
)

3
25) Let
. If we can show that there is a c such that g (c )  0 then g (c)  f (c )  3  0

1
2
1
f (c)  3 . g ( x)  f ( x)  3  3x 2  2 x  2 is a polynomial, thus is continuous.
f (10) is negative and f (10) is positive. By the intermediate value theorem, there is a c between 10 and 10 such that g (c )  0 or f (c )  3
which implies that
7
26) First find all the critical points in (-1, 4).

1
16  2 x 2
.
f ( x)  x 16  x 2  f ' ( x)  (16  x 2 ) 2  x( (16  x 2 )( 2 x))  (16  x 2 ) 2 (16  x 2  x 2 ) 
2
16  x 2
2
2
This is undefined at x = 4, which is an end point. This is 0 when 16  2 x  0  x  8  x  2 2 .
Observe that  2 2 is not in the domain, so do not include this value in the table. Thus the absolute
maximum is 8 at x = 2 2 , the absolute minimum is  15 at x = -1
1
X
y  x 16  x 2
1
 15
8
0
2 2
4
1
27)
The constraint is its volume, which can be computed as V  x h  8 . The optimization function is
the surface area of the open top box, which consists of 4 side faces and one bottom face. Thus its
2
surface area is
write s as
s'  
s  4 xh  x 2 .
s  4 xh  x 2  s 
Use the constraint to eliminate h:
32
 x2 .
x
x2h  8  h 
8
x2
. Now we can
Differentiate this and set it 0 to find the critical point:
32
 2 x  0  x 3  16  x  3 16 . Using a sign test, this is the absolute minimum.
2
x
3
16 is the
length of the base that minimizes the surface area.
28) The domain is all real numbers. This is an odd function, thus it is symmetric about the origin. The xintercept and y-intercept are both 0.
f ( x) 
x
1( x 2  1)  x(2 x)  x 2  1

f
'
(
x
)

 2
 f ' ( x)  0
x2 1
( x 2  1) 2
( x  1) 2
when x = -1 and x = 1. A sign test shows a local max is attained at x = -1, local min at x = 1. For HA,
lim
x 
8
x
x2
x
  lim 2
0
x  x
x 1
1

x2 x2
2
and ,
lim
x  
x
0
x 1
2
Thus y=0 is a HA.
For the concavity,
f " ( x) 
 2 x( x 2  1) 2  ( x 2  1)2( x 2  1)( 2 x) ( x 2  1)[( 2 x( x 2  1)  4 x( x 2  1)] 2 x 3  6 x

 2
.
( x 2  1) 4
( x 2  1) 4
( x  1) 3
2 x 3  6 x  2 x( x 2  3)  0  x  0, 3, 3 .
X
Solving
They are all inflection points.
x
x 1
 3 / 4  .43
f ( x) 
 3
-1
2
-1/2
0
1
0
½
0.43
3
29)
s  4r 2 
ds
dr
 8r
dt
dt
. It is given that
ds
dr
3
 3, r  2 

dt
dt 16
ft/sec
30) The domain is all real numbers, no symmetry, no asymptotes. X-int is (1,0), y-int is (0,1) To find the
critical point,
2
f ' ( x) 
3( x  1)
1
3
. This is never 0, but it is undefined at x = 1. Making a sign graph , x =
1is a local min. It is a cusp since it is minimum with undefined first derivative. For points of
inflection,
2
f " ( x)  
9( x  1)
4
3
. This is never 0 but is undefined at x = 1. The sign test shows f”(x) is
always negative by picking x = 0 (concave down). No points of inflection Increasing on
decreasing on
X
1
9
(,1)
, concave down on
( , , )
Y
0
(1, , ) ,
31)
Let b be the half of the measurement of the base (the base is 2b) . Then the height is
The area is
A  b(6  36  b 2 )  A'  6  36  b 2 
we obtain, after simplifying,
b3
b
2
36  b
2
0,
6  36  b 2
Multiply by the LCD
6 36  b 2  2b 2  36  b 4  33b 2  216  0  b 2  9, b 2  24
36  b 2
.
,
. Thus
b  24 . So the base could be 6 or 2 24 .
32) It is given that a (t )  20 . We may set s (0)  0 . Let t 0 be the time took for the car to come to stop.
or
a(t )  20  v(t )  20t  c (we need to find this c) v(t )  20t  c  s(t )  10t 2  ct  d . But d = 0
since s(0)=0. It is also given that s(t 0 )  160, v(t 0 )  0 . Thus we have two equations and two
unknowns: solve the first equation for t 0 and substitute into the second to solve for c:
 20t 0  c  0,10t 0  ct 0  160  t 0 
2
c
c
c
 10( ) 2  c( )  160  c  80 ft / sec
20
20
20
33) Let x be the distance between the launch site and the rocket, z be the distance between the radar
and the rocket. We know that the distance between the radar and the launch site is always 3. Thus
x 2  32  z 2  2 x
dx
dz
 2z
dt
dt
The speed is 6250 mi/hr.
10
. It is given that
z  5,
dz
 5000 .
dt
First compute x, then compute
dx
.
dt
34) Since f(x) is continuous as it is a polynomial, f(-10) is negative, f(10) is positive, f has at least one zero
between 10 and 10. Since f is differentiable for all x, the Rolle’s theorem applies. If f has two real
zeros, then there must be a point that f’(x)=0. But
f ' ( x)  x 6  1
is never 0. Thus it has exactly one
real root.
200  x 6
  200 y  xy  1200 Differentiate both sides with respect to t,
200
y
dy
dy
dx
dx
 10 . Since x = 80, go back to the ratio to
we obtain 200
x y
 0 . It is given that
dt
dt
dy
dt
200  80 6
dy
dy
dx
dy
dy
find y:
  y  7.5 . Then 200  x  y
 0  200  (80)  7.5(30)  0 .
200
y
dt
dy
dt
dt
dt
dy 225

ft / sec
Therefore,
dt 120
3
36) Consider f ( x)  x  sin x This is a continuous function since a polynomial is continuous, a
35) Set up similar triangles:
trigonometric function is continuous and a difference of continuous functions is continuous.
f (10)  0
and
f (10)  0 .
Thus by the IVT, there is a zero in
37) The domain of the function is
f ( x) 
(,1)  (1, ) , no x-intercept, the y-int is 1.
e
e (1  x)  e x
xe x
 f ' ( x) 

1 x
(1  x) 2
(1  x) 2
x
(10,10)
x
.
The numerator is 0 when x is 0. The denominator is 0 when x = -1. But x = -1 is not in the domain so it is not
a critical point. The critical point is thus x = 0 (but x = -1 must be included in the sign graph) . By making a
sign graph, x = 0 is a local minimum. Now concavity:
f ' ( x) 

xe x
(e x  xe x )(1  x) 2  ( xe x )2(1  x)

f
'
(
x
)

(1  x) 2
(1  x) 4
e x (1  x)[(1  x)(1  x)  2 x(1  x)] e x ( x 2  1) e x ( x  1)


.
(1  x) 4
(1  x) 3
(1  x) 2
The numerator is 0 when x =1, denominator
is 0 when x = -1. But x = -1 is not in the domain so it cannot be a point of inflection, although it must be
included in the sign graph.
Increasing in
(0, ) , decreasing in (,0)
concave up
(,1) , concave down (1,  ) .
x
maximum, x = 1 is a point of inflection. Vertical asymptote at x = 0.
ex

x  1  x
lim
by using L’Hopsital.
X
ex
1 x
0
1
1
1.35
11
lim
x 
Thus x = 0 is a local
e
 0 , so this is a one sided HA.
1 x
38) a)
e3x
 (use Lopital)
lim
x  ln x
b
ln y  ln( 1  ax) x 
3e 3 x

lim
1
x 
x
b
3x
lim 3xe   b) let y  (1  ax) x . Then
x 
b ln( 1  ax)
b ln( 1  ax)
ab
 lim
 ab . Since ln y  ab , by taking e
. lim
x

0
x

0
x
x
1  ax
b
of both sides, lim (1  ax) x  e ab
x 0
39) ( f
1
)' (7) 
1
1
f ' ( f (7))

1
1
 4
f ' (2) 1
4
40)
A) Since the index is 3, the domain is all real numbers.\
B)
2
3
x (5  x)  0  x  0, x  5
C) Use f’(x) to make a sign graph\
Inc on
(,2)  (0, ) , dec on (2,0) Local max is attained at x=-2. Local min is attained at x=0
D)
Concave down (,0)  (0,1) , (2,0) Concave up on (1,  ) , x=1 is a poi
12
2 x 1
41) A) let y  (1  ) . Then y 
x
2
2
(1  ) x
ln( 1  )
2
2
x .. ln( 1  ) x  x ln( 1  ) 
x (first work only on the
2
1
x
x
1
x
x
1
(
2
2
1
ln( 1  )
x
x  lim
numerator). lim
x 
x 
1
1
 2
x
x
2
2
lim (1  ) x  e 2 . Thus lim ln( 1  ) x 1
x 
x


x
x
2
)
x2
 2 . Since ln y  2 , by taking e of both sides,
2
(1  ) x
e2
x
 lim

 e2
x 
2
1
1
x
b) First combine the fractions: then apply L’Hopital twice
1
x 0 
cos x  1
 sin x
0
lim sin x  x cos x  lim cos x  cos x  x sin x  2  0
x 0
42) A) Apply log diff:
13
x 0
ln y  ln x x  x ln x 
1
lim ( x  sin x ) = lim (
1
y '  ln x  1  y '  x x (ln x  1)
y
x 0
sin x  x
)
x sin x
B) Apply log diff:
( x  1) 3
y
x2 1
1
3
x
1
y' 
 2

 y' 
y
x  1 x  1 3x  5
y  e tan
C) Chain rule:
1
x2
3
3x  5
( x  1) 3
x2 1
3
 y '  e tan
1
40)
1
1
ln( x 2  1)  ln( 3x  5) 
2
3
3
x
1
 2

)
x

1
3
x

5
x

1
3x  5
x2
2x
d) log diff:
 ln y  3 ln( x  1) 
(
1
 2x
1  x4
y  2  ln y  2 x ln 2 
x
1
y '  2 x (ln 2)(ln 2)  y '  22 2 x (ln 2) 2
y
Side View:
First construct the optimization function, which is the volume of the cylinder: the radius is y, the height is
2x:
V  r 2 h   ( y 2 )(2 x)  2 xy 2 .
Thus
0:
Using the Pythagorean Theorem, we know that
V  2 xy 2  2 x(5  x 2 )  2 (5x  x 3 ) .
5
.
3
V '  2 (5  3x 2 )  0  x 
Thus
Now find the critical point by setting the derivative equals
V  2 (5 x  x )  2 (
41) Combine the terms before applying L’hopital

 lim (
x 0
1
2 1 x
x
1 x 
2 1 x
lim (
x 0
5 10
)( )
3 3
1
1 1 x
 )  lim (
)
x 0
x 1 x x
x 1 x
1
1
1
) 2 
1
2

42) First, observe that the volume of the water is V=(base)(height)
14
x2  y2  5  y2  5  x2 .
The volume of the water is
1
V  bh(20)  10bh .
2
However, one of the variables must be eliminated.
Eliminate b since the questions asks for the rate of change of h. Using similar triangles,
b 2
2
 b h
h 3
3
2
20 2
40
dh
dh 3
V  10bh  10( h)h 
h  10  (2)

 m / sec
3
3
3
dt
dt 8
43)
a)
3
2

1
2
x ( x  1)dx   ( x  x )dx :
now use the power rule: add 1 to the exponent, divide by the
new exponents( or multiply by the reciprocal of the exponent):
b)
c)

3
1
5
( x 2  x 2 )dx 
3
2 2 2 2
x  x c
5
3
2 1
1
1
 )dx  2 e  4 x dx   dx   e  4 x  ln x  c
4x
x
x
4
e

(

x2  1
dx :
x

x2 1
1 1
1
1
dx   ( 2  )dx   ( x  2  )dx    ln x  c
x
x
x
x
x
first separate the fraction.
d) You must find the constant:
y (
dy
1

 e x , y (0)  2
2
dx 1  x
1
 e x )  tan 1 x  e x  c .
x 1
2
Substitute x=0, y=2.
2  tan 1 0  e 0  c  c  1
 y  tan 1 x  e x  1
44) By the mean value theorem,
sin x  cos c( x)  1( x)  x
45)
15
f ( x)  f (0)  f ' ( x)( x  0)  sin x  cos c( x) .
But
cos c  1.
Thus
f ( x)  3 x  f ' ( x) 
a) Pick a=1000,
1
3x
1
L( x)  3 1000 
3(1000)
2
3
( x  1000)  10 
b) Use the area formula:
2
3
. Using the formula
L( x)  f (a)  f ' (a)( x  a)
1
1
20
.
( x  1000) 
x
300
300
3
A  r 2  dA  2r dr 
3
998  L(998) 
dA 2r
dr
 2 dr  2 .
A r
r
998 20

300 3
Since
dr
dV
 3%,
 (2)(3%)  6%
r
V
46)
f ( x)  x  8  f ' ( x) 
1
2 x8
L( x)  f (a)  f ' (a)( x  a)  3 
 f (1)  3, f ' (1) 
1
.
6
Thus
1
( x  1)
6
1
47)
lim (1 
x 0
1 x2
)
x2
approaches
Sketches of graphs
10)
14)
16
‘: Let
e1
ln y  x 2 ln( 1 
1
)
x2
ln( 1 
1
x2
1
1
)
1 2
2
x 
x
(
2
( 3 )
x
2
)
x3

1
1
1 0
Taking e, y
28)
30)
37)
17
18