Download Math 180 Exam I Solutions

`Math 180 Exam I Solutions
1)
f (0  h)  f (0)
f (h)  f (0)
lim
 lim
 lim
h0
h 0
h 0
h
h
h sin
1
0
1
h
 lim sin .
h 0
h
h
But this limit does not exist (you
can see this from the graph below). Thus the function is not differentiable at x=0
The graph of
sin
1
x
:
2)
a) f has a horizontal tangent at -1, 1,2. They are x-intercepts of the derivative graph. f is increasing
in (-1,0), (1,2) so the derivative graph is positive. Otherwise negative.
B) f has a vertical tangent ay x = 0. Thus its derivative has a vertical asymptote at x = 0. f is
increasing when x is negative. Thus its derivative graph is positive.
C) The graph of f consists of line segments. Thus the slope is constant at each piece. Its derivative
is undefined at each sharp turn.
3)
x  4 x  1

f (x)  
3,
 2x  3,

x 1
x 1
a) At x = 1, the function is defined : f(1) = 3
b)
lim f ( x)  lim ( x  4)  5
x 1
x 1
and
lim f ( x)  lim (2 x  3)  5 , thus the limit exists.
x 1
x 1
1
lim f ( x)  5, f (1)  3 , thus lim f ( x)  f (1)
c)
x 1
x 1
Therefore, the function is not continuous at x = 1. It has a removable discontinuity at x=1.
B) When x  1, f is continuous since it is a polynomial. When x  1, it is continuous since it is a polynomial.
4)
sec( y)  x 2  xy  sec y tan y
5) To show
lim 9  x  2 , first express f ( x)  L
x 5
a  5, f ( x)  9  x .
f ( x)  L 
52
9x 2
1
in the denominator of
9 x 2 
1
Now we find the bound for
1
dy
dy
dy
y  2x
 2x  x  y 

dx
dx
dx sec y tan y  x
9x 2
, which implies that
as a function of
9 x 2
9 x 2
We may assume
xa
9 x 2
  1.
Then
. In this problem,

5 x
x5

9 x2
9 x 2
x 5 1 4  x  6.
, this is the largest when x = 4. Thus the maximum of
f ( x)  f (5) 
x4
Let
52
  min{ 1,

1
}.
.
Since x is
1
9x 2
is
Then when
52
0  x  5    f ( x)  f (5)  
)
6)
( x  h) 2  4  x 2  4
( x  h) 2  4  x 2  4
f ( x  h)  f ( x )
lim
 lim
 lim
h 0
h 0
h 0
h
h
h
 lim
h 0
( x  h) 2  4  ( x 2  4)
h ( ( x  h)  4  x  4 )
7) A)
B)
2
2
 lim
h 0
h( 2 x  h)
h( ( x  h )  4  x  4 )
2
2
 (sub 0 into h) 
( x  h) 2  4  x 2  4
( x  h) 2  h  x 2  4
x
x 4
2
f ( x)  e  x sin 3x  f ' ( x)  3e  sin 3x  3x cos 3x
3x
3x
f ( x)  tan 2 (2 x  5)  f ' ( x)  2 tan( 2 x  5)  sec 2 (2 x  5)  2
C)
f ( x)  e 4 sec 3 x  f ' ( x)  e 4 sec 3 x sec 3x tan 3x(12)
D)
Use the quotient rule:
f  2x  f '  2
1
1
g  ( x  1) 2  g ' 
f ( x) 
2x
x 1

1
1
( x  1) 2 
2
2 x 1
(2 x  1  2 x
 f ' ( x) 
1
2 x 1
( x  1) 2
2 x 1 
)

( x  1)
x
x 1
2 x  1( x  1) 

x
x 1
( x  1) x  1
x 1
2

2( x  1)  x
( x  1)
3
2
x2

( x  1)
3
2
Note that the second to the last equality was obtained by multiplying each
term of the complex fraction by
x 1
Another way to simplify the expression is to factor out
1
2

1
2
:
1

2
1
1
1
1


(2( x  1)  2 x ( x  1) )
2
2
2
2
(
x

1
)

x
(
x

1
)
(
x

1
)
[2( x  1)  x]
2
 f ' ( x) 


( x  1)
( x  1)
x 1
( x  1) 2
2x
f ( x) 

( x  1)
x2
( x  1)
3
2
8) A)First substitute 2 into the expression, obtaining
3
.
0
The limit is
which one. But since the denominator is a small negative number,
B)
Divide the top and bottom by x. Inside the radical, divide by
 lim
x 
4
x
 or   .
lim
x 2
x2
.
You need to determine
x5 3


2  x 0
5x  4
lim
x 
x2 1
.
4
x  50  5
 lim
2
x 
1
1 0
x
1
1


2
x
x2 x2
5
5
 x  2, x  2
.
x2  
x2 x  2
 ( x  2), x  2
x2
x2
x2
 ( x  2)
 lim
 lim 1  1 and lim
 lim
 lim (1)  1 . Thus the limit
Thus lim
x2 x  2
x 2 x  2
x2
x2 x  2
x2
x 2
x2
x5
0
0
does not exist. B) lim
. For this ratio to be 2, it must be an indeterminate form
,

x 5 g ( x)
0
g (5)
otherwise the ratio would be 0. Therefore, lim g ( x )  0 c)
9) a) to compute
lim
x2
, recall that
x 5
x sec x 1
5x
1
1
1
lim
 lim
 (1)(1)  d) Use the squeeze theorem. First observe that since
x 0 sin 5 x
5 x 0 sin 5 x cos x 5
5
1
1
 1  sin  1 , we have  x 2  x 2 sin  x 2 . We know that lim x 2  0 and lim ( x 2 )  0 . Thus by the
x0
x 0
x
x
1
2
 0 e)
sandwich theorem, lim x sin
x 0
x
tan 3x
sin 3x cos 5 x
sin 3x 3x 5 x 1 cos 5 x 3
lim
 lim
 lim

x 0 tan 5 x
x 0 cos 3 x sin 5 x
x 0
3x 1 sin 5 x 5 x cos 3x 5
3
2
2

f (2  h)  f (2)
10) First compute the slope: m  lim
 lim 2  h  1 3 > This is a complex fraction.
h 0
h 0
h
h
2(3)  2(h  3)
 2h
2
2
Multiply each term by the LCD 3(h  3) :  lim
 lim
 lim
  . The
h 0
h

0
h

0
h(3)( h  3)
3h(h  3)
3(h  3)
9
2
2
2
point is ( 2, ) . Thus the tangent line has an equation y 
  ( x  2)
3
3
9
11) Recall that a discontinuity is removable if the limit exists but the function is not continuous (the
graph has a hole). Since
x = 1.
x = 1.
f ( x) 
x 1
x2 1
is not continuous at x = 1 since the function is not defined at
x 1
1
1
 lim
 , thus the limit exists. Therefore, f has a removable discontinuity
2
x

1
x 1 2
x 1
 x 1
 x 2  1 x  1
We can define g ( x )  
(this process closes the hole). This is a continuous
1

 2 x  1
lim
x 1
at
function that agrees with f when x is not 1.
12) a)
lim csc x  
x  
(the graph of cscx can be sketched based on the graph of sinx: you can see the
limit from the graph)
3 x
3

1
x
x
x
 lim
 lim
 1 since x 2  x when x is positive. C)
b) lim
2
2
x  
x 
x 
2
2 x
2 x
1
 2
2
x2
x
x
3
x

3 x
lim
 lim  x  x  1 since x 2   x when x is negative d)
2
x  
x 
2 x
2 x2

x2 x2
x 2 3x


2
 x  3x
x
x  lim  x  3   (divide each term by the highest power of the
lim
 lim
x  
x


x 
x 3
3
x3

1
x x
x
3 x
4
denominator) e)Multiply by the conjugate:
x  x 2  12 x  x 2  12
 12
lim ( x  x  12 )  lim
 lim
 0 f)Factor out x4
2
2
x 
x 
x


1
x  x  12
x  x  12
1
lim ( x 2  x 4 )  lim x 4 ( 2  1)  (1)   g)Use the continuous function theorem to take the limit
x  
x 
x
2
inside of the exponential function:
lim e
x
tan 1 x
e
lim tan 1 x
x 
e


2
i)
lim ln(
x 
x
)   : Use the
x 1
2
continuous function theorem to take the limit inside of ln: inside goes to zero after dividing each
term by x2
lim ln(
x 
sin 4 x
:
x0 3 x 4
j)
lim
x
x
)  ln( lim 2 )  (ln 0)  
x  x  1
x 1
2
first take the constant out. Use the continuous function theorem to take the limit
inside the power:
h)
k)
sin 4 x 1
sin x 4 1
 (lim
) 
4
x 0 3 x
3 x 0 x
3
lim
1  2x 3
1  2x 3
) (use the continuous function theorem to take the limit inside) ( lim
)  23  8
x


1 x
1 x
lim x  2
x
lim    x3
 (when approaches 3 from the left side, e.g. 2.9, 2.99, 2.9999…, x remains
x 3
x
lim x
3
lim (
x 
x 3
2.)
13)
a) For the function to be continuous the left limit and the right limit must agree at x = 1.
lim f ( x)  lim (1  cx)  1  c,
x 1
b)
lim f ( x)  lim (4 x  c)  4  c .
x 1
x 1
f ( x)  e x 2 x1
is continuous since
3
x 1
e x is continuous
Thus 1+c = 4 – c. c=3/2.\
as all exponential functions are continuous,
x 3  2 x  1 is continuous since all polynomial functions are continuous.
Thus
f ( x)  e x 2 x1
3
is
continuous as the composition of continuous functions is continuous.
c) 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.
f ( x)  x 5  2 x . Since f is a polynomial, it is continuous. f (0)  0 , which is greater than 1 and
f (2)  28 , which is less than 1. Thus by the intermediate value theorem, there is a number c
between 0 and 2 such that f (c)  1
14) Let
dy
dy
dy 2 x  y 2 sec 2 ( xy 2 )
2
2
2

15)
y  tan( xy )  x  2 y  sec ( xy )( y  2 xy )  2 x 
Implicit diff:
dx
dx
dx 2 y  2 xy sec 2 ( xy 2 )
d
3
2
16) A) sin x  2 x sin x cos x b) c)
d)
(sec 3x )  sec( 3x )(tan 3x )
dx
2 x
d
(csc 2 (tan x))  2 csc(tan x))( 1) csc(tan x) cot(tan x) sec 2 x e)
dx
2
2
2
5
1
1
1


1 2
( x  1) 2 (2 x)  x( x 2  1) 2
2
f  ( x 2  1) 2  f ' 
g  ( x  1)  g '  1
f ' ( x)  6
1
2
1
2

1
2
( x( x  1) ( x  1)  (1)( x  1)
( x  1) [ x( x  1)  ( x 2  1)]

6

( x  1) 2
( x  1) 2
2
2
2
6( x  1)
( x  1) ( x  1)
2
2
1
2
6( x  1)

5
( x 2  1) 2
f) Use the product rule
f ( x)  x
3
x3 1 :
f  x  f ' 1
1
2
2


1
g  ( x 3  1) 3  g '  ( x 3  1) 3 (3 x 2 )   x 2 ( x 3  1) 3
3
1
Using the product rule,
f ' ( x)  ( x 3  1) 3  x  x 2 ( x 3  1)

2
3

2
1
 ( x 3  1) 3 ( x 3  1  x 3 )  
( x  1)
3
2x
1
, then f ' ( x )  
x 1
(1  x 2 ) 2
f ( x) 
g) If
2
2
3
. Thus
d
1
 2 tan x
4 tan x sec 2 x
2
2
2
( f (tan x))  2( f (tan x))  f ' (tan x)  sec x  2
sec x  
dx
1  tan 2 x (1  tan 2 x) 2
(1  tan 2 x) 3
17) Since the velocity v(t) = s ' (t )  64  32t  0  t  2 , the object reaches its maximum height at t = 2.
Thus from t=0 to t = 2, the object travels upward (positive displacement) and downward from t = 2 to t
= 3 (negative displacement). Thus we need to break s(t) at t=2. The total distance traveled is
s(0)  s(2)  s(2)  s(3)  64  16  80
18) Let
  0 be given.
Express
f ( x)  L  x 2  9
may assume
Then
Need to find
  1.
Then
  0 such that if 0  x  3  
as a function of
x a  x 3 .
x  3 1 2  x  4.
x2  9  x  3 x  3  7 x  3 .
Let
then
( x 2  4)  5  x 2  9  
f ( x)  L  x 2  9  x  3 x  3 .
For this range of x,
x3  7
We
since x is at most 4.

  min{ 1, } Then when 0  x  3  
7
.
,
( x 2  4)  5  x 2  9  
19)
a)
lim
h
f (0  h)  f (0)
 lim
h 0 h
h
and
lim
h 0
h 0
h
h
 lim
h 0
h
 1 .
h
. Recall that
Since lim
h 0
h h  0
h 
.
 h h  0
Thus
lim
h 0
h
h
 lim
h 0
h
 lim 1  1
h h 0
f (0  h)  f (0)
does not exist, f is not differentiable at x
h
=0
6
b)
x 2 is continuous, since it is a polynomial. cos x
is a polynomial. Thus their composition
continuous functions is continuous,
20)
21)
is continuous and
cos( x 4  1) is continuous.
x 4  1 is continuous since it
Since the product of
f ( x)  x 2 cos( x 4  1) is continuous.
d
1
4 cos 4 x
f (sin 4 x)  [ f ' (sin 4 x)](cos 4 x)( 4) 
(cos 4 x)( 4) 
2
dx
1  (sin 4 x)
1  sin 2 4 x
1
1  (1  h) 2

1
f (1  h)  f (1)
(1  h) 2
(1  h) 2
lim
 lim
 lim
h 0
h 0
h 0
h
h
h
First compute the slope:
 2h  h 2
2h
 lim
 lim
 2 . When x = 1, y = 1. Thus an equation of the tangent line is
h 0 h(1  h) 2
h 0 (1  h) 2
y  1  2( x  1)
d
( f ( f ( x)  2 g ( x))), x  0
22)
dx
d
( f ( f ( x))  2 g ( x))  f ' ( f ( x))  f ' ( x)  2 g ' ( x) x 0  f ' ( f (0))  f ' (0)  2 g ' (0)  f ' (1)(5)  2(7)
dx
 (2)(5)  14  24
d
( f ( x) g ( x)  2 )  f ' ( x) g ( x)  2  f ( x)( 2 g ( x) 3 ) g ' ( x) x 0  f ' (0) g (0)  2  f (0)( 2) g (0) 3 g ' (0)
b)
dx
1
1
1
(7 ) 
= (5)  1(2)
9
27
27
d
( f (4 x)) 2 , x  0  [2 f (4 x)] f ' (4 x)( 4), x  0  2 f (0) f ' (0)( 4)  2(1)(5)( 4)  40
dx
c)
c)
d)
d
f ( g (3x 2  x))  f ' ( g (3x 2  x))  g ' (3x 2  x)  (6 x  1)
dx
x 0
 f ' ( g (0))  g ' (0)  1  f ' (3)  7  1  14
23)
At x = 1, the function is defined : f(1) = 4
a)
lim f ( x)  lim ( x  1)  0
x 1
x 1
and
lim f ( x)  lim ( x 2  1)  0 , thus the limit does not exist
x1
x1
Therefore, the function is not continuous at x = 1. It has a jump discontinuity,
b)when x is not 1, it is continuous since f is a polynomial on (1,  ) and on (,1)
24) To show
lim ( x 2  6 x)  16
x 2
( x 2  6 x)  16   .
Express
Let
  0 be given.
Need to find
f ( x)  L  x 2  6 x  16
f ( x)  L  x 2  6 x  16  x  2 x  8
 0
as a function of
. We may assume
  1.
Then
such that if
0 x2 
then
xa  x2 .
x  2  1  1  x  3 . When x is
7
x  8  11 since s is at most 3.
chosen in this range,
  min{ 1,

}
11
Then when
0 x2 
26) a)1) at x = 2, f is defined (f(2)=6), 2)
4  lim f ( x)  f (2)  6
x2
,
x 2  6 x  16  x  2 x  8  11 x  3 .
Thus
Let
( x 2  6 x)  16  
lim
x 2
x2  4
 lim ( x  2)  4 , the limit exist, but 3)
x  2 x 2
. Thus it is not continuous at x = 2. It has a removable singularity.
b) If x is not 2, f is continuous since a rational function is continuous whenever the denominator is not 0.
26)
lim
h 0
f (0  h)  f (0)
 lim
h 0
h
1
0
1
h
 lim h 2 cos
h

0
h
h
h 3 cos
lim h 2  0, lim  h 2  0 using the squeeze theorem, lim h 2 cos
h0
h0
 h 2  h 2 cos
Since
h 0
1
 0.
h
1
 h2
h
and
Thus it is differentiable at x = 0
using the definition of derivative.
27)
f ( x  h)  f ( x )
( x  h) 3  (3( x  h))  x 3  3x
x 3  3x 2 h  3xh2  h 3  3x  3h  x 3  3x
 lim
 lim
h 0
h 0
h 0
h
h
h
2
2
2
 lim (3x  3xh  h  3)  3x  3
lim
h 0
28)
d
( f ( g (h( x)))
dx
x2
 f ' ( g (h(2))  g ' (h(2))  h' (2)  f ' (5)  g ' (3)  h' (2)  (12)(11)(10)  1320
30)
A)
1
1
1
1
x 2 ( x  h) 2  2 x 2 ( x  h) 2
 2
2
2
( x  h)
x
( x  h)
x
f ( x  h)  f ( x )
lim
  lim

  lim
2
x 0
h 0
h 0
h
h
hx ( x  h) 2
 lim
h 0
x 2  ( x  h) 2
h(2 x  h)
 2x
2
 lim
 2 2  3
2
2
2 2
h 0 h( x  h) x
x x
x
h( x  h) x
B)
lim
h 0
( x  h) 2 1  x 2  1
( x  h) 2  1  x 2  1
f ( x  h)  f ( x )
 lim
 lim
h 0
h 0
h
h
h
 lim
h 0
 lim
h 0
 lim
h 0
( x  h) 2  1  x 2  1 ( x  h) 2  1  x 2  1
h
( x  h) 2  1  x 2  1
( x  h) 2  1  ( x 2  1)
h( ( x  h) 2  1  x 2  1)
2x  h
( x  h)  1  x  1
2
2

 lim
h 0
2 xh  h 2
h( ( x  h) 2  1  x 2  1)
2x
x 1  x 1
2
2

x
x 1
2
8
31) Since x is positive, we know that
lim
x 
x2 1
 lim
x 
3x  1
c)
x2
, the denominator by
x
,
x2
1
 2
2
1
x
x

3x 1
3

x x
b) Since x is negative, we know that
,
x 2  x . Dividing the numerator by
x 2  x .
Divide the numerator by
x2
, the denominator by
x:
x2
1
1
 2
1 2
2
2
x 1
1
x
x
x
 lim

lim
 lim
x   3 x
x  
x  3 x  1
1
1
3

3
x x
x
lim
x 11
x3 2
x 1 2
    the limit is positive infinity since there two negative signs) d) lim1

 
x

1
x 1 0
1  x 0
(the limit is negative infinity since there is only one negative sign.)
32) This must be an indeterminate form: otherwise the limit would be undefined, infinity or negative infinity.
Thus
e)
lim g ( x)  0
x4
lim  tan x   (sketch the graph of tanx from
x

)
2
33) Use the third condition of continuity:
limit value must be f(5).
Then

3
to
2
2
lim f ( x)  lim
x 5
x 5
lim f ( x)  f (5)
x 5
. Thus compute the limit as x approaches 5. The
( x  5)( x  5)
x 2  25
x5
lim
2
 lim
2
x 5
x
x  5 x x5 x( x  5)
.
Thus we define
f (5)  2 .
lim f ( x)  f (5)  2
x 5
9
34) The limit exits if its left limit and right limit are equal: Use kx-1 for the right limit since f(x) is kx-1
lim f ( x)  lim (kx  1)  k  1
x 1
when x is greater than or equal to 1, the right side of 1.
2kx in computing its left limit:
lim f ( x)  lim (3  2kx)  3  2k
x 1
x 1
x 1
. Now solving
. Similarly, use 3-
3  2k  k  1  k 
4
.
3
35)
a) First find t when the height s is 64:
function by computing the derivative:
velocity),
v  32(4)  128
64  16t 2  64t  64  t  0,4
v  s'  32t  64
.
. Next compute the velocity
Finally substitute t=4 (t=0 gives upward
ft/sec.
b) You must first find out the time the projectile reaches its maximum height: the projectile changes
its direction after reaching the maximum height.
v(t )  32t  64  0  t  2
The distance traveled is
s(t )
over the intervals of same direction. Thus you must break up
s(t )
as
s(0)  s(2)  s(2)  s(3)  64  128  128  102  90 ft
36)
a)
h' (3)  f ' ( g (3))  g ' (3)  f ' (1) g ' (3)  6(2)  12
b)
d f (cos 4 x)
(
):
dx g 2 ( x)
use the quotient rule:
f  f (cos 4 x)  f '  f ' (cos 4 x)(  sin 4 x)( 4)
g  ( g ( x)) 2  g '  2( g ( x))  g ' ( x)
d f (cos 4 x)
(4 sin 4 x) f (cos 4 x)( g 2 ( x))  2( g ( x)) g ' ( x) f (cos 4 x)
(
)

dx g 2 ( x)
( g ( x)) 4
 2(2)( 4)(5)
 5
24
x 0

 2( g (0)) g ' (0) f (1)

g (0) 4
37) a) false: without the limit, it is the slope of a secant line
b) True. If the limit exits, the two sided limit equals the one sided (left/right) limit.
c) The limit is 0. Notice that as x approaches 5, the denominator approaches 0. Yet the limit exits:
which means limit is indeterminant (0 over 0)
10
d)
f ( x)  x
is continuous, but not differential at x = 0.
e) x>0
38) Observe the following:
a) In
(,1) , the function is increasing and the graph is getting steeper:
thus the derivative is positive
and increasing.
b) At x=-1, there is a share corner. Thus the derivative is undefined,
c) In
(1,1) the function is decreasing at a constant rate.
Thus the derivative is negative and constant.
d) At x=1, there is a sharp corner. The derivative is undefined.
e) In
(1,2) , the function is constant:
the derivative is 0
39)
11
v(t )  s' (t )  32t  16 .
a) The velocity function is
0.
 32t  16  0  t 
1
.
2
b) The distance traveled is
The maximum height is attained when the velocity is
Thus the maximum height is
s(t )
Note that the direction changes at
1
1
1
s( )  16( ) 2  16( )  32  36 ft .
2
2
2
over the intervals of same direction.
t
1
:
2
The projectile begins its decent after attaining its maximum
height. Next find the time that the projectile reaches the ground:
s(t )  0  16t 2  16t  32  0  t  2
The distance is
1
1
s(0)  s( )  s( )  s(2)  32  36  36  0  40 ft
2
2
40)
a) Use the quotient rule:
1
1 2
x
2
g  x 2  1  g'  2x
f  x  f '
1
1
1 2 2
x ( x  1)  2 x( x) 2
2
f ' ( x) 
( x 2  1) 2
c)
1
(factor out
1 2 2
x [( x  1)  4 x 2 ]
1
 3x 2  1
1 2

x ) 2
2
( x 2  1) 2
2 x ( x 2  1) 2
Use the product rule:
f  x 2  f '  2x
1
1
g  (1  x 2 ) 2  g ' 
1
2
1


1
(1  x 2 ) 2 (2 x)   x(1  x 2 ) 2
2

1
2
f ' ' ( x)  2 x(1  x )  x ( x(1  x ) )  (factor (1  x )
2
2
2
2)

1
2
)

1
2
(1  x ) [2 x(1  x )  x ] 
2
2
3
 3x 3  2 x
1 x2
41)
a)
lim
x2
x3
5
  
2x 0
12
x 3
x 3
 lim
x 3
x 3
x3
x3
cos x  sin x
lim
 0 since 
x 
x2
x 3
lim
b)
c)
2  x  x2
 x2
 lim
x 
x   x
1 x
 lim x  
lim
d)
 lim
1
x  3 x3 x  3
2 cos x  sin x
2

 2
2
2
x
x
x

1
2 3
and
lim
x 
2
0
x2
(the highest term of the numerator and the denominator dominate the limit)
x 
lim
e)
x 
1  4x 2
1  2x 2
(use the continuous function theorem to take the limit in the square root)
1  4x 2
 lim (
) 2
x  1  2 x 2
Use the squeeze theorem to compute
f)
lim e  x  0 .
x 
a)
b)
c)
Thus
lim e  x cos x  0
x 
lim
x 
x 
3( x  h)  1
f ( x  h)  f ( x )
 lim
x 
h
h
 lim
x 
 lim
x 
 lim
x 
 e  x  e  x cos x  e  x
and
by the squeeze theorem.

1
3x  1
1
the LCD
. Observe that
d
g (4 xf ( x))  g ' (4 xf ( x))[ 4( f ( x)  xf ' ( x)]  g ' (12)[ 4(1  3(4))]  (2)(52)  104
dx
d 3
[ x ( f ( x)) 2 ]  3x 2 ( f ( x) 2 )  x 3 (2 f ( x) f ' ( x))  3(9)(1)  27(2)(1)( 4)  243
dx
2
d f ( x2 ) g ( x)
(e
)  e f ( x ) g ( x ) [2 xf ' ( x 2 ) g ( x)  f ( x 2 ) g ' ( x)]  e1(10) [2(3)(1)(10)  1(4)]  64e10
dx
1
43)
lim e  x cos x
3( x  h)  1 3x  1 )  lim
3( x  h)  1
 (first simplify the complex fraction by multiplying by
3( x  h)  1 3x  1 
h 3( x  h)  1 3 x  1

3( x  h)  1 3x  1

(now multiply by the conjugate)
3x  1  3( x  h)  1 3x  1  (3( x  h)  1
h 3( x  h)  1 3 x  1
3x  1
h 3( x  h)  1 3x  1
x 
3 x  1  3( x  h)  1
1
3 x  1  3( x  h)  1
 3h
h 3( x  h)  1 3x  1( 3x  1  3( x  h)  1)
 lim
x 
 lim
x 
(3x  1)  (3( x  h)  1)
h 3( x  h)  1 3x  1( 3x  1  3( x  h)  1)
3
3( x  h)  1 3x  1( 3x  1  3( x  h)  1)
(after
canceling h, we no longer have indeterminant form: we can substitute 0 into h and drop the limit)

3
3x  1 3x  1( 3x  1  3x  1)

3
(3x  1)( 2 3x  1)

3
3
2(3x  1) 2
13
44)
x 4  y 4  1  4x 3  4 y 3
dy
dy
x3
0
 3
dx
dx
y
x3
1
2 3
3 2
dy
3
x
y

3
x
y
(

)
3x 2 y 3  3x 6
3x y  3x y
2
3
d y
y
y
dx  


2
6
6
6
dx
y
y
y
2
remembering
x 4  y 4  1, to get

f ( x)  x 3  2  sin x .
. Then differentiate both sides with respect to x,
3
3
2
3x 2 y 4  3x 6
3x 2 ( y 4  x 4 )
3x 2




y7
y7
y7
x 3  2 is continuous as it is a polynomial, sinx is
continuous, and the difference of continuous functions is continuous. f (2) is negative and f (1) is positive.
45) Let
f is continuous since
Thus by the IVT, there is a number c between -2 and 1 such that
c 3  2  sin c .
Thus the equation
x 3  2  sin x
f (c )  0 .
i.e.
f (c)  c 3  2  sin c  0 , i.e.
has a solution.
14