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Math 265 Final Exam (Simulated) – Solutions
Take note of how the points are assigned in each question.
a) By the Product and chain Rules.
y'  cos x cos(sin x 2 )  sin x( sin(sin x 2 ) cos( x 2 )(2 x))
(4 pts)
y'  cos x cos(sin x 2 )  2 x sin x cos( x 2 ) sin(sin x 2 )
b) By the general power and quotient rules.
1  1  x3 

y '  
3  1  x 2 
2 / 3
1 1 x2 

y '  
3  1  x 3 
2/3
3 x 3 (1  x 2 )  2 x(1  x 3 )
(1  x 2 ) 2
(4 pts)
x(3x  x 3  2)
(1  x 2 ) 2
b) By the General power rule.


y' 
1
d
(1  1  x ) 1 / 2
1  (1  x)1 / 2
2
dx
y' 
 1 
1

 
2 1 1 x  2 1 x  4 1 1 x 1 x
1
(2 pts)
(2 pts)
d) By the quotient and general power rules.


1 2
 2
1 / 2
2
 x  1) 2 x ( x  2 x  8)  2 x  1( x  1)
2

y'  
( x 2  2 x  8) 2
(4 pts)
e) By the chain and quotient rules
 x  1   x  1  ( x  2)  ( x  1)
y'  2 sec 2 
 tan 

( x  2) 2
 x 2  x 2
 x 1   x 1   6
y'  sec 2 
 tan 

2
 x  2   x  2  ( x  2)
(4 pts)
  
2. a) cos 62 = cos(60   2  )  cos  
 3 90 

2

Let f ( x)  cos x , a  60  , a  2 
=
3
180 90
 
    
cos 62 0  f (a)  f ' (a)x  cos   sin   
3
 3  90 
cos 62 0 
1
3 
1
3

 
2 2 90 2 180
(1 pt)
(1 pt)
(1 pt)
(1 pt)
  
 
    
cos 62 0  cos    cos   sin   
 3 90 
3
 3  90 
#2. b) 16.4 = 16  0.4
Let f ( x)  x , a  16, a  0.4
16.4  f (a)  f ' (a)x  16 
16.4  4 
(1 pt)
(1 pt)
4
 
2 16  10 
1
(2 pts)
1
1
81
 4

5(4)
20 20
3. By definition of differentiation
cos( x  h) cos( x)

sin( x  h) sin( x)
 lim
h 0
h
1  cos( x  h) cos( x) 
 lim 


h 0 h sin( x  h)
sin( x) 

(1 pt)
1  cos( x  h)sin( x)  cos( x)sin( x  h) 
 lim 

h 0 h
sin( x  h)sin( x)



1
1  cos( x  h)sin( x)  cos( x)sin( x  h) 
lim 

h

0
sin( x)
h
sin( x  h)


1
1
1  cos( x  h) sin( x)  cos( x) sin( x  h) 
lim
lim 

sin( x) h0 sin( x  h) h0 h 
1

(1 pt)
1
1
1
(1 pt)
*
lim  cos( x  h)sin( x)  cos( x)sin( x  h) 
sin( x) sin( x  0) h0 h
1
1

lim  cos( x  h)sin( x)  cos( x)sin( x  h) 
2
h

0
sin ( x)
h
1
1

lim   cos( x) cos(h)  sin( x) sin( h)  sin( x)  cos( x)  sin( x) cos( h)  cos( x) sin(h)  
2
sin ( x) h0 h


1
1   cos( x) sin( x) cos(h)  sin( x) sin( x) sin( h)  
lim 

2
sin ( x) h0 h    cos( x) sin( x) cos(h)  cos( x) cos( x) sin( h)  
1
1  cos( x)sin( x) cos(h)  sin( x)sin( x)sin(h) 
lim


sin 2 ( x) h0 h   cos( x)sin( x) cos(h)  cos( x) cos( x)sin(h) 
1
1

lim   sin( x)sin( x)sin(h)  cos( x) cos( x)sin(h) 
2
sin ( x) h0 h
1
1

lim   sin( x)sin( x)  cos( x) cos( x)  sin(h)
2
sin ( x) h0 h
1
1
 2
lim  sin 2 ( x)  cos 2 ( x)  sin(h)
h

0
sin ( x)
h
1
1
 2
lim 1 sin(h)
h

0
sin ( x)
h
1
sin(h)
 2
lim
sin ( x) h0 h
1
  2 *1
sin ( x)
1
 2
sin ( x)
  csc2 ( x)

4
1) We find the domain and range of the function f ( x) 
x is in domain  x 2  1  0
 ( x  1)( x  1)  0  x  1  0 and x  1  0
 x  1 and x  1
Domain is all numbers except x  1 and x  -1
(1 pt)
(1 pt)
(1 pt)
x2
.
x2 1
(1 pt)
2) (0,0) is the x and y intercepts
(1 pt)
2x
3) f ' ( x)   2
, hence f ( x)  0 if x = 0, and the critical points are 0, -1, 1. We
( x  1) 2
have that f ' ( x)  0 for x < -1 and -1 < x < 0; f ' ( x)  0 for 0 < x < 1 and x > 1.
On the real line, we have
(2 pts)
U
0
U
-1
0
1
4) We find f (x) .
x
f ( x)  2
 2
2
2
 1  x  2( x 2  1)  2 x
2
x
 1
4
2
( x 2  1)( 3x 2  1)
x
2
 1
3x 2  1
4
( x  1)( x  1) 3
2
2
 2
( x 2  1)( x 2  1)  x  2  2 x 
x
2
 1
4
3x 2  1
x
2
 1
3
3x 2  1
( x  1) 3 ( x  1) 3
U
U
1
-1
Critical points of f (x) are 1 and –1. We put the critical points on a real line below.
When x is on right most interval 1,  , all the factors of f (x) are positive.
So the sign of f (x) is positive.
Since ( x  1)3 and ( x  1)3 have odd degree, if x crosses critical points, the sign of f (x)
changes. Thus we sketch the sign of f (x) on real below.
Sign of f 
+
U
-1
-
U
+
1
Thus we have the diagram below.
(2 pts)
the real line indicating where function f is concave downward or upward
U
U
-1
1
5) We use all the information in the above steps to sketch the graph of y 
below.
x2
as
x2 1
(4 pts)
#4b.
1) domain all real numbers
2) f ( x)  ( x)  sin(  x)
  x   sin( x)
(1 pt)
   x  sin( x) 
  f ( x)
The function is odd function so the graph of y  x  sin x is symmetric with respective to
origin (0, 0) . (1 pt)
2) f ( x)  x  sin x
f ( x)  1  cos x  0
Thus the graph is increasing on  ,   .
(1 pt)
3) f ( x)   sin x
f ( x)   sin x  0
 sin x  0
   2n  x  2  2n
y  x  sin x is concave upward on the intervals   2n , 2  2n  for integer n .
y  x  sin x is concave downward on the intervals  2n ,   2n  for integer n .
(2 pts)
The graph is
(5 pts)
5. Find all maxima and minima of the following functions on the indicated intervals.
5 5 1
4 4 1
a) f’(x)  2* x 3  5* x 3
3
3
1
10 2 20 13 10 23
10 1 1
 x3 
x 
x  2x 3  x 3 x 3  2
3
3
3
3
f ( x)  0
10 1 1
 x3 x3  2  0
3

1


1





 x3 x3  2  0
1
1
 x 3  0 or x 3  2
 x  0 or x  8
(3 pts)
To find maximum or minimum value we check x  1, x  0, x  8, x  20
5
4
f (1)  2(1) 3  5(1) 3
5
4
 2(1) 3  5(1) 3
 2  5  7
5
4
f (0)  2(0) 3  5(0) 3  0
5
4
f (8)  2(8) 3  5(8) 3
 2(2)5  5(2)4
 (2)4  2(2)  5
 (2)4  4  5
 16
5
4
f (20)  2(20) 3  5(20) 3  23.2807582317404
Thus the minimum value is f (8)  16
5
(1 pt)
(1 pt)
4
and the maximum value is f (20)  2(20) 3  5(20) 3  23.2807582317404 .
#5b f’(x)  1 sin x  0 and sin x  1 , this is true for any real x.
(2 pt)
The function is continuous and increasing on the real line.
Hence
Maximum value is f (2 )  2  cos( 2 )  2  1
Minimum value is f ( )    cos( )    1
6. Compute the following integrals:
a)  ( x 2  x) 3x dx = 3  x 5 / 2  x 3 / 2 dx 
 2x 7 / 2 
 2x 5 / 2 
  3
  C
 3
 7 
 5 
d
sin x  cos x
b) since
dx
sin 2 x
sin
x
cos
xdx
=
C

2
b
x2 1
b
 sin( 2 x) a
b)  x  cos(2 x)dx =
a
2 2
2
2
x  sin( 2 x) b b  sin( 2b) a 2  sin( 2a)



a
2
2
2
(1 pt)
(1 pt)
(1 pt)
(1 pt)
(2 pts)
(1 pt)
(2 pts)
(2 pts)
(2 pts)
(1 pt)
6. a) 1) We sketch the graph of y  x and y  2  x 2 .
(2 pts)
2) We find the x values where two curves meet.
y  x and y  2  x 2  x  2  x 2
(2 pts)
 x 2  x  2  0  ( x  2)( x  1)  0  x  2  0 or x  1  0
 x  2 or x  1
3) We find the area of the region.
Let A be the area of the region. Then we know:
1




A   (2  x 2 )  x dx
2
1
1
1
 1

   x 2  x  2 dx   x 3  x 2  2 x 
2
2
 3
 2
1
 
1
 
1
1
1
1
 1 
 1 
1
1
  x 3    x 2   2 x  2   x 3  2   x 2  2  2x  2
3
2
 3  2  2  2
1
1
1
1
  13  (2) 3   12  (2) 2  21  (2)    9    (3)  2  3
3
2
3
2


A  3 
3
9
6
2
2

8. Perimeter P  2 x  2 y and
P( x)  2 x 

( 4pts)
xy = 220
thus
y
440
440
and P ' ( x)  2  2 = 0 for x = 2 55
x
x
220
x
(2 pts)
(2 pt)
880
 0 for any positive x. Hence the perimeter is minimum when the
x3
dimensions are
x = 2 55 and y  220 = 2 55 . The perimeter is 8 55
(2 pts)
P' ' ( x) 

9.
x
0
d
d
d
sin( t 2 ) dt =
sin( t 2 )dt 
sin( t 2 )dt



dx 2 x
dx 0
dx 2 x
x
(1 pt)
2x
d
d
sin( t 2 )dt 
sin( t 2 )dt 


dx 0
dx 0
(1 pt)
= sin( x 2 )  2 sin(( 2 x) 2 )  sin( x 2 )  2 sin( 4 x 2 )
(2 pts)
10. x1  1 ,
2
1 4
x2  1 
 1  ,
6
3 3
x3 
4 287(310)

 41.9312
3
81(27)
(1 pt)
(2 pts)
(2 pts)