PRACTICETEST#3
1
2
MACH5
dy
 (1)(sin x)  (x)(cos x)  sin x  x cos x
dx
f (x)  300x  x 3  increasing when f '(x)  0
y  x sin x 
SOLUTIONS
B
f '(x)  300  3x 2  3(10  x)(10  x)  0 when  10  x  10 or 10,10 
3
4
5
 d

 sec x  sec x  tan x  A
dx
1
1
f (x)  7x  3  ln x  f '(x)  7 
f '(1)  7   8 E
x
1
lim f (x)  2  lim f (x)  4 hence lim f (x) does not exist C
 sec x tan x dx  sec x  C
x 4
6
B
x 4
x 4
v(t)  6t  t  0 on 0, 3
2
Total Distance 


3
0
3
v(t) dt   6t  t dt  
2
0
  y'  5 x
3
0
 3x


3
t3
6t  t dt  3t 
 27  9  18 D
30
2

2
7
y  x 3  cos x
8
right Riemann sum  50   R(t)dt  50  3(6.2)  5(5.9)  3(5.6)  114.9 liters C
9
(2x  1)(x  2)
lim
 lim(2x  1)  5  k E
x2
x2
x2
10
y  e x 2  0 on 0,2  so area 
5
3
 cos x
4
2
 sin x
E
15
0
11
f (x) 

2
0

2
0
x 2
  1 (2  x)1 2 , x  2
 2
 f '(x)   1
 lim f (x)   so f not differentiable
1 2
x2
(x

2)
,
x

2
x  2, x  2
 2
1
1
u  x, du 
dx  2du 
dx and when x  4,u  4  2, when x  1,u  1  1
2 x
x

f (x)  


4
1
13
A
x  2  f (2)  0 so C and E are false. lim f (x)  lim f (x)  0 so D is false and f is continuous
x2
12

e x 2 dx  2e x 2   2 e1  e0  2e  2
e
x
x
2  x, x  2
2
dx  2  eu du
A
C
1
5
14

x2
15 3
 25
 9

f
(x)dx

2
dx

(x

1)dx

2x

 x   6  2  
 5     3   4    10 D
1
1
1
3
 2

2
2 2
 2
3
d
d
7
7
A
3 
 f (g(x)  f 'g(x)g '(x) so  f (g(x)x  3  f 'g(3)g '(3)  f '(7)g '(3) 
dx
dx
3 5
5
1 2
7
7
f '(x)  12 x 2  4  (2x)  f '(7) 

and g '(x)  g '(3)  3
45 3 5
15
h(x) 

x
h(6) 

6
5
3
0
0
5
3
f (t)dt  h '(x)  f (x)
f (t)dt  0 since below the x  axis; h'(6)  f (6)  0; h"(6)  f '(6)  0 since f increasing around 6.
So, h(6)  h '(6)  h"(6)
A
PRACTICETEST#3
MACH5
SOLUTIONS
16
x(t)  (t  a)(t  b)  t 2  at  bt  ab  x '(t)  v(t)  2t  a  b  0  t 
17
f (x) 
18
19

x
2
ab
2
B
g(t)dt  f '(x)  g(x)  a positive constant since f is increasing at a constant rate.
A
1
ln(4  h)  ln(4)
 f '(4) 
B
h0
4
h
x
(x  2)(1)  x(1)
2
1
f (x) 
 f '(x) 

 
2
2
x2
(x  2)
(x  2)
2
f (x)  ln x  lim
(x  2)2  4  x  2  2  x  0,4  f (0)  0, f (4)  2 C
20
f (x)  (2x  1)3; f (0)  1 so g(1)  0; f '(x)  3(2x  1)2 2  6(2x  1)2  f '(0)  6
g '(x) 
21
1
1
1
1
 g '(1) 


f 'g(x)
f 'g(1) f '0  6
D
20x 2  x
5 E
x 1  4x 2
horizontal asymptote: lim y  5 so take the limit of each: lim
x
22
23
24
25
x 1x  ln x 1  ln x
ln x
f (x) 
, x  0  f '(x) 

 0  ln x  1  x  e; f '  0, 0  x  e; f '  0, x  e
x2
x
x2
ln e 1

B
So relative maximum and absolute maximum value is f (e) 
e
e
dP
When
is constant, then growth is constant and linear. A
dt
 2
 2
 2
g(x)  x 2 ekx  g '(x)  2xekx k  ekx 2x  g '    2   e(2 3) k k  e(2 3) k 2    0 
 3
 3
 3
4
4
4 4
4 4 
e(2 3) k   k   0   k  0  k    k  3 A
9
3
3 9
3 9 
 dy   2 sin x dx y  2 cos x  C 
y( )  1, 1  2 cos   C  1  2(1)  C  C  1  y  2 cos x  1 E
26
Since et is always positive, then g '(2) 
2
g"(x)  e
27
 x2
0
et dt  0 so g is increasing on the interval.
2
 0 so g is concave up on the interval.
A
dy 2 x  y
d y ( x  2 y )(2  )  (2 x  y )(1  2 )
dy

 2 

2
2
dx x  2 y
dx
dx (3,0)
( x  2 y)
2
dy
dx
d2y
(3)(0)  (6)(5)
30
10



2
2
dx (3,0)
(3)
9
3
28

2
dy
dx
A
3
;
4
2
2
 3 

 2 A
a(t)  v'(t)   sin t  cost  a    
 4 
2
2
v(t)  x '(t)  cost  sint  0  tan t  1  t 
PRACTICETEST#3
29
MACH5
f '  f increasing  c, e; f 'increasing  f " concave up  d, e E
SOLUTIONS
30
From the IVT, f (x)  13 for some x on 2, 4  since f (2)  10 and f (4)  20. A
31
y  e tan x  2  0  e tan x  2  tan x  ln 2  x  tan 1 ln 2   0.60611193  store as "a"
Using calculator: y'(a)  2.9609  2.961
32
33
D
8
1
Average value of the velocity on 0, 8  
v(t)dt B

80 0
I. f ' changes from  to  at x  3 : true II. f ' changes from inc to dec or dec to inc at x  2 : false
III. f ' decreases on 0  x  4 : true
34
E
20
Water in the tank = initial amount + amount pumped in  800   r(t)dt  1220gal
0
35
36
D
Graph f '(x) and find that the relative maximum occurs at x  0.350 C
An estimate for the distance traveled is an estimate of the area below the curve on the interval 0  x  8.
One estimate would be to connect v(0) and v(8) with a straight line and use the
1
area of the triangle as the estimate: (8)(50)  200 close to 210 D
2
37
38
39
x
f "(x)  (4x 3  4x)e
4
. Graph f "(x) and find where it is below the x  axis : (1.5, 1) and 0,1 D
 2 x 2 1
f has a relative maximum at x  1 because f ' changes from + to - there. C
f '  0 so f increases for all x and

7
4
f (t)dt  0 so f is both negative and positive on 4, 7   so not A, D, or E
41
3
3 6
   0 So, B
2
75
f is concave down when f "  0 or when its graph is below the x  axis :  2  x  1 and 1  x  3 E
6
2
15
  15s  6x  6s  9s  6x  s  x 
3
xs s
8 ft
ft
ds 2 dx 2

 (4) 
 2.667
B
3 sec
sec
dt 3 dt 3
42
v(3)  v(0)   v '(t)dt  v(0)   a(t)dt  5  6.710  11.710
For C: slope
40
43
3
0
0
Let t  2x, so dt  2dx  dx 
E
1
dt and x  12  t  2(12)  24; x  6  t  2(6)  12
2
24
1 24
f (t)dt  10   f (t)dt  20 B

6
12
2 12
f ' changed from decreasing to increasing somewhere on  2  x  3, and increasing to decreasing somewhere

44
3
12
f (2x)dx  10 
on 3  x  6. So f " changed signs at least once on these intervals  f has at least 2 inflection points
B
PRACTICETEST#3
45
MACH5
Area of the cross sectional square = s 2 
Volume 
 xx 
  x  x  dx  0.12857  0.129
1
0
2
2
2
A
2
SOLUTIONS