l =i.*r=, r = 0
\o
Calculators
SOLUTIONS-Exaw
Vl Section I
Part A
81
Exam VI Section I
PartA-NoCalculators 1.
A
p.121.
Using the product rule we obtain the first and second derivatives of y
\ ,xr --e
:
xsx
+):e
.t.rx | :e +e
+xe
Thus y"changes signat
= Zex + 2ex
x:
-2.
=ex(2+x)
C
p.727 4
Since H(4) = ! f (t)
dt, then
B
I
- area triangle DEF 11 = - (2)(1 + 2) - -(3)(2) : 3 - 3 22 :0
H(1) =
B
area trapezoid ABCD
A
ill tll -!--!--L--t--
rn
p.722 Solution I. 2
/1, - tl dx
:
areaof 2 equal triangles
0
='(1) (t)(1): \2/
t'
Solution II.
.rlr-rl dx-- I-x+tdx+lx-tdx= 00t20_[
E
;*rl^ *;--l =i.*r=,
p.722
(A), (B), and (D) are part of the definition of continuity and hence true. (C) is a statement equivalent to (A). (E) is the only one that could be false, for example, consider y = lxl at = 0.
r
F
2 t) SOLUTIONS-Exaw
82
5. C
Part
A
Multiple-Choice
t) " = tu,(zt. ?)|o = on(r* . t) - o,t = .*(r* . 1) -,
p.123 Using implicit differentiation on ry + x2 = 6 we obtain
) ) ) ) ,)
Atx=l
y+xy'+2x=0 xy':-2x-y -2x-t y' =------:x-l
-y+1=6 andy=-5 Thus
-) ) ,) ,) ,) ,) ,) ,) ,) ,)
p.122
'gz*"2(r, .
6. A
Vl Section I
2+5 y'=-=-7
')
7. C
p.123
) ) -) *)
Solution I.
ii;
*:
ii*
ex
dx) =
f,,nr*'.,l[ = }[*'o -
tns]=
)nz
) ),)
Solution II.
Let u = x2 + l, ttten du = 2xdx. Then
,) ')
310
t?;* [: =
8. E
Lr'u =
=
]r'n'o -
tnst =
)tnz'
p.r23
f'(x) = e"in* I. f"(x) = e'i"
II. III.
9. D
)^'l'!
cos
x
f"(0) = 1'1 = 1
Slope of A = x + 1 is 1, and the line goes through (Q
1)
t)
h'(x) = f'(*3 - t).(gxz) = g*2rsin(x3 -11 Since h'(x) > 0 for all r, then the graph of h(x) is increasing.
f(*s -
h(x) =
TRUE TRUE
TRUE
p.124 6
The total flow is
wlichis I Xrl dt
equal to the area under the curve. Each square under the
0
curve represents 10 gallons. There are 11 full squares and 2.5 part squares. Hence, the
approximate area
is
13.5 x 10 = 135 gallons.
) ) ) ) ) ) ) ) ) ') ') .) ') ) ) ) ) ) )
-t
\o
Calculqtors
10. B
SOLUTIONS
-Exam
VI
Section
I
part A
p.724 The instantaneous rate of change
f'\xl:2r2*
= e2x -3sinr is
-3rorr.
Thus, /'(0) =2eo -3cos0 =
11. E
of f ( x)
2-3=
-1.
p.724
/'
is negativ e on (a,
b) so f is decreasing there and (A) is false. ,[' increases on (a, c), hence, /,, is positive and the graph of/is concave up. As a result, (B), (C), and (D) are also false. (E) is true because/ is an antiderivative of
r:. E
/,.
p.125 Using the Fundamental Theorem and the Chain Rule on
Fr.r)=
';1 I ^.,
J
Then,
i,r D
dr,weobtain F'tx1=
) LI
1 _.)f
2+qx2s3
F'r-1,=#.e2)=-1
p.725
f /
is not differentiabre at x = 0 because
,r',\_
f
't* t -
*1T*
/,,r,.
is not differentiable at .r = 3 because/ is not continuous there.
Hence the answer is (D).
11. D
p.725 The velocity v(t) = y'11;
-
(t2
For / > 0, v(t) = 0 when / = 2.
++)(t)-t(2t)
(r'*+)'
=
4-tz
(z+t)(z-t)
V;ry=
UiT
83
84
SOLUTIONS-Exaw
Vl
Section
I
t-.4 r-J Part A
Multiple-Choice
-1
aJ -J t-'
15. D
p.L26
rt__
To find themaximumvalue of f(x) = 2x3 + 3xz -Izx+ 4 on the closed interval [O2], we evaluate the function/at the critical numbers and endpoints and take the largest resulting value.
,J
.
L
Thus in the closed interval [O2], the only critical number the one critical number and at the endpoints are
is x = 1. The values of the function at
From this list we see that the maximum value is
--J !
-J a z 1 -J 1 1
critical value: f(l): -3 endpoints: /(0)= 4 andf(2):8 8.
-J
p.126
..J
Using the Chain Rule twice on /(x) = ln(cos2x), we obtain
f '(x)
:-f
.
..J
(-sin2x) . (2).
-J a
cos2x
--J a
: _) _sin2;r : _f tan2.r
--1 1 --4 t_
.
cos2x
---1
1,7. A p.126 The slopes of the segments in the slope field vary
form
equation is not of the
ff
=
S(il.
as
*
changes. Thus the differential
That eliminates B, C and E of the five choices.
If the differential equation were # = *' + y2, then slopes would always be at least They aren't in Quadrant III. That eliminates choice (D).
18. E
I
t_ L
f' (*) = 6x2 + 6x -12 = 6(x + 2)(x -l)
16. A
-a
rt-
p.127
-1
0.
b A 3 .Z 3 --1 !a a -1
I
--'1
y=4;+3=1x+3)'
3
-J
1
-
n'=!(*+!i '2'
,aa --a
At ('J.,2), r'
=
:#=
When x = 0, then
y
i
*O
-2:i,,
-
the equation of the line is y
1) and
y
=i
-2 = ir- -U.
.J -a ,-J Z
-a
= :
t
h l-
s ; ): ):
l;
No
SOLUTIONS-Exam
Calculators
19. B
Solution I. Using the Product Rule twice on f (x) = (.r - 1)(x + 2)z gives
t;
; ; ; ; ;
1 L-
Pqrt A
p.127
);
; ; ; ;. ; ;. ; ;. ;, ;. ;. ;. ;. ;. ;;;" ;. ;. ;. ;. ;. ;. ;. ;. ; ; ;
Vl Section I
f ' (x)
= (x + 2)z + (x
l"@)
= 2(x + 2)
*
- l)'
2(x + 2)
2(x + 2) + 2(x
-l)
=6x+6 Thus,
/"(x)
= 0 when x
= -1.
Solution II.
:
Multiplying out 7(x)
(x _ 1Xx +
f(x) = (x -l)(x
2)2
gSves
+2)2
: (x -l)(x2 + 4x + 4) = *' +3x2 - 4 f'(x)=3x2 +6x
f"(*)
20. C
6,
thus
/"(x)
=
o
when x = -1.
p.127 b
in*
- *') dx =
0
ry +1,'.=
rn , $ =:o =+ 21.. C
= 6x +
b3
=27
zb3
a,,d b
-+
=
+
=3.
p.128 Separating variables
n ff
=
Integratinggives y3 = x + C
Thus,
!3=x +
gives:
3y2dy =
contains (-1,-1)
!=xtt3 x < 0..
tdx.
At(-1,-1), -1=-1+C +
!=x113
The largest domain that
Thus,
irt,
is
x < 0.
C=0.
85
SOLUTIONS-Exaw
86
22. C
Vl Section I
Part A
p.728
f\xt=rt-5*'*3* f'(x) =3xz -l}x*3 = (3-r - lXx - 3) and f'(x)
Calculators
SOLUTIONS-Exaw
Vl Section I
Part A
81
Exam VI Section I
PartA-NoCalculators 1.
A
p.121.
Using the product rule we obtain the first and second derivatives of y
\ ,xr --e
:
xsx
+):e
.t.rx | :e +e
+xe
Thus y"changes signat
= Zex + 2ex
x:
-2.
=ex(2+x)
C
p.727 4
Since H(4) = ! f (t)
dt, then
B
I
- area triangle DEF 11 = - (2)(1 + 2) - -(3)(2) : 3 - 3 22 :0
H(1) =
B
area trapezoid ABCD
A
ill tll -!--!--L--t--
rn
p.722 Solution I. 2
/1, - tl dx
:
areaof 2 equal triangles
0
='(1) (t)(1): \2/
t'
Solution II.
.rlr-rl dx-- I-x+tdx+lx-tdx= 00t20_[
E
;*rl^ *;--l =i.*r=,
p.722
(A), (B), and (D) are part of the definition of continuity and hence true. (C) is a statement equivalent to (A). (E) is the only one that could be false, for example, consider y = lxl at = 0.
r
F
2 t) SOLUTIONS-Exaw
82
5. C
Part
A
Multiple-Choice
t) " = tu,(zt. ?)|o = on(r* . t) - o,t = .*(r* . 1) -,
p.123 Using implicit differentiation on ry + x2 = 6 we obtain
) ) ) ) ,)
Atx=l
y+xy'+2x=0 xy':-2x-y -2x-t y' =------:x-l
-y+1=6 andy=-5 Thus
-) ) ,) ,) ,) ,) ,) ,) ,) ,)
p.122
'gz*"2(r, .
6. A
Vl Section I
2+5 y'=-=-7
')
7. C
p.123
) ) -) *)
Solution I.
ii;
*:
ii*
ex
dx) =
f,,nr*'.,l[ = }[*'o -
tns]=
)nz
) ),)
Solution II.
Let u = x2 + l, ttten du = 2xdx. Then
,) ')
310
t?;* [: =
8. E
Lr'u =
=
]r'n'o -
tnst =
)tnz'
p.r23
f'(x) = e"in* I. f"(x) = e'i"
II. III.
9. D
)^'l'!
cos
x
f"(0) = 1'1 = 1
Slope of A = x + 1 is 1, and the line goes through (Q
1)
t)
h'(x) = f'(*3 - t).(gxz) = g*2rsin(x3 -11 Since h'(x) > 0 for all r, then the graph of h(x) is increasing.
f(*s -
h(x) =
TRUE TRUE
TRUE
p.124 6
The total flow is
wlichis I Xrl dt
equal to the area under the curve. Each square under the
0
curve represents 10 gallons. There are 11 full squares and 2.5 part squares. Hence, the
approximate area
is
13.5 x 10 = 135 gallons.
) ) ) ) ) ) ) ) ) ') ') .) ') ) ) ) ) ) )
-t
\o
Calculqtors
10. B
SOLUTIONS
-Exam
VI
Section
I
part A
p.724 The instantaneous rate of change
f'\xl:2r2*
= e2x -3sinr is
-3rorr.
Thus, /'(0) =2eo -3cos0 =
11. E
of f ( x)
2-3=
-1.
p.724
/'
is negativ e on (a,
b) so f is decreasing there and (A) is false. ,[' increases on (a, c), hence, /,, is positive and the graph of/is concave up. As a result, (B), (C), and (D) are also false. (E) is true because/ is an antiderivative of
r:. E
/,.
p.125 Using the Fundamental Theorem and the Chain Rule on
Fr.r)=
';1 I ^.,
J
Then,
i,r D
dr,weobtain F'tx1=
) LI
1 _.)f
2+qx2s3
F'r-1,=#.e2)=-1
p.725
f /
is not differentiabre at x = 0 because
,r',\_
f
't* t -
*1T*
/,,r,.
is not differentiable at .r = 3 because/ is not continuous there.
Hence the answer is (D).
11. D
p.725 The velocity v(t) = y'11;
-
(t2
For / > 0, v(t) = 0 when / = 2.
++)(t)-t(2t)
(r'*+)'
=
4-tz
(z+t)(z-t)
V;ry=
UiT
83
84
SOLUTIONS-Exaw
Vl
Section
I
t-.4 r-J Part A
Multiple-Choice
-1
aJ -J t-'
15. D
p.L26
rt__
To find themaximumvalue of f(x) = 2x3 + 3xz -Izx+ 4 on the closed interval [O2], we evaluate the function/at the critical numbers and endpoints and take the largest resulting value.
,J
.
L
Thus in the closed interval [O2], the only critical number the one critical number and at the endpoints are
is x = 1. The values of the function at
From this list we see that the maximum value is
--J !
-J a z 1 -J 1 1
critical value: f(l): -3 endpoints: /(0)= 4 andf(2):8 8.
-J
p.126
..J
Using the Chain Rule twice on /(x) = ln(cos2x), we obtain
f '(x)
:-f
.
..J
(-sin2x) . (2).
-J a
cos2x
--J a
: _) _sin2;r : _f tan2.r
--1 1 --4 t_
.
cos2x
---1
1,7. A p.126 The slopes of the segments in the slope field vary
form
equation is not of the
ff
=
S(il.
as
*
changes. Thus the differential
That eliminates B, C and E of the five choices.
If the differential equation were # = *' + y2, then slopes would always be at least They aren't in Quadrant III. That eliminates choice (D).
18. E
I
t_ L
f' (*) = 6x2 + 6x -12 = 6(x + 2)(x -l)
16. A
-a
rt-
p.127
-1
0.
b A 3 .Z 3 --1 !a a -1
I
--'1
y=4;+3=1x+3)'
3
-J
1
-
n'=!(*+!i '2'
,aa --a
At ('J.,2), r'
=
:#=
When x = 0, then
y
i
*O
-2:i,,
-
the equation of the line is y
1) and
y
=i
-2 = ir- -U.
.J -a ,-J Z
-a
= :
t
h l-
s ; ): ):
l;
No
SOLUTIONS-Exam
Calculators
19. B
Solution I. Using the Product Rule twice on f (x) = (.r - 1)(x + 2)z gives
t;
; ; ; ; ;
1 L-
Pqrt A
p.127
);
; ; ; ;. ; ;. ; ;. ;, ;. ;. ;. ;. ;. ;;;" ;. ;. ;. ;. ;. ;. ;. ;. ; ; ;
Vl Section I
f ' (x)
= (x + 2)z + (x
l"@)
= 2(x + 2)
*
- l)'
2(x + 2)
2(x + 2) + 2(x
-l)
=6x+6 Thus,
/"(x)
= 0 when x
= -1.
Solution II.
:
Multiplying out 7(x)
(x _ 1Xx +
f(x) = (x -l)(x
2)2
gSves
+2)2
: (x -l)(x2 + 4x + 4) = *' +3x2 - 4 f'(x)=3x2 +6x
f"(*)
20. C
6,
thus
/"(x)
=
o
when x = -1.
p.127 b
in*
- *') dx =
0
ry +1,'.=
rn , $ =:o =+ 21.. C
= 6x +
b3
=27
zb3
a,,d b
-+
=
+
=3.
p.128 Separating variables
n ff
=
Integratinggives y3 = x + C
Thus,
!3=x +
gives:
3y2dy =
contains (-1,-1)
!=xtt3 x < 0..
tdx.
At(-1,-1), -1=-1+C +
!=x113
The largest domain that
Thus,
irt,
is
x < 0.
C=0.
85
SOLUTIONS-Exaw
86
22. C
Vl Section I
Part A
p.728
f\xt=rt-5*'*3* f'(x) =3xz -l}x*3 = (3-r - lXx - 3) and f'(x)
