J#+ I#:?",h* I
L
e
u_
ra
35
lL---
Exam III Section I
lt-,--
lt--
v_
ra
PartA-NoCalculators 1.
A
u. v_
P.49
B'(x) : cos(sin x) B"(x) : -sin(sin x) ' cos x g'(0) : cos(sin 0) : cos 0: 1. Since g'(0) > 0, g is increasing at x : 0. g"(0) : -sin(sin 0) ' cos 0 : 0. Then g is not concave down at x: O because g'is not decreasing at x :0. g is increasing at x : 0, so g cannot have a relative maximum there.
lt-* ,t--
lr--
l;_
The only true statement is (I).
1L---
u. ,, __
p.49
2.
55
L-_
J#+ I#:?",h*
?.
I : ;@_t) :2
p.50
J.
An equation of the tangent line that passes through the point (4 1) with slope /'(4) : 5 is y-l:5(x-4). Wheny:0, 0-1=5(;r-4)--l:5x-20 => 5x=19 + x=3.8
4.
E
p.50 2
2
dx: -e-*l :-\-1e' /"-" 'o 5.
1-12 e-
p.50
:
x+cosx By definition. .g(x)
ltT h-0
**d
Hence the value of the limit
6.
=
: is
g'(x)
B'(x)
: L - sin x.
P.51
iXdx
: hl *2 +sl], :r, %-tn : *[3]
) ) 36
7.
SOLUTIONS
D
Exam
-
lll
Part
A
Multiple-Choice
-) -)
:
1i1n
e(x+h)-e(x)
n h-0 qvhjzh2 lim --T- h-0 - lim (+x+zh):
h-0'
D
I
p.51
Bv definition. o, s'(x)
8.
Section
rreinnc g(x+h) (srnce
- s(x) :
-./ )
nrjr.+zti)
+x
) ) ) )
p.s1
I. II. II
^4 f(4)=/(0)*Jof'Q)dx=3+0=3 On the interval (0, 5), the graph of f
positive and negative slope.
I.
/'
True
has both
a
False
versa
changes from increasing to decreasing or vice 0 and r 5. Thus, the graph ofl has points of inflection points at r 0 and r 5. at
r
:
:
:
l
-i Ll
) / )
True
:
J al
9.
C
p.52
f on [a,b]
The average value of
averasevalue:
10. E
L1. A
) ) )1l ) )-l )il-l )rl )Ll ) )'0 if sint
e
u_
ra
35
lL---
Exam III Section I
lt-,--
lt--
v_
ra
PartA-NoCalculators 1.
A
u. v_
P.49
B'(x) : cos(sin x) B"(x) : -sin(sin x) ' cos x g'(0) : cos(sin 0) : cos 0: 1. Since g'(0) > 0, g is increasing at x : 0. g"(0) : -sin(sin 0) ' cos 0 : 0. Then g is not concave down at x: O because g'is not decreasing at x :0. g is increasing at x : 0, so g cannot have a relative maximum there.
lt-* ,t--
lr--
l;_
The only true statement is (I).
1L---
u. ,, __
p.49
2.
55
L-_
J#+ I#:?",h*
?.
I : ;@_t) :2
p.50
J.
An equation of the tangent line that passes through the point (4 1) with slope /'(4) : 5 is y-l:5(x-4). Wheny:0, 0-1=5(;r-4)--l:5x-20 => 5x=19 + x=3.8
4.
E
p.50 2
2
dx: -e-*l :-\-1e' /"-" 'o 5.
1-12 e-
p.50
:
x+cosx By definition. .g(x)
ltT h-0
**d
Hence the value of the limit
6.
=
: is
g'(x)
B'(x)
: L - sin x.
P.51
iXdx
: hl *2 +sl], :r, %-tn : *[3]
) ) 36
7.
SOLUTIONS
D
Exam
-
lll
Part
A
Multiple-Choice
-) -)
:
1i1n
e(x+h)-e(x)
n h-0 qvhjzh2 lim --T- h-0 - lim (+x+zh):
h-0'
D
I
p.51
Bv definition. o, s'(x)
8.
Section
rreinnc g(x+h) (srnce
- s(x) :
-./ )
nrjr.+zti)
+x
) ) ) )
p.s1
I. II. II
^4 f(4)=/(0)*Jof'Q)dx=3+0=3 On the interval (0, 5), the graph of f
positive and negative slope.
I.
/'
True
has both
a
False
versa
changes from increasing to decreasing or vice 0 and r 5. Thus, the graph ofl has points of inflection points at r 0 and r 5. at
r
:
:
:
l
-i Ll
) / )
True
:
J al
9.
C
p.52
f on [a,b]
The average value of
averasevalue:
10. E
L1. A
) ) )1l ) )-l )il-l )rl )Ll ) )'0 if sint
