Study Resources
Main Menu
by School
by Textbook
by Literature Title
Study Guides
Infographics
by Subject
Expert Tutors
Contributing
Main Menu
Earn Free Access
Upload Documents
Refer Your Friends
Earn Money
Become a Tutor
Scholarships
For Educators
Log in
Sign up
Find
Study Resources
by School
by Textbook
by Literature Title
Study Guides
Infographics
by Subject
Ask
Expert Tutors
You can ask
!
Earn by
Contributing
Earn Free Access
Learn More >
Upload Documents
Refer Your Friends
Earn Money
Become a Tutor
Scholarships
Learn More >
Are you an educator?
Log in
Sign up
Test 2 Review 2.pdf  MATH 1432 EXAM 1 REVIEW Answers 1...
School
No School
Course Title
AA 1
Uploaded By
AgentDugong953
Pages
2
This
preview
shows page
1  2
out of
2
pages.
MATH 1432
EXAM 1 REVIEW:
Answers
1. Inverse Functions
(a)
(
i
)
Yes;
f
(
x
) = 3
x
2
+ 4
>
0
for all
x
=
⇒
f
is onetoone.
(
ii
)
No;
for example
g
(

2) =
g
(0) =
g
(2) = 3;
g
is not onetoone.
(
iii
)
Yes;
h
(
x
) =

2
e

2
x
<
0
for all
x
.
(b) (
i
)
domain:
(0
,
∞
)
(
ii
)
domain:
(
∞
,
∞
)
(
iii
)
domain:
[

1
,
1]
range:
(
∞
,
∞
)
range:
(0
,
∞
)
range:

π
2
,
π
2
(c)
(
i
)
f

1
(
x
) =
3
1

x
x
(
ii
)
range:
all
x
such that
x
=

1.
(d)
(
f

1
)
(3) =
1
f
(1)
=
1
2
/
3
=
3
2
.
(e)
(
G

1
)
(

1) =
1
G
(
π
)
=
1

1
=

1
Log and exponential functions
(a)
ln 2
(b)
x
=
1
2
e
(c)
1
2
(d)
y
=
x
+ 2
(e)
ln 2

7
24
(f)
1
1
2
3
x
1
0.4
y
(g)
(
i
)
points of inflection:
(
1
, e

1
/
2
)
,
(

1
, e

1
/
2
)
(
ii
)
V
=
1
0
2
πx e

x
2
/
2
dx
= 2
π
1

e

1
/
2
3. Derivatives transcendental functions
(a)
1
(1 +
x
2
) tan

1
x
+
2

2
x
+ 1

(2
x
+ 1)
2

1
(b)
g
(
x
) =
e
x
2
x
1

(ln
x
)
2
+ 2
x e
x
2
sin

1
(ln
x
)
(c)
h
(
x
) =
5
x
sinh
(
e
2
x
)
e
2
x
(2)

cosh
(
e
2
x
)
5
x
ln 5
5
2
x
(d)
y
= (sinh 2
x
)
x
3
2
x
3
cosh 2
x
sinh 2
x
+ 3
x
2
ln (sinh 2
x
)
4. Integrals of transcendental functions
(a)
√
cosh 2
x
sinh 2
x dx
=
1
3
(cosh 2
x
)
3
/
2
+
C
(substitution:
u
= cosh 2
x
)
1
(b)
e
ln(
x
2
+1
dx
=
(
x
2
+ 1
)
dx
=
1
3
x
3
+
x
+
C
(c)
[ln(
x
+ 2)]
2
x
+ 2
dx
=
1
3
[ln(
x
+ 2)]
3
+
C
(substitution:
u
= ln(
x
+ 2))
(d)
sec(ln
x
)
x
dx
= ln [sec(ln
x
) + tan(ln
x
)] +
C
(substitution:
u
= ln
x
)
(e)
ln
√
2
0
e
x
√
4

e
2
x
dx
=
sin

1
e
x
2
ln
√
2
0
=
π
12
(substitution:
u
=
e
x
)
(f)
tan

1
x
1 +
x
2
dx
=
1
2
(
tan

1
x
)
2
+
C
(substitution
u
= tan

1
x
)
5. Exponential growth and decay
(a)
P
(
t
) = 2000
e
t
2
ln(3
/
2)
= 2000
(
3
2
)
t/
2
(
i
)
P
(8) = 2000
(
3
2
)
4
∼
= 10
,
125
(
You've reached the end of your free preview.
Want to read both pages?
TERM
Fall '19
TAGS
Exponential Function,
0 g,
ELN,
17.86 grams,
6.84 years,
237.84 grams
Share this link with a friend:
Copied!
Stuck? We have tutors online 24/7 who can help you get unstuck.
Ask Expert Tutors
You can ask
You can ask
You can ask
(will expire
)
Answers in as fast as
15 minutes