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University of Michigan **We aren't endorsed by this school

Course

MATH 115

Subject

Mathematics

Date

Nov 20, 2023

Type

Other

Pages

2

Uploaded by MegaTank11706 on coursehero.com

Prepared by Ryuichi Man
1
MATH115 Calculus I (Section 129)
Fall 2023 Section 3.1 Worksheet
Problem 1
:
A portion of the graph of the function
(
)
m x
, which
is defined on
(
)
,
−
, is shown on the right.
Note that
(
)
m x
is linear on the intervals
(
)
4, 1
−
−
,
(
)
1,2
−
and
(
)
2,6
.
Find the
exact
values, or write
DNE
if the value does
not exist.
a.
Let
(
)
(
)
3
2
P x
m x
x
=
−
. Find
(
)
'
3
P
−
.
b.
Let
( )
( )
3
2
Q b
b
m b
=
+
. Find
( )
" 5
Q
.
c.
Let
( )
( )
6
R t
t
t
m t
=
+
−
. Find
( )
' 2
R
.
d.
Let
(
)
(
)
2
5
3
u
u
S u
m u
u
−
=
−
. Find
( )
' 1
S
.
Solution
:
a.
( )
( )
2
3
'
2
'
P
x
m
x
x
= −
−
➔
(
)
(
)
(
)
2
3
1
13
'
3
2
'
3
2 2
3
3
3
P
m
−
= −
−
−
= −
−
= −
−
b.
( )
( )
2
'
3
2
'
Q
b
b
m
b
=
+
➔
( )
( )
"
6
2
"
Q
b
b
m
b
=
+
➔
( )
( )
( )
( )
" 5
6 5
2
" 5
30
2 0
30
Q
m
=
+
=
+
=
[Since
(
)
m x
is linear on
(
)
2,6
, its second derivative is always zero on
(
)
2,6
.]
c.
( )
( )
1
1
'
6
'
2
R
t
t
m
t
t
−
=
+
−
Since
(
)
m x
is
not
differentiable at
2
x
=
, we conclude that
( )
' 2
R
does not exist (
DNE
).
d.
(
)
(
)
3
1
2
5
3
S u
u
u
m u
−
−
=
−
−
➔
(
)
(
)
5
2
2
3
'
5
3
'
2
S
u
u
u
m
u
−
−
−
=
+
−
➔
(
)
( )
( )
( )
5
2
2
3
3
2
3
'
1
5 1
3
' 1
5
3
2
2
3
2
S
u
m
−
−
−
=
+
−
= −
+
−
=

Prepared by Ryuichi Man
2
Problem 2
:
Let
(
)
p x
be a differentiable function defined on
(
)
,
−
. Define
(
)
(
)
2
3
q x
x
p x
=
−
.
If the equation of the tangent line to
( )
y
p x
=
at the point
1
x
=
is
2
5
y
x
=
−
, find the equation of
the tangent line to
(
)
y
q x
=
at the point
1
x
=
.
Solution
:
Note that
( )
( )
1
2 1
5
3
p
=
−
= −
and
( )
' 1
2
p
=
.
Thus, we have
( )
( )
2
1
1
3
1
10
q
p
=
−
=
and
( )
( )
( )
' 1
2 1
3
' 1
4
q
p
=
−
= −
.
Thus, the required equation is given by
(
)
4
1
10
y
x
= −
−
+
, or
4
14
y
x
= −
+
.
Problem 3
:
[Modified from 2018 Fall Exam 2 Problem 2]
Let
(
)
2
3
3
0
1
1
B
x
x
k x
x
Bx
Ax
x
+
=
+
where
A
and
B
be constants.
a.
Find the values of
A
and
B
such that
( )
k x
is differentiable on
(
)
0,
. If there are no such
values of
A
and
B
, write
NONE
.
b.
Is
(
)
'
k
x
differentiable at
1
x
=
? Explain your answer.
Solution
:
a.
Since
( )
k x
is differentiable at
1
x
=
,
( )
k x
is continuous at
1
x
=
.
Thus, we have
( )
( )
( )
2
3
3 1
1
1
1
B
B
A
+
=
+
➔
3
A
=
.
Note that
(
)
2
2
3
0
1
'
2
9
1
B
x
k
x
x
Bx
x
x
−
=
+
.
Since
( )
k x
is differentiable at
1
x
=
, we have
( )
( )
( )
2
2
3
2
1
9 1
1
B
B
−
=
+
➔
2
B
= −
.
b.
Note that
(
)
3
4
0
1
"
4
18
1
x
k
x
x
x
x
−
=
−
+
.
Since
( )
( )
3
4
4
14
4
18 1
1
−
= −
= −
+
, we conclude that
(
)
'
k
x
is
not
differentiable at
1
x
=
.