vasquez (mpv244) – HW09 – Schultz – (56445)
1
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001
10.0 points
Determine if the function
f
(
x
) =
x
√
x
+ 12
satisfies the hypotheses of Rolle’s Theorem on
the interval [
−
12
,
0], and if it does, find all
numbers
c
satisfying the conclusion of that
theorem.
1.
c
=
−
8
correct
2.
c
=
−
9
3.
c
=
−
8
,
−
9
4.
c
=
−
5
5.
c
=
−
8
,
8
6.
hypotheses not satisfied
Explanation:
Rolle’s Theorem says that if
f
is
1.
continuous on
[
a, b
]
,
2.
differentiable on
(
a, b
)
, and
3.
f
(
a
) =
f
(
b
) = 0,
then there exists at least one
c
,
a < c < b
,
such that
f
′
(
c
) = 0.
Now the given function
f
(
x
) =
x
√
x
+ 12
,
is defined for all
x
≥ −
12, is continuous on
[
−
12
,
∞
), and differentiable on (
−
12
,
∞
). In
addition
f
(
−
12) =
f
(0) = 0
.
In particular, therefore, Rolle’s theorem ap
plies to
f
on [
−
12
,
0].
On the other hand, by the Product and
Chain Rules,
f
′
(
x
) =
√
x
+ 12+
x
2
√
x
+ 12
=
3
x
+ 24
2
√
x
+ 12
.
Thus there exists
c,
−
12
< c <
0, such that
f
′
(
c
) =
3
c
+ 24
2
√
c
+ 12
= 0
,
in which case
c
=
−
8
.
002
10.0 points
Let
f
be a function defined on [0
,
1] such
that
f
(0) = 1
,
f
(1) = 2
.
Consider
the
following
properties
that
f
might have:
A.
f
(
x
) =

3
x
−
1

;
B.
f
is cont. on [0
,
1]
,
;
C.
f
(
x
) =
braceleftBigg
x
2
+ 1
,
x
negationslash
= 1
/
2,
1
,
x
= 1
/
2 ;
which properties ensure that
f
′
(
c
) = 1 for
some
c
in (0
,
1)?
1.
A and B only
2.
B only
3.
C only
4.
A only
5.
none of them
correct
6.
B and C only
7.
all of them
8.
A and C only
Explanation:
The Mean Value Theorem (MVT) says:
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vasquez (mpv244) – HW09 – Schultz – (56445)
2
If
f
is a function continuous on
[
a, b
]
and
differentiable on
(
a, b
)
, then
f
′
(
c
) =
f
(
b
)
−
f
(
a
)
b
−
a
for at least one
c
in
(
a, b
)
.
If one or more of the hypotheses fail, then
there need not exist any such
c
.
A. No
c
:
f
′
(
x
) =
braceleftBigg
−
3
,
x <
1
/
3,
3
,
x >
1
/
3.
B. No
c
:
f
(
x
) =
braceleftBigg
1
,
0
≤
x <
1
/
2
2
x,
1
/
2
≤
x
≤
1.
C. No
c
: (
f
′
(
x
) = 2
x, x
negationslash
= 1
/
2).
003
10.0 points
How many real roots does the equation
x
5
+ 5
x
+ 3 = 0
have?
1.
exactly four real roots
2.
exactly two real roots
3.
exactly three real roots
4.
exactly one real root
correct
5.
no real roots
Explanation:
Define
f
by
f
(
x
) =
x
5
+ 5
x
+ 3
.
Then the roots of the equation
x
5
+ 5
x
+ 3 = 0
are the
x
intercepts of the graph of
f
.
Now
f
(
x
)
∼
x
5
for

x

large, so
f
(
x
)
→ ∞
as
x
→ ∞
, while
f
(
x
)
→ −∞
as
x
→ −∞
.
Thus the graph of
f
must cross the
x
axis at
least once.
Suppose the graph crosses the
x
at values
x
=
a
and
x
=
b
with
a < b
,
i.e.
,
f
(
a
) =
f
(
b
) = 0
,
(
a < b
)
.
On the other hand,
f
is a polynomial function,
it is continuous and differentiable for all
x
.
Hence Rolle’s Theorem applies, so there exists
some
c
,
a < c < b
, at which
f
′
(
c
) = 0. But
f
′
(
x
) = 5
x
4
+ 5
>
0
for all
x
which isn’t consistent with
f
′
(
c
) = 0
for some
c
. Consequently, the graph of
f
can’t
have
x
intercepts at both
x
=
a
and
x
=
b
, so
the equation has
exactly one root
.
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 Spring '09
 RUTH
 Chemistry, Derivative, Vasquez

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