the minimum value of e^mx + e^-mx is
Answers
Answer:
The minimum is either
f
(
0
)
=
2
or
f
(
1
)
=
e
+
e
−
2
=
e
+
1
e
2
=
e
3
+
1
e
2
or
f
(
c
)
for some critical number of
f
in
[
0
,
1
]
.
So we need to find critical numbers for
f
f
'
(
x
)
=
e
x
−
2
e
−
2
x
=
e
x
−
2
e
2
x
=
e
3
x
−
2
e
2
x
Since the denominator
e
2
x
is never
0
(it is always positive), we see that
f
'
is never undefined.
Solving
f
'
(
x
)
=
0
gets us
e
3
x
=
2
so
3
x
=
ln
2
and
x
=
ln
2
3
=
ln
3
√
2
Because
ln
is an increasing function, and
1
<
3
√
2
<
e
,
we see that
ln
3
√
2
is in
[
0
,
1
]
So
c
=
ln
3
√
2
=
ln
2
3
is the only critical number.
We need to determine whether
f
(
c
)
is a minimum, a maximum or neither.
The denominator of
f
'
(
x
)
is always positive, so the sign of
f
'
(
x
)
will be the same as the sign of the numerator:
e
3
x
−
2
.
For
x
<
ln
2
3
, we have
3
x
<
ln
2
, and
e
3
x
<
e
ln
2
=
2
so
e
3
x
−
2
. and
f
'
(
x
)
is negative.
By similar reasoning, for
x
>
ln
2
3
, we have
f
'
(
x
)
>
0
Therefore
f
(
ln
2
3
)
is a local minimum.
Furthermore,
f
(
ln
2
3
)
<
f
(
0
)
since
f
is decreasing on
[
0
,
ln
2
3
)
and
f
(
ln
2
3
)
<
f
(
1
)
since
f
is increasing on
(
ln
2
3
,
1
]
So on the interval
[
0
,
1
]
the value
f
(
ln
2
3
)
is the absolute minimum.
f
(
ln
3
√
2
)
=
e
ln
3
√
2
+
e
−
2
(
ln
(
2
1
3
)
)
=
e
ln
3
√
2
+
e
ln
(
2
−
2
3
)
=
3
√
2
+
1
3
√
4
The minimum value of
f
(
x
)
=
e
x
+
e
−
2
x
on [0,1] is
3
√
2
+
1
3
√
4
(Rewrite using algebra until you're happy with the way it looks. Personally, I like:
3
3
√
4
)
Now that we're finished, it might be nice to see the graph of
f
graph{(y - e^x-e^(-2x))=0 [-0.965, 2.453, 1.486, 3.195]}
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