Respuesta :
What we are being asked here is to simply minimize distance. Also, note that we can write
f
(
x
)
=
√
x
as
y
=
√
x
.
Now, what is this "distance?" How do we find it? Well, if you think back to Algebra I or Geometry, you'll remember that the distance between two points
(
x
1
,
y
1
)
and
(
x
2
,
y
2
)
is given by:
√
(
y
2
−
y
1
)
2
+
(
x
2
−
x
1
)
2
. For example, the distance between the points
(
4
,
0
)
and
(
0
,
3
)
would be:
√
(
3
−
0
)
2
+
(
4
−
0
)
2
=
√
9
+
16
=
√
25
=
5
Ok, so what is
(
x
1
,
y
1
)
and
(
x
2
,
y
2
)
in our example?
(
x
1
,
y
1
)
is simple - it's just the point given in the problem,
(
4
,
0
)
. Because we don't know what
x
2
is, we'll just call it
x
for now. As for
y
2
, we don't know that either; and since
y
=
√
x
, we'll call it
√
x
.
Our formula then becomes:
√
(
√
x
−
0
)
2
+
(
x
−
4
)
2
=
√
(
√
x
2
)
+
x
2
−
8
x
+
16
=
√
x
+
x
2
−
8
x
+
16
=
√
x
2
−
7
x
+
16
We are being asked to minimize this distance, which we'll call
s
to make the following calculations easier. To minimize something, we have to take its derivative, so let's start there:
s
=
√
x
2
−
7
x
+
16
=
(
x
2
−
7
x
+
16
)
1
2
d
s
d
x
=
(
2
x
−
7
)
⋅
1
2
(
x
2
−
7
x
+
16
)
1
2
→
Using power rule and chain rule
d
s
d
x
=
2
x
−
7
2
√
x
2
−
7
x
+
16
Now we set this equal to
0
and solve for
x
:
0
=
2
x
−
7
2
√
x
2
−
7
x
+
16
0
=
2
x
−
7
x
=
7
2
This is known as the critical value, and it represents the
x
-value for which the function is minimized. All we need to do now is find the corresponding
y
-value, using the definition of
y
:
y
=
√
x
. Substituing
7
2
for
x
:
y
=
√
7
2
y
≈
1.87
And voila, the
y
-value. We can now say that the minimum distance between
f
(
x
)
=
√
x
and the point
(
4
,
0
)
(the place where these two are closest) occurs at
(
7
2
,
1.87
)
. For a little extra fun, we can use the distance formula to see what the actual distance between the points is:
s
=
√
(
1.87
−
0
)
2
+
(
7
2
−
4
)
2
≈
1.8
units
f
(
x
)
=
√
x
as
y
=
√
x
.
Now, what is this "distance?" How do we find it? Well, if you think back to Algebra I or Geometry, you'll remember that the distance between two points
(
x
1
,
y
1
)
and
(
x
2
,
y
2
)
is given by:
√
(
y
2
−
y
1
)
2
+
(
x
2
−
x
1
)
2
. For example, the distance between the points
(
4
,
0
)
and
(
0
,
3
)
would be:
√
(
3
−
0
)
2
+
(
4
−
0
)
2
=
√
9
+
16
=
√
25
=
5
Ok, so what is
(
x
1
,
y
1
)
and
(
x
2
,
y
2
)
in our example?
(
x
1
,
y
1
)
is simple - it's just the point given in the problem,
(
4
,
0
)
. Because we don't know what
x
2
is, we'll just call it
x
for now. As for
y
2
, we don't know that either; and since
y
=
√
x
, we'll call it
√
x
.
Our formula then becomes:
√
(
√
x
−
0
)
2
+
(
x
−
4
)
2
=
√
(
√
x
2
)
+
x
2
−
8
x
+
16
=
√
x
+
x
2
−
8
x
+
16
=
√
x
2
−
7
x
+
16
We are being asked to minimize this distance, which we'll call
s
to make the following calculations easier. To minimize something, we have to take its derivative, so let's start there:
s
=
√
x
2
−
7
x
+
16
=
(
x
2
−
7
x
+
16
)
1
2
d
s
d
x
=
(
2
x
−
7
)
⋅
1
2
(
x
2
−
7
x
+
16
)
1
2
→
Using power rule and chain rule
d
s
d
x
=
2
x
−
7
2
√
x
2
−
7
x
+
16
Now we set this equal to
0
and solve for
x
:
0
=
2
x
−
7
2
√
x
2
−
7
x
+
16
0
=
2
x
−
7
x
=
7
2
This is known as the critical value, and it represents the
x
-value for which the function is minimized. All we need to do now is find the corresponding
y
-value, using the definition of
y
:
y
=
√
x
. Substituing
7
2
for
x
:
y
=
√
7
2
y
≈
1.87
And voila, the
y
-value. We can now say that the minimum distance between
f
(
x
)
=
√
x
and the point
(
4
,
0
)
(the place where these two are closest) occurs at
(
7
2
,
1.87
)
. For a little extra fun, we can use the distance formula to see what the actual distance between the points is:
s
=
√
(
1.87
−
0
)
2
+
(
7
2
−
4
)
2
≈
1.8
units
