Consider the function f(x) = -3x² + 2x – 11. Find the critical point, A, of the function.
A=
At x = A, does f(x) have a local min, a local max, or neither? Type in your answer as LMIN, LMAX, or NEITHER.
Answer=
Answers
Answer:
The critical points are
(
x
,
y
)
=
(
1
,
−
2
)
,
(
x
,
y
)
=
(
1
,
2
)
, and
(
x
,
y
)
=
(
1
3
,
0
)
.
Explanation:
The partial derivatives of
z
=
f
(
x
,
y
)
=
x
y
2
−
3
x
2
−
y
2
+
2
x
+
2
are
∂
z
∂
x
=
y
2
−
6
x
+
2
and
∂
z
∂
y
=
2
x
y
−
2
y
=
2
y
(
x
−
1
)
.
Setting these equal to zero gives a system of equations that must be solved to find the critical points:
y
2
−
6
x
+
2
=
0
,
2
y
(
x
−
1
)
=
0
.
The second equation will be true if
y
=
0
, which will lead to the first equation becoming
−
6
x
+
2
=
0
so that
6
x
=
2
and
x
=
1
3
, making one critical point
(
x
,
y
)
=
(
1
3
,
0
)
.
The second equation of the system above will also be true if
x
=
1
, which will lead to the first equation becoming
y
2
−
4
=
0
and
y
2
=
4
, making
y
=
±
2
and leading to two critical points
(
x
,
y
)
=
(
1
,
2
)
,
(
x
,
y
)
=
(
1
,
−
2
)
.
You didn't ask for this, but we can also classify these critical points as follows:
1) Find the second-order partials:
∂
2
z
∂
x
2
=
−
6
,
∂
2
z
∂
y
2
=
2
x
−
2
, and
∂
2
z
∂
x
∂
y
=
∂
2
z
∂
y
∂
x
=
2
y
.
2) Find the discriminant
D
=
∂
2
z
∂
x
2
⋅
∂
2
z
∂
y
2
−
(
∂
2
z
∂
x
∂
y
)
2
=
12
−
12
x
−
4
y
2
3) Plug the critical points into the discriminant:
D
(
1
,
2
)
=
12
−
12
−
16
=
−
16
,
D
(
1
,
−
2
)
=
12
−
12
−
16
=
−
16
, and
D
(
1
3
,
0
)
=
12
−
4
−
0
=
8
.
4) Since
D
(
1
,
±
2
)
=
−
16
<
0
, the critical points at
(
1
,
±
2
)
are saddle points.
5) Since
D
(
1
3
,
0
)
=
8
>
0
and
∂
2
z
∂
x
2
∣
(
1
3
,
0
)
=
−
6
<
0
, the critical point at
(
1
3
,
0
)
is a local maximum.
Here's a contour map of this function in the
x
y
-plane along with its critical points.