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RE: Last to post wins - Kanjuura - 03-22-2023
A critical point of a continuous function
�
f is a point at which the derivative is zero or undefined. Critical points are the points on the graph where the function's rate of change is altered—either a change from increasing to decreasing, in concavity, or in some unpredictable fashion. Critical points are useful for determining extrema and solving optimization problems.
Contents
Definition
Types of Critical Points
See Also
Definition
A continuous function
�
f with
�
x in its domain has a critical point at that point
�
x if it satisfies either of the following conditions:
�
′
(
�
)
=
0
;
f
′
(x)=0;
�
′
(
�
)
f
′
(x) is undefined.
A critical point of a differentiable function
�
f is a point at which the derivative is 0.
Find all critical points of
�
(
�
)
=
�
4
−
4
�
3
+
16
�
f(x)=x
4
−4x
3
+16x.
The derivative of
�
f is
�
′
(
�
)
=
4
�
3
−
12
�
2
+
16
=
4
(
�
+
1
)
(
�
−
2
)
2
,
f
′
(x)=4x
3
−12x
2
+16=4(x+1)(x−2)
2
,
so the derivative is zero at
�
=
−
1
x=−1 and
�
=
2
x=2. Since
�
′
f
′
is defined on all real numbers, the only critical points of the function are
�
=
−
1
x=−1 and
�
=
2.
□
x=2.
□
Types of Critical Points
A local extremum is a maximum or minimum of the function in some interval of
�
x-values. An inflection point is a point on the function where the concavity changes (the sign of the second derivative changes). While any point that is a local minimum or maximum must be a critical point, a point may be an inflection point and not a critical point.
A critical point is a local maximum if the function changes from increasing to decreasing at that point and is a local minimum if the function changes from decreasing to increasing at that point.
A critical point is an inflection point if the function changes concavity at that point.
A critical point may be neither. This could signify a vertical tangent or a "jag" in the graph of the function.
The first derivative test provides a method for determining whether a point is a local minimum or maximum. If the function is twice-differentiable, the second derivative test could also help determine the nature of a critical point. However, if the second derivative has value
0
0 at the point, then the critical point could be either an extremum or an inflection point.
Classify the critical points of the following function:
�
(
�
)
=
{
1
−
(
�
+
1
)
2
�
<
0
2
�
0
≤
�
≤
1
3
−
(
�
−
2
)
2
1
<
�
≤
2
3
+
(
�
−
2
)
3
�
>
2.
f(x)=
⎩
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎧
1−(x+1)
2
2x
3−(x−2)
2
3+(x−2)
3
x<0
0≤x≤1
1<x≤2
x>2.
The derivative of
�
f is
�
′
(
�
)
=
{
−
2
(
�
+
1
)
�
<
0
2
0
≤
�
≤
1
−
2
(
�
−
2
)
1
<
�
≤
2
3
(
�
−
2
)
2
�
>
2.
f
′
(x)=
⎩
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎧
−2(x+1)
2
−2(x−2)
3(x−2)
2
x<0
0≤x≤1
1<x≤2
x>2.
Note that the derivative has value
0
0 at points
�
=
−
1
x=−1 and
�
=
2
x=2. At
�
=
0
x=0, the derivative is undefined, and therefore
�
=
0
x=0 is a critical point. At
�
=
1
x=1, the derivative is
2
2 when approaching from the left and
2
2 when approaching from the right, so since the derivative is defined
(
(and equal to
2
≠
0
)
,
2
=0),
�
=
1
x=1 is not a critical point.
The critical point
�
=
−
1
x=−1 is a local maximum.
The critical point
�
=
0
x=0 is a local minimum.
The critical point
�
=
2
x=2 is an inflection point.
□
□
RE: Last to post wins - OnTheHorizon143 - 03-22-2023
That post is so long I’m calling it spam
RE: Last to post wins - Kanjuura - 03-22-2023
maths
RE: Last to post wins - OnTheHorizon143 - 03-22-2023
Algebruh
RE: Last to post wins - New makasta - 03-22-2023
ok weebs
RE: Last to post wins - Kanjuura - 03-22-2023
(03-22-2023, 02:23 AM)OnTheHorizon143 Wrote: Algebruh
no, calculus
RE: Last to post wins - OnTheHorizon143 - 03-22-2023
(03-22-2023, 02:59 AM)New makasta Wrote: ok weebs
*nerd
Also someone who has nyaaaa in their discord bio can’t call anyone a weeb
RE: Last to post wins - Spode - 03-22-2023
nyaaaaaaaaaaaaa
RE: Last to post wins - OnTheHorizon143 - 03-22-2023
Cope harder, Spode
RE: Last to post wins - Quebecshire - 03-22-2023
seethe