Extremum & Inflection Finding and Confirming the Points of Extremum & Inflection.

Extremum & Inflection

Finding and Confirming the Points of Extremum & Inflection

Extremum Inflection

ICritical Points

Points at which the first derivative is equal 0 or does not exist.

Candidate Points for InflectionPoints at which the second derivative is equal 0 or does not exist.

IIThe First Derivative Test for Local ExtremumTesting the sign of the first derivative about the point

The second Derivative Test for inflectionTesting the sign of the second derivative about the point

IIIThe Second Derivative Test for Local ExtremumTesting the sign of the second derivative at the point

The Third Derivative Test for Inflection

Testing the sign of the third derivative at the point

Extremum

I

II.

Critical PointsWe find critical points, which include any point ( x0 , f((x0) ) of f

at which either the derivative f’ (x0) equal 0 or does not

exist.

The First Derivative Test for Local ExtremumFor each critical point ( x0 , f((x0) ) we examine the sign of the

first derivative f’ (x) on the immediate left and the immediate right of this point x0.

If there is a change of sign at x0, then the point ( x0 , f((x0) ) is

a point of local extremum, with the extremum being:

(1) A local maximum if the sign of f’ (x) is positive on the immediate left and negative on the immediate right of the critical point x0

(2) A local minimum if the sign of f’ (x) is negative on the immediate left and positive on the immediate right of the critical point

Extremum

III.The Second Derivative Test for Local Extremum

Only for critical points ( x0 , f((x0) ) for which the

first derivative f’ (x0) = 0

(1) If f’’ (x0) is negative, then the point ( x0 , f(x0) )

is a point of local maximum

(2) If f’’ (x0) is positive, then the point ( x0 , f(x0) )

is a point of local minimum

Inflection

I

II.

Candidate PointsWe find candidate points, which include any point (x0,y0) of f

at which either the second derivative f’’ (x0) equal 0 or does

not exist

The Second Derivative Test for InflectionFor each candidate point ( x0 , f((x0) ) we examine the sign of the

second derivative f’’ (x) on the immediate left and the immediate right of this point x0.

If there is a change of sign at x0, then the point ( x0 , f((x0) ) is a point

of inflection, at which:

(1) The graph of f changes from concave upward to concave downward if the sign of second derivative f’’ (x) changes from positive on the immediate left to negative on the immediate right of the candidate point x0

(2) The graph of f changes from concave downward to concave upward if the sign of second derivative f’’ (x) changes from negative on the immediate left to positive on the immediate right of the candidate point x0

Inflection

III.The Third Derivative Test for InflectionOnly for candidate points ( x0 , f((x0) ) for which

the second derivative f’’ (x0) = 0

(1) If f’’’ (x0) is negative, then the point

( x0 , f((x0) ) is a point of inflection at which the

graph of f changes from concave upward to concave downward

(2) If f’’ (x0) is positive, , then the point

( x0 , f((x0) ) is a point of inflection at which the

graph of f changes from concave downward to concave upward

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Example

Let:

f(x) = 2x3 – 9x2 +12x + 1

Determine all points of extremum and inflection of the

function f and use this information and other information to

sketch its graph.

Solution

mumilocaloftpoinaisf

imumlocaloftpoinaisf

extremunlocalfortestderivativeondThe

ff

toequalisxfwhichattspointspoinCritical

xxxf

xxxxxxxf

xxxxf

haveWe

min)5,2(0)2

32(12)2(

max)6,1(0)2

31(12)1(

:sec

)5,2())2(,2(&)6,1())1(,1(

0)(:

)2

3(121812)(

)2)(1(6)23(612186)(

11292)(

:

22

23

upwardconcavetodownwardconcavefrom

changesfofgraphthewhichatlectionoftpoinais

f

lectionfortestderivativethirdThe

f

toequalisxfwhichattspointspoinCandidate

xf

xxxf

inf)2

11,2

3(

012)2

3(

:inf

)2

11,2

3())

2

3(,

2

3(

0)(:

2)(

)2

3(121812)(

Graphing f

)2

11,2

3(inf.5

)5,2(min.4

)6,1(max.3

)1,0(int.2

.1

:

?)()(

)1,0())0(,0(:int

11292)(

:

lim

23

atlectionanhasf

atlocalahasf

atlocalahasf

ataxisythecsersef

boundnowithdecreasesxwhen

boundnowithdecreasesandboundnowith

increasesxwhenboundnowithincreasesf

haveweNow

Whyxf

faxisythewithfofctionerseThe

xxxxf

haveWe

x

Graph of f

f’

RonupwardconcaveisgRinxallforxgceSin

mumilocaloftpoinaisg

extremunlocalfortestderivativeondThe

g

toequalisxgwhichattspointspoinCritical

xg

xxxg

xxxx

xxxfxg

Let

,)(

min)2

3,2

3(012)

2

3(

:sec

)2

3,2

3())

2

3(,

2

3(

0)(:

12)(

)2

3(121812)(

)2)(1(6)23(6

12186)()(2

2

Graphing g=f’

Ronupwardconcaveisg

atlocalahasg

ataxisythecserseg

boundnowith

decreasesorincreasesxwhenboundnowithincreasesg

haveweNow

Whyxg

gaxisythewithfofctionerseThe

xxxxxg

haveWe

x

.4

)2

3,2

3(min.3

)12,0(int.2

.

.1

:

?)()(

)12,0())0(,0(:int

)2)(1(612186)(

:

lim

2

Graph of g=f’

Question?Home Quiz (1)

Can you graph g without using the derivatives?

f’’

)0,2

3(&)18,0(

)2

3(12

1812)()(

throughlinestraigttheisThis

x

xxfxh

Let

f’’’

)12,0(

12)()(

throughlinehorizontaltheisThis

xfxv

Let

The relation between f and f’

minint))1(,1(

2

2)(.

maxint))1(,1(

1

1)(.

int))2(,2())1(,1(

)1()1(.3

)2,1(

)2,1(.2

),2()1,(

),2()1,(.1

localofpoaisf

xofrightimmediatethetopositiveand

xofleftimmediatethetonegativeisxfb

localofpoaisf

xofrightimmediatethetonegativeand

xofleftimmediatethetopositiveisxfa

fforspocriticalarefandf

zeroarefandf

ingdecreasisfOn

negativeisfOn

ingincreasisfandOn

positiveisfandOn

The relation between f and f’’

.

intinf))2

3(,

2

3(

2

32

3)(.4

.infint))2

3(,

2

3(

)2

3(.3

),2

3(

),2

3(.2

),2

3(

)2

3,(.1

upwardconcavetodownwardconcavefromchangesfof

graphthewhichatfforpolectionanf

xofrightimmediatethetopositiveand

xofleftimmediatethetonegativeisxf

fforlectionforpocandidateisf

zeroisf

upwardconcaveisfOn

positiveisfOn

downwardconcaveisfOn

negativeisfOn

The relation between f and f’’’

upwardconcaveto

downwarconcavefromchangesfofgraphthe

whichatctionleofpoaisf

f

ctionleofpocandidateaisf

infint))2

3(,

2

3(

0)2

3(

&

infint))2

3(,

2

3(

The Four Graphs graphs

Homework (1)

12)(.2

52)(.1

infint

42

24

xxxf

xxxf

itgraph

andfoflectionandextremumofspoallFind

Homework (2)

For each of the functions f of the previous homework (1)

a. Determine the intervals on which f is increasing or decreasing.

b. Determine the intervals on which f is concave upward or concave downward.