m53 lec3.1 mvt and relative extrema

8/18/2019 M53 Lec3.1 MVT and Relative Extrema

1/218

The Mean Value Theorem and Relative Extrema

Mathematics 53

Institute of Mathematics (UP Diliman)

Institute of Mathematics (UP Diliman) MVT and Relative Extrema Mathematics 53 1 / 42


2/218

For today

1 The Mean Value Theorem

2 Critical Numbers

3 Increasing/Decreasing Functions

4 The First Derivative Test For Relative Extrema



3/218

For today


2 Critical Numbers





4/218

Rolle’s Theorem

Michel Rolle (1652-1719)



5/218

Rolle’s Theorem

Theorem

Let a , b ∈ such that a < b .

a b



6/218

Rolle’s Theorem

Theorem

Let a , b ∈ such that a < b . If a function f is1 continuous on [a , b ]

a b



7/218

Rolle’s Theorem

Theorem


2 differentiable on (a , b ) and

a b



8/218

Rolle’s Theorem

Theorem



3 f (a )

= f (b )

=0,

a b



9/218

Rolle’s Theorem

Theorem



3 f (a )

= f (b )

=0,

a b



10/218

Rolle’s Theorem

Theorem



3 f (a )

= f (b )

=0,

then there exists c ∈ (a , b ) such that f (c ) = 0.

a b

c



11/218

Rolle’s Theorem: The Assumptions

Continuity on [a , b ]

a

b

satisfies conditions (2) and (3) but not (1)



12/218


Continuity on [a , b ]

b a




13/218


Differentiability on (a , b )

a b




14/218


Differentiability on (a , b )

a b

c

satisfies conditions (1),(2) and (3), not differentiable at x = a and x = b



15/218


c may not be unique



16/218


c may not be unique

b a


ll ’ h l


17/218

Rolle’s Theorem: Applications

Example

Determine if Rolle’s Theorem is applicable to

f (x ) = x 3 − 4x 2 + 5x − 2 on [1,2].


R ll ’ Th A li i


18/218


Example


f (x ) = x 3 − 4x 2 + 5x − 2 on [1,2].

Solution:


R ll ’ Th A li i


19/218


Example


f (x ) = x 3 − 4x 2 + 5x − 2 on [1,2].

Solution:

f (x ) = x 3 − 4x 2 + 5x − 2


R ll ’ Th A li ti


20/218


Example


f (x ) = x 3 − 4x 2 + 5x − 2 on [1,2].

Solution:

f (x ) = x 3 − 4x 2 + 5x − 2

continuous on [1,2]?

differentiable on (1,2)?


R ll ’ Th A li ti


21/218

Rolle s Theorem: Applications

Example


f (x ) = x 3 − 4x 2 + 5x − 2 on [1,2].

Solution:

f (x ) = x 3 − 4x 2 + 5x − 2 is a polynomial

continuous on [1,2]?

differentiable on (1,2)?


R ll ’s Th A li ti s


22/218


Example


f (x ) = x 3 − 4x 2 + 5x − 2 on [1,2].

Solution:


continuous on [1,2]? (Yes.)

differentiable on (1,2)? (Yes.)




23/218


Example


f (x ) = x 3 − 4x 2 + 5x − 2 on [1,2].

Solution:




f (1) = 0 = f (2)




24/218


Example


f (x ) = x 3 − 4x 2 + 5x − 2 on [1,2].

Solution:




f (1) = 0 = f (2) (yes!)




25/218


Example


f (x ) = x 3 − 4x 2 + 5x − 2 on [1,2].

Solution:




f (1) = 0 = f (2) (yes!)

∴ By Rolle’s Theorem, there is a c ∈ (1,2) such that f (c ) = 0.




26/218


Example


f (x ) = x 3 − 4x 2 + 5x − 2 on [1,2].

Solution:




f (1) = 0 = f (2) (yes!)


f (x )=

3x 2

−8x

+5

=(3x

−5)(x

−1)




27/218


Example


f (x ) = x 3 − 4x 2 + 5x − 2 on [1,2].

Solution:




f (1) = 0 = f (2) (yes!)


f (x )=

3x 2

−8x

+5

=(3x

−5)(x

−1)

⇒ c

= 53




28/218


Example

Determine if RT applies to f (x ) =

x 2

, x ≤ 1

2

x − 1 , x > 12

on [0,1].




29/218


Example


x 2

, x ≤ 1

2

x − 1 , x > 12

on [0,1].

Solution:




30/218


Example


x 2

, x ≤ 1

2

x − 1 , x > 12

on [0,1].

Solution:

Check continuity of f on [0,1].




31/218


Example


x 2

, x ≤ 1

2

x − 1 , x > 12

on [0,1].

Solution:

Check continuity of f on [0,1]

.

f

1

2

= 1

4




32/218

pp

Example


x

2

, x ≤ 1

2

x − 1 , x > 12

on [0,1].

Solution:


.

f

1

2

= 1

4

limx → 1

2− f (x )




33/218

pp

Example

Determine if RT applies to f (x ) = x 2 , x ≤ 12x − 1 , x > 1

2

on [0,1].

Solution:


.

f

1

2

= 1

4

limx → 1

2− f (x )

= limx → 1

2−

x 2




34/218

pp

Example


2

on [0,1].

Solution:


.

f

1

2

= 1

4

limx → 1

2− f (x )

= limx → 1

2−

x 2

=1

4




35/218

pp

Example


2

on [0,1].

Solution:


.

f

1

2

= 1

4

limx → 1

2− f (x )

= limx → 1

2−

x 2

=1

4

limx → 1

2

+ f (x )




36/218

Example


2

on [0,1].

Solution:


.

f

1

2

= 1

4

limx → 1

2− f (x )

= limx → 1

2−

x 2

=1

4

limx → 1

2

+ f (x ) = lim

x → 12

+x − 1




37/218

Example


2

on [0,1].

Solution:

Check continuity of f

on [0,1]

.

f

1

2

= 1

4

limx → 12 −

f (x )=

limx → 12 −

x 2

=1

4

limx → 1

2

+ f (x ) = lim

x → 12

+x − 1 = −1

2




38/218

Example


2

on [0,1].

Solution:

Check continuity of f

on [0,1]

.

f

1

2

= 1

4

limx → 12 −

f (x )=

limx → 12 −

x 2

=1

4

limx → 1

2

+ f (x ) = lim

x → 12

+x − 1 = −1

2

∴ f is discontinuous at x =

12

and RT cannot be applied.Institute of Mathematics (UP Diliman) MVT and Relative Extrema Mathematics 53 10 / 42

Mean Value Theorem


39/218

Theorem

Let a , b ∈ R such that a < b . If a function f is

a b


Mean Value Theorem


40/218

Theorem

Let a , b ∈ R such that a < b . If a function f is(i) continuous on [a , b ] and;

(ii) differentiable on (a , b ),

a

(a , f (a ))

b

(b , f (b ))


Mean Value Theorem


41/218

Theorem

Let a , b ∈R

such that a < b . If a function f is(i) continuous on [a , b ] and;


then there exists c

∈(a , b ) such that

a

(a , f (a ))

b

(b , f (b ))


Mean Value Theorem


42/218

Theorem

Let a , b ∈R



then there exists c

∈(a , b ) such that f (c )

=

f (b ) − f (a )

b −a .

a

(a , f (a ))

b

(b , f (b ))


Mean Value Theorem


43/218

Theorem

Let a , b ∈R



then there exists c ∈ (a , b ) such that f (c ) = f (b ) − f (a )

b −a .

a

(a , f (a ))

b

(b , f (b ))

c


Mean Value Theorem: Applications


44/218

Example

Let f (x ) = x + 2x + 1 . Show that there is a c ∈ (1,2) such that f

(c ) = −16

.

.




45/218

Example


(c ) = −16

.

Solution:

.




46/218

Example


(c ) = −16

.

Solution:

Is f is continuous on [1,2]?

.




47/218

Example


(c ) = −16

.

Solution:

Is f is continuous on [1,2]? (Yes.)

.




48/218

Example


(c ) = −16

.

Solution:


Is f is differentiable on (1,2)?

.




49/218

Example


(c ) = −16

.

Solution:


Is f is differentiable on (1,2)? (Yes.)

.




50/218

Example


(c ) = −16

.

Solution:



By the Mean Value Theorem, there is a c ∈ (1,2) such that

f (c ) = f (2) − f (1)2 − 1

.




51/218

Example


(c ) = −16

.

Solution:




f (c ) = f (2) − f (1)2 − 1 =

4

3− 3

2

.




52/218

Example


(c ) = −16

.

Solution:




f (c ) = f (2) − f (1)2 − 1 =

4

3− 3

2= −1

6

.




53/218

Example

Suppose that f (x ) is continuous on the interval [6,15], differentiable on(6,15) and f (x ) ≤ 10 for all x . If f (6) = −2, what is the largest possiblevalue for f (15)?




54/218

Example


By MVT, there is a c ∈ (6,15) such that




55/218

Example



f (c ) = f (15) − f (6)15− 6




56/218

Example



f (c ) = f (15) − f (6)15− 6

= f (15) + 29




57/218

Example



f (c ) = f (15) − f (6)15− 6

= f (15) + 29

Then




58/218

Example



f (c ) = f (15) − f (6)15− 6

= f (15) + 29

Then f (15) = 9 · f (c ) − 2




59/218

Example



f (c ) = f (15) − f (6)15− 6

= f (15) + 29

Then f (15) = 9 · f (c ) − 2

≤ 9(10) − 2




60/218

Example



f (c ) = f (15) − f (6)15− 6

= f (15) + 29

Then f (15) = 9 · f (c ) − 2

≤ 9(10) − 2∴ f (15)

≤88.


For today


61/218


2 Critical Numbers




Relative Extrema


62/218

Definition

A function f is said to have a relative maximum at x = c


Relative Extrema


63/218

Definition

A function f is said to have a relative maximum at x = c if there is anopen interval I , containing c , such that


Relative Extrema


64/218

Definition

A function f is said to have a relative maximum at x = c if there is anopen interval I , containing c , such that f (x ) is defined for all x ∈ I and


Relative Extrema


65/218

Definition


f (x ) ≤ f (c ) for all x ∈ I .


Relative Extrema


66/218

Definition


f (x ) ≤ f (c ) for all x ∈ I .

Definition

A function f is said to have a relative minimum at x = c if there is anopen interval I , containing c , such that f (x ) is defined for all x ∈ I and

f (x ) ≥ f (c ) for all x ∈ I .


Relative Extrema


67/218

Definition


f (x ) ≤ f (c ) for all x ∈ I .

Definition

A function f is said to have a relative minimum at x = c if there is anopen interval I , containing c , such that f (x ) is defined for all x ∈ I and

f (x ) ≥ f (c ) for all x ∈ I .

DefinitionWe say f has a relative extremum at x = c if f has either a relativemaximum at x = c or a relative minimum at x = c .


Relative Extrema


68/218

c 1 c 2 c 3 c 4 c 5 c 6


Relative Extrema


69/218

c 1 c 2 c 3 c 4 c 5 c 6

relative MAX at x =

c 3, x =

c 5


Relative Extrema


70/218

c 1 c 2 c 3 c 4 c 5 c 6

relative MIN at x =

c 4


Relative Extrema


71/218

c 1 c 2 c 3 c 4 c 5 c 6

no relative MAX at x =

c 1, no relative MIN at x =

c 6


Relative Extrema


72/218

c 1 c 2 c 3 c 4 c 5 c 6

no relative MIN at x =

c 2


Relative Extrema


73/218

(c , f (c )

c

f (c )

Terminologies:

1 f has a relative extremum at x = c .


Relative Extrema


74/218

(c , f (c )

c

f (c )

Terminologies:


2 (c , f (c )) is a relative extremum point of f .


Relative Extrema


75/218

(c , f (c )

c

f (c )

Terminologies:


2 (c , f (c )) is a relative extremum point of f .

3 f (c ) is a relative extremum value of f .


Critical Numbers


76/218

Example

y

= −x 2

f (x ) = −x 2


Critical Numbers


77/218

Example

y

= −x 2

f (x ) = −x 2

f (x ) = −2x


Critical Numbers


78/218

Example

y

= −x 2

f (x ) = −x 2

f (x ) = −2x

f (0) = 0


Critical Numbers


79/218

Example

y = |x − 3|

f (x ) = |x − 3|


Critical Numbers


80/218

Example

y = |x − 3|

f (x ) = |x − 3|

f (x ) =

1 , x > 3−1 , x < 3


Critical Numbers


81/218

Example

y = |x − 3|

f (x ) = |x − 3|

f (x ) =

1 , x > 3−1 , x < 3

f (3) dne


Critical Numbers


82/218

Example

y = |x − 3|

f (x ) = |x − 3|

f (x ) =

1 , x > 3−1 , x < 3

f (3) dne

Theorem

If a function f has a relative extremum point at x = c , then either f (c ) = 0or f (c ) does not exist.


Critical Numbers

Definition


83/218

A number c ∈ dom f is said to be a critical number of f if either f (c ) = 0or f (c ) is undefined.


Critical Numbers

Definition

i i l b f f


84/218


Is c a criticalnumber of f ?


Critical Numbers

Definition

A b d f i id b i i l b f f if i h f


85/218



no


Critical Numbers

Definition

A b d f i id b i i l b f f if i h f ( ) 0


86/218

A number c ∈ dom f is said to be a critical number of f if either f (c ) = 0

or f (c ) is undefined.


f has no relativeextremum at x = c .

no


Critical Numbers

Definition

A b d f i id t b iti l b f f if ith f ( ) 0


87/218





yes

no


Critical Numbers

Definition

A b ∈ d f i id t b iti l mbe f f if ith f ( ) 0


88/218




Does f have a relative

extremum at x = c ?


yes

no


VENN DIAGRAM


89/218

domain of f

critical numbers of f

x -coordinates of relative extrema of f


Critical Numbers


90/218

Example

Find all critical numbers of f (x ) = x 3

9− 3x .


Critical Numbers


91/218

Example


9− 3x .

Solution: f (x )


Critical Numbers


92/218

Example


9− 3x .

Solution: f (x ) = x

2

3− 3


Critical Numbers


93/218

Example


9− 3x .


2

3− 3 = 1

3(x 2 − 9)


Critical Numbers


94/218

Example


9− 3x .


2

3− 3 = 1

3(x 2 − 9) = 1

3(x − 3)(x + 3)


Critical Numbers


95/218

Example


9− 3x .


2

3− 3 = 1

3(x 2 − 9) = 1

3(x − 3)(x + 3)

f (x )=0 when x = ±3


Critical Numbers

E l


96/218

Example


9− 3x .


2

3− 3 = 1

3(x 2 − 9) = 1

3(x − 3)(x + 3)

f (x )=0 when x = ±3

f (x ) is always defined


Critical Numbers

E l


97/218

Example


9− 3x .


2

3− 3 = 1

3(x 2 − 9) = 1

3(x − 3)(x + 3)

f (x )=0 when x = ±3


CN : ±3

Instit te of Mathematics (UP Diliman) MVT and Relati e E t ema Mathematics 53 22 / 42

Critical Numbers


98/218

ExampleFind all critical numbers of f (x ) = x 4 + 2x 3.

I tit t f M th ti (UP Dili ) MVT d R l ti E t M th ti 53 23 / 42

Critical Numbers


99/218


Solution:


Critical Numbers


100/218


Solution:

f (x ) = 4x 3 + 6x 2


Critical Numbers


101/218


Solution:

f (x ) = 4x 3 + 6x 2 = x 2(4x + 6)


Critical Numbers


102/218


Solution:

f (x ) = 4x 3 + 6x 2 = x 2(4x + 6) f (x ) = 0 when x = 0 or x = − 3

2


Critical Numbers


103/218


Solution:


2



Critical Numbers


104/218


Solution:


2


CN : 0,−32


Critical Numbers

Example


105/218

Example


9− x 2 .


Critical Numbers

Example


106/218

Example


9− x 2 .

Solution:

f (x ) =


Critical Numbers

Example


107/218

Example


9− x 2 .

Solution:

f (x ) = 18x (9 −x 2)2


Critical Numbers

Example


108/218

a p e


9− x 2 .

Solution:

f (x ) = 18x (9 −x 2)2

f (x ) = 0 when x = 0


Critical Numbers

Example


109/218

p


9− x 2 .

Solution:

f (x ) = 18x (9 −x 2)2

f (x ) = 0 when x = 0

f (x ) is undefined when x = ±

3


Critical Numbers

Example


110/218

p


9− x 2 .

Solution:

f (x ) = 18x (9 −x 2)2

f (x ) = 0 when x = 0


3 but ±

3

∉dom f !


Critical Numbers

Example


111/218

p


9− x 2 .

Solution:

f (x ) = 18x (9 −x 2)2

f (x ) = 0 when x = 0


3 but

±3

∉dom f !

Thus, CN : 0


Critical Numbers

Example


112/218

Example

Find all critical numbers of f (x ) = −x 4/3 + 4x 1/3.


Critical Numbers

Example


113/218

Example


Solution:

f (x ) =


Critical Numbers

Example


114/218

Example


Solution:

f (x ) = −4

3 x 1/3


Critical Numbers

Example


115/218

Example


Solution:

f (x ) = −4

3 x 1/3

+4

3 x −

2/3


Critical Numbers

Example


116/218

p


Solution:

f (x ) = −4

3 x 1/3

+4

3 x −

2/3

= −4(x

−1)

3x 2/3


Critical Numbers

Example


117/218

p


Solution:

f (x ) = −4

3 x 1/3

+4

3 x −2/3

= −4(x

−1)

3x 2/3

f (x ) = 0 when x = 1


Critical Numbers

Example


118/218

p


Solution:

f (x ) = −4

3 x 1/3

+4

3 x −2/3

= −4(x

−1)

3x 2/3

f (x ) = 0 when x = 1

f (x ) is undefined when x = 0


Critical Numbers

Example


119/218


Solution:

f (x ) = −4

3 x 1/3

+4

3 x −2/3

= −4(x

−1)

3x 2/3

f (x ) = 0 when x = 1

f (x ) is undefined when x = 0 and 0 ∈ dom f


Critical Numbers

Example


120/218


Solution:

f (x ) = −4

3 x 1/3

+4

3 x −2/3

= −4(x

−1)

3x 2/3

f (x ) = 0 when x = 1

f (x ) is undefined when x = 0 and 0 ∈ dom f

Thus, CN : 0, 1


For today


121/218


2 Critical Numbers




Increasing/Decreasing Functions


122/218

Definition

Let f be a function defined on an interval I .




123/218

Definition


1 f is said to be (strictly) increasing on I if f (a )


124/218

Definition


1 f is said to be (strictly) increasing on I if f (a ) f (b ) for all a , b ∈ I such that a < b .




125/218

c 1 c 2 c 3 c 4 c 5 c 6




126/218

c 1 c 2 c 3 c 4 c 5 c 6

f is DECREASING on




127/218

c 1 c 2 c 3 c 4 c 5 c 6

f is DECREASING on [c 1, c 2)




128/218

c 1 c 2 c 3 c 4 c 5 c 6

f is DECREASING on [c 1, c 2), [c 3, c 4]




129/218

c 1 c 2 c 3 c 4 c 5 c 6

f is DECREASING on [c 1, c 2), [c 3, c 4] and [c 5, c 6].




130/218

c 1 c 2 c 3 c 4 c 5 c 6

Incorrect to say f is decreasing on [c 1, c 2) ∪ [c 3, c 4] ∪ [c 5, c 6]




131/218

c 1 c 2 c 3 c 4 c 5 c 6

f is increasing on




132/218

c 1 c 2 c 3 c 4 c 5 c 6

f is increasing on (c 2, c 3]




133/218

c 1 c 2 c 3 c 4 c 5 c 6

f is increasing on (c 2, c 3] and [c 4, c 5].


Derivative of Increasing/Decreasing Functions

Theorem

L f b f i h i i i l [ b] d


134/218

Let f be a function that is continuous on an interval [a , b ] anddifferentiable on (a , b ).



Theorem

L t f b f ti th t i ti i t l [ b] d


135/218


1 If f (x )>

0 for all x

∈(a , b ), then f is increasing on [a , b ].



Theorem

Let f be a f ctio that is co ti o s o a i te al [ b] a d


136/218


1 If f (x )>

0 for all x


2 If f (x ) < 0 for all x ∈ (a , b ), then f is decreasing on [a , b ].



Theorem

Let f be a function that is continuous on an interval [a b] and


137/218


1 If f (x )>

0 for all x


2 If f (x ) < 0 for all x ∈ (a , b ), then f is decreasing on [a , b ].

3 If f (x ) = 0 for all x ∈ (a , b ), then f is constant on [a , b ].



Example

f (x) x2


138/218

y = x 2

f (x ) = x 2



Example

f (x) = x2


139/218

y = x 2

f (x ) = x

f (x ) = 2x



Example

f (x) = x2


140/218

y = x 2

f (x ) = x

f (x ) = 2x x (−∞, 0) 0 (0,∞)

f (x ) − 0 +



Example

f (x) = x2 ( 0) 0 (0 )


141/218

y = x 2

f (x ) = x

f (x ) = 2x x (−∞, 0) 0 (0,∞)

f (x ) − 0 +

⇒ f is increasing on [0,

∞)



Example

f (x) = x2 ( 0) 0 (0 )


142/218

y = x 2

f (x ) = x

f (x ) = 2x x (−∞, 0) 0 (0,∞)

f (x ) − 0 +

⇒ f is increasing on [0,

∞)

⇒ f is decreasing on (−∞, 0]



Example

f (x) = x3


143/218

y = x 3

f (x ) = x



Example

f (x) = x3


144/218

y = x 3

f (x ) = x

f (x ) = 3x 2



Example

f (x) = x3


145/218

y = x 3

f (x ) = x

f (x ) = 3x 2

f (x ) > 0 whenever x = 0.



Example

f (x) = x3


146/218

y = x 3

f (x ) x

f (x ) = 3x 2

f (x ) > 0 whenever x = 0.⇒ The graph of f is increasing on !


For today



147/218

2 Critical Numbers




Theorem (First Derivative Test for Relative Extrema)Let f be a function continuous on the open interval (a , b ) which containsthe number c . Suppose that f is also differentiable on the the interval(a , b ), except possibly at c .


148/218

NOTE: f (c ) has to be defined but not necessarily f (c )



1 If f (x ) > 0 for all x ∈ (a , c ) and f (x ) < 0 for all x ∈ (c , b ),


149/218


a c b x

sign of f (x ) + –



1 If f (x ) > 0 for all x ∈ (a , c ) and f (x ) < 0 for all x ∈ (c , b ), then f has arelative maximum at x = c.


150/218

relative maximum at x c .


a c b x

sign of f (x ) + – f has a relativemaximum at x

=c



1 If f (x ) > 0 for all x ∈ (a , c ) and f (x ) < 0 for all x ∈ (c , b ), then f has arelative maximum at x = c .


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e at e a u at x c

2 If f (x ) < 0 for all x ∈ (a , c ) and f (x ) > 0 for all x ∈ (c , b ),


a c b x

sign of f (x ) – +





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2 If f (x ) < 0 for all x ∈ (a , c ) and f (x ) > 0 for all x ∈ (c , b ), then f has arelative minimum at x = c .


a c b x

sign of f (x ) – + f has a relativeminimum at x

=c





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3

If f

(x )

does not change sign from (a , c )

to (c , b )

, then there is norelative extremum at x = c .


a c b x

sign of f (x ) + + f has no relativeextremum at x

=c





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3 If f (x ) does not change sign from (a , c ) to (c , b ), then there is no

relative extremum at x = c .


a c b x

sign of f (x ) – – f has no relativeextremum at x

=c


Finding the Relative Extrema of f


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table of signs for f





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table of signs for f

First Derivative Test



Example

Find all relative extrema of f (x

)=

x 3

9 −3

x .


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Example

Find all relative extrema of f (x

)=

x 3

9 −3

x .

S l i


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Solution:



Example

Find all relative extrema of f (x )

=x 3

9 −3x

.

S l ti


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Solution:

f (x ) = x 2 − 9

3



Example


=x 3

9 −3x

.

Solutio


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Solution:

f (x ) = x 2 − 9

3CN : ±3



Example

Find all relative extrema of f (x )=

x 3

9 −3x .

Solution:


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Solution:

f (x ) = x 2 − 9

3CN : ±3

−3 3 f (x ) + − +



Example


x 3

9 −3x .

Solution:


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Solution:

f (x ) = x 2 − 9

3CN : ±3

−3 3 f (x ) + − +

CONCLUSIONS

f has a rel. max. at x = −3 OR (−3,6) is a rel. max. pt of f .



Example


x 3

9 −3x .

Solution:


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Solution:

f (x ) = x 2 − 9

3CN : ±3

−3 3 f (x ) + − +

CONCLUSIONS

f has a rel. max. at x = −3 OR (−3,6) is a rel. max. pt of f . f has a rel. min. at x = 3 OR (3,−6) is a rel. min. pt of f .



Example


x 4

+2x 3.


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Example


x 4

+2x 3.

Solution:


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Example


x 4

+2x 3.

Solution:


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f (x ) = 4x 3 + 6x



Example


x 4

+2x 3.

Solution:


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f (x ) = 4x 3 + 6x CN : 0,−32



Example


=x 4

+2x 3.

Solution:


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f (x ) = 4x 3 + 6x CN : 0,−32

−3

2 0

f (x ) − + +



Example


=x 4

+2x 3.

Solution:


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f (x ) = 4x 3 + 6x CN : 0,−32

−3

2

0

f (x ) − + +

CONCLUSIONS

f has a relative minimum at x = − 32

.

f has no relative extremum at x = 0.



Example

Find all relative extrema of f (x ) =x 2

9 − x 2 .


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Example


9 − x 2 .

Solution:


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Solution:



Example


9 − x 2 .

Solution:


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Solution:

f (x ) = 18x (9

−x 2)2



Example


9 − x 2 .

Solution:


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Solution:

f (x ) = 18x (9

−x 2)2

CN : 0



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Example


9 − x 2 .

Solution:


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Solution:

f (x ) = 18x (9

−x 2)2

CN : 0

−3 0 3 f (x ) −

− +

+

CONCLUSION

f has a relative minimum at x = 0.



Example


= −x 4/3

+4x 1/3.


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Example


= −x 4/3

+4x 1/3.

Solution:


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Example


= −x 4/3

+4x 1/3.

Solution:

f −4(x − 1)


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f (x ) = (x )3x 2/3



Example


= −x 4/3

+4x 1/3.

Solution:

f ( )−4(x − 1)

CN 0 1


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f (x ) = ( )3x 2/3

CN : 0, 1



Example


= −x 4/3

+4x 1/3.

Solution:

f ( )−4(x − 1)

CN 0 1


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f (x ) =3x 2/3

CN : 0, 1

0 1 f (x ) + + −



Example


= −x 4/3

+4x 1/3.

Solution:

f ( )−4(x − 1)

CN 0 1


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f (x ) =3x 2/3

CN : 0, 1

0 1 f (x ) + + −

CONCLUSIONS

f has a relative maximum at x

=1.

f has no relative extremum at x = 0.


Exercises

1 Use the Mean Value Theorem to show that

cosb − cosa ≤ b −a

f ll b h h b


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for all a , b ∈ R such that a < b .

2 Find the coordinates of the relative extremum point/s of f (x ) = x 43 −6x 13 .

3 Show using FDT that if a , b , c are constants and a = 0, then the graphof y

=ax 2

+bx

+c has a relative maximum at the vertex when a

<0

and a relative minimum at the vertex when a > 0.


Exercises: Solutions

Use MVT to show that if a , b ∈ R with a < b , then

cos b − cos a ≤ b − a


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Solution:


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Solution:Let f (x ) = cos x ,


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Solution:Let f (x ) = cos x , w/c is continuous and differentiable on R.


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Solution:Let f (x ) = cos x , w/c is continuous and differentiable on R.f is continuous on [a b] and differentiable on (a b) for any a b ∈ R with


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f is continuous on [a , b ] and differentiable on (a , b ) for any a , b ∈ R witha < b .





Solution:Let f (x ) = cos x , w/c is continuous and differentiable on R.f is continuous on [a b] and differentiable on (a b) for any a b ∈ R with


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By MVT,





Solution:Let f (x ) = cos x , w/c is continuous and differentiable on R.f is continuous on [a, b] and differentiable on (a, b) for any a, b ∈ R with


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By MVT,there is a c ∈ (a , b ) such that





Solution:Let f (x ) = cos x , w/c is continuous and differentiable on R.f is continuous on [a, b] and differentiable on (a, b) for any a, b ∈ R with


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By MVT,there is a c ∈ (a , b ) such thatcos b − cos a

b −a = f (c )






f is continuous on [a , b ] and differentiable on (a , b ) for any a , b ∈ R with


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f s co t uous o [a, b] a d d e e t ab e o (a, b) o a y a, b ta < b .


b −a = f (c ) = −sinc








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f [ , ] ( , ) y ,a < b .


b −a = f (c ) = −sinc ≤ 1








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f , , y ,a < b .


b −a = f (c ) = −sinc ≤ 1

Thus, cos b − cos a b −a ≤ 1








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f ya < b .


b −a = f (c ) = −sinc ≤ 1

Thus, cos b − cos a b −a ≤ 1 ⇒ cos b − cos a ≤ b − a



Find the coordinates of the relative extremum point/s of f (x ) = x 43 �

m53 lec3.1 mvt and relative extrema

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