Bab v Aplication of Derifative

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5. APPLICATION OF5. APPLICATION OFDIFFERENTIATIONDIFFERENTIATION



2

5.1 Curve Sketching

Information which is needed :

A. The x-intercepts and the y-intercepts

B. AsymptotesDefinition: A straigth line that the curve approaches.Three kind of asymptotes

(i) Vertical asymptoteA line x = c is called a vertical asymptote if

(ii) Horizontal asymptoteA line y = b is called a horizontal asymptote if

(iii) Oblique AsymptoteA line y = ax + b is called oblique asymptote if

±∞=→

)(lim x f c x

b x f x

=±∞→

)(lim

a x

x f

x=

±∞→

)(lim lim ( )

xand f x ax b

→ ±∞

− =



3

x=a is a vertical asymptote

a

∞=−→

)(lim x f a x

∞=+

→

)(lim x f a x

because

and

because

−∞=−→

)(lim x f a x

∞=+

→

)(lim x f a x

and

a

Vertical asymptote

x=a is a vertical asymptote



4

y= b

The line y = b is a horizontal asymptote because

Remark :

It is possible for the graph of a function to intersect a horizontalasymptote, but it cannot intersect a vertical asymptote

b x f x

=+∞→

)(lim



5

bax y +=

y=f(x)

The line y = ax + b is asymptote miring

It is possible for the graph of a function to intersect a oblique

asymptote.



6

Example Find Asymptotes of

Solutions:

(i) Vertical Asymptote : x = 2, because

and

(ii) Horizontal Asymptote :

∞=−

+−

+→ 2

42lim

2

2 x

x x

x

horizontal Asymptote doesn’t exist

2

42)(

2

−

+−=

x

x x x f

−∞=−

+−

−→ 2

42lim

2

2 x

x x

x

)(

)1(lim

2

42lim)(lim

2

2

212

4222

x x

x x

x x x x

x

x

x x x f

−

+−=

−

+−=

∞→∞→∞→

∞=−

+−=

∞→ )()1(lim

2

2

21

42

x x

x x

x



7

x x

x x

x

x f a

x x

1.

2

42lim

)(lim

2

−

+−==

±∞→±∞→ x x

x x

x 2

42lim

2

2

−

+−=

±∞→

1)1(

)1(lim

)1(

)1(lim

2

42

22

42222

=−

+−=

−

+−=

±∞→±∞→ x

x x

x x

x x

x x

x

(iii) Asymptote miring

02

4lim =−

=±∞→ x x

2

)2(42lim

2

−

−−+−=

±∞→ x

x x x x

x

x x

x x

x −−

+−

=±∞→ 2

42lim

2

ax x f b x −= ±∞→ )(lim

Oblique asymptote y = x

2

242lim

22

−

+−+−=

±∞→ x

x x x x

x



8

1

1)(

−

=

x

x f

3

1)(

−+=

x x x f

1

2

)( 2

2

−

+

= x

x x

x f

3

2)(

−=

x

x x f

Find asymptotes of

Problems

1

2)(

2

2

+

+=

x

x x x f

1.

2.

3.

4.

5.



9

C. Intervals of Increase and Decrease

Definition Function f (x ) is called

increasing on the interval I if

( ) ( )1 2 1 2 f x f x whenever x x I < < ∈

x1

f(x1)

x2

f(x2)

I

Function f(x) is increasing on interval I



10

Function f is decreasing on interval I

f(x1)

f(x2)

x1 x2

Decreasing on the interval I if

I

( ) ( )1 2 1 2 f x f x whenever x x I > < ∈



11

Theorem : Let f is differentiable on interval I, then

f (x ) is increasing on the interval I if

f(x ) is decreasing on the interval I if

Example Find the interval on which f(x) is increasing and decreasing, if

Solution :

f (x ) is increasing on the interval

f (x ) is decreasing on the interval (0,2) and (2,4).

I x x f ∈∀> 0)('

I x x f ∈∀< 0)('

2

42)(

2

−

+−=

x

x x x f

( , 0), (4, )and −∞ +∞

2

2

)2(

)42(1)2)(22()('

−

+−−−−=

x

x x x x x f 2

22

)2(

42462

−

−+−+−=

x

x x x x

22

2

)2(

)4(

)2(

4

−

−=

−

−=

x

x x

x

x x

0 2 4

++++++---------------------+++++++



12

D. Maximum and Minimum Values of A Function

Definition Let function f (x ) is continuous on interval Iand c I ,

f (c ) is called absolute value of f if

f (c ) is called local value of f if for x

sufficiently near c

Maximum and Minimum values of f is called Extreme Value

maxmin

imumimum

I x x f c f x f c f ∈∀

≤≥

)()()()(

)()(

)()(

x f c f

x f c f

≤

≥

Extreme value occurs at a critical point of f

∈

max

min

imum

imum



13

Maxlocal

Min

local

Maxglobal

Minglobal

Maxlocal

Minlocal

a b c d e f

Extreme values of f on interval I=[a,f]



14

Critical points :The end of interval I

Stationary point

x = c at which

geometrically : Tangent line is horizontal atpoint (c ,f (c ))

Singular point

x =

c at which does not exist

0)(' =c f

)(' c f



15

Theorem : First derivative test for local extreme

If 0)('

0)('

<

>

x f

x f ),( cc δ −

0)('

0)('

>

<

x f

x f on and on ),( δ +cc

then f(c) is a local value

c

On the left of c f is increasing(f ’>0) and on the right of c f isdecreasing (f’<0)

f(c) is a local max value

c

f(c) is a local min value

f(c)

f(c)

maxmin

imumimum

On the left of c f is decreasing(f ’<0) and on the right of c f isincreasing (f’>0)



16

Thorem Second derivative test for extreme value

Let . If , then f (c ) is local

value of f

Example: Find extreme value of

Solution :

0)(' =c f 0)(''

0)(''

>

<

c f

c f

2

42)(

2

−

+−=

x

x x x f

2)0( −= f

6)4( = f

2)2(

)4()(' −

−

= x

x x x f

0 2 4

++++++---------------------+++++++

By First derivative test :

There is local maximum at x = 0 with value

There is local minimum at x = 4 with value

max

min

imum

imum



17

Problems

630152)(

345−+−=

x x x x f

3

13)(

2

−

+−=

x

x x x f

212)(

2

−

+−= x

x x x f

x

x x f

2)1(

)(+

=

Find the interval on which the following function is increasingand decreasing

1.

2.

3.

4.



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E. Concavity

Definition Let f be differentiable on interval Ia. f is called concave up on interval I if is increasing on interval I

b. f is called concave down on interval I if is decreasing on interval I

Theorem Second derivative test for concavity1. If , then f is concave up on I.

2. Jika , then f is concave down on I .

)(' x f

)(' x f

I x x f ∈∀> ,0)("

I x x f ∈∀< ,0)("

Concave up Concave down

x

y

x

y



19

2

42)(

2

−

+−=

x

x x x f

Find interval on whichis concave up and concave downExampleSolution:

2

2

)2(

4)('

−

−=

x

x x x f

4

22

)2(

)4)(2(2)2)(42()(''

−

−−−−−=

x

x x x x x x f

4

2

)2())4(2)2)(42)((2(

−

−−−−−= x

x x x x x

3

22

)2(

82882

−

+−+−=

x

x x x x3

)2(

8

−

=

x

f is concave up on ),2( ∞

f is concave down on )2,(−∞



20

F. Inflection Point

Definition If f is continuous on an open interval containingx = b, and if f changes the direction of its concavity at

x = b, then point (b,f(b)) on the graph of f is called an

inflection point off .

x = b is an absist of inflection point if or

does not exist

f b"( ) = 0 )(" b f



21

c

f(c)

(c,f(c)) is an inflection point

c

f(c)

(c,f(c)) is an inflection point



22

c

f(c)

(c,f(c)) is not aninflection point

c

f is not continuous at c ,there is not an inflectionpoint



23

12)(.13

−= x x f

4)(.2 x x f =

Find inflection point of the following function

2

6)(' x x f = x x f 12)('',=

●

0

+++++++-------------

At x = 0 f changes the direction of its concavity andf is continuous at x = 0 then (0,f(0))=(0,-1) is an inflection point

212)(''

x x f =

●

0++++++++++++++

Because f does not changes the direction of its concavity theninflection point does not exist



24

242)(.3

2

−

+−=

x x x x f

3

)2(

8)(''

−=

x

x f

●

2

+++++++--------------

Although at x = 2 f changes the direction of its concavity

but f is not continuous at x = 2, then inflection pointdoes not exist



25

Problems

630152)(

345−+−=

x x x x f

3

13)(

2

−

+−=

x

x x x f

212)(

2

−+−=

x x x x f

x

x x f

2)1(

)(+

=

Find the interval concavity and inflection point of

1.

2.

3.

4.

3 / 1)( x x f =5.



26

2

42)(

2

−

+−=

x

x x x f Example: Let

a. Find interval on which f is increasing and decreasing, andextreme value of f

b. Find interval concavity of f and inflection point

c. Find asymptotes of f d. Sketch graph of f(x)

a. f is increasing on ),4(,)0,( +∞−∞

f is decreasing on (0,2) and (2,4).

f(0) = -2 is a local maximum valuef(4) = 6 is a local minimum value

b. f is concave up on ),2( ∞ and f is concave down on )2,(−∞

inflection point does not exist

c. Vertical asymptote x = 2, oblique asymptote y = x, horizontal asymptotedoes not exist



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d. The graph of f(x)

2

y=x

0 2 4++++++----------++++++ ' f

2--------------------- +++++++++++ '' f

-24

6



28

21

2)(

x

x x f

+=

x x x f

1)( +=

134

)(2

34

+−−= x x x

x f

1)(

+=

x

x x f

4)(

2

2

−=

x

x x f

A. Sketch the graph of given function

Problems

1.

2.

3.

4.

5.



29

)(' x f y =

B. Let f is a continuous function and f(-3)=f(0)=2. If graph of

is

a. Find interval on which f is increasing and decreasingb. Find interval concavity of f

c. Sketch graph of function f (x).

j



37

5.5 Applied Maximum and Minimum Problems

Application of differentiation to solve applied optimization problems.Steps :

1. Find a formula for the quantity to be maximized or minimized

2. Express the quantity to be maximized or minimized as a functionof one variable

3. Find the maximum or minimum value of this function



38

Example

Let we have a 100 cm of wire . What dimensionsof a rectangle which is built by this wire in orderarea maximumsolution

Let y = length of rectangle , x = width of rectangley

x

Area A = x y, Because 2x + 2y = 100 y = 50 - x

Thus A = L(x) = x(50-x) ,502

x x −= 500 ≤≤ x

x x L 250)(' −= x = 25

02)25('' <−= Lbecause then at x = 25 there is max value

L(0) = 0, L(25) = 625, L(50) = 0 In order area maximum thenx = 25 and y = 25



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Problems

1. Express the number 10 as a sum of two nonnegative numberswhose product is as large as possible

2. A rectangular field is to be bounded by a fence on three sidesand a straight stream on the fourth side. Find the dimensions of thewith maximum area that can be enclosed with 1000 feet of fence

3. Find the dimensions of the rectangle with maximum area that can beinscribed in a circle of radius 10 cm

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Bab v Aplication of Derifative