Fundamental Engineering Maths 4

8/9/2019 Fundamental Engineering Maths 4

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Chapter 4: APPLICATIONS of DIFFERENTIATION

Duong T. PHAM - EEIT2014

Fundamental Engineering Mathematics for EEIT2014

Vietnamese German UniversityBinh Duong Campus

Duong T. PHAM - EEIT2014 October 22, 2014 1 / 60


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Outline

1 Maximum and minimum values

2 Derivative and shape of a graph

3 Indeterminate forms and LHopital Rule

4 Optimization problems

5

Newtons method

6 Antiderivatives



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Applications in Optimization

Important optimization problems require differential calculus, e.g.

What is the shape of a can that minimizes manufacturing costs?

What is the maximum acceleration of a space shuttle?

What is the interest rate that the banks earn most in the market?

what is the shape of a racing car or a aircraft to minimize drag?

etc.

= finding the minimum and maximum of a function.



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UMaximum and minimum of a function

Def: Let f :D R be a function and let c D.f has an absolute maximum(global maximum) at c if

f(x) f(c) x D.

Then, f(c)is called the maximum value of f.f has an absolute minimum (global minimum) at c if

f(x) f(c) x D.

Then, f(c)is called the minimum value of f.

The maximum and minimum values of fare called the extremevaluesof f.



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UMaximum and minimum of a function

x

y

y = f(x)

a b c d e

f(a)

f(d)

c

f(c)

c1 c2

f(a)is the absolute minimum of f;

f(d)is the absolute maximum of f

f(c)

f(x)

x

(c1, c2)=

f(c)is a local minimum of f



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ULocal maximum and minimum

Def: Let f :D R be a function and let c D.f has an local maximum (relative maximum) at c if there exists aninterval (c1, c2) D such that c (c1, c2) and

f(x)

f(c)

x

(c1, c

2).

f has an local minimum (relative minimum) at c if if there existsan interval (c1, c2) D such that c (c1, c2) and

f(x)

f(c)

x

(c1, c2).



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ULocal maximum and local minimum

x

y

y = f(x)

a b c d e

f(a)

f(d)

c

f(c)

c1 c2

f(c)is a local minimum

f(b)is the local maximum of f;

f(d)is the local maximum of f (also an absolute maximum)

f(e)is a local minimum of f



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UThe Extreme value Theorem

Theorem: Let f :[a, b] R be a continuous function on[a, b]. Then fattains an absolute maximum value f(c)and an absolute minimum valuef(d)at some numbers c and d in[a, b].

Remark: The theorem is not true if we replace the closed interval [a, b]by the open interval (a, b) or other interval (a, b] or [a, b).

Ex: The function f(x) = 1x

does not attain an absolute maximum value

on (0, 5].



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UThe Extreme value Theorem

y

x

y= 1

x

1

2

34

5

6

7

8

9

10

11

12

13

14

15

16

17

1819

1 2 3 4



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UThe Fermats Theorem

x

y

y = f(x)

(c, f(c)

c d

(d, f(d)

fattains local maximum at cand local minimum at d = What is specialabout f at x=c and x=d?

The Fermats Theorem: Iffattains local maximum or local minimum at candif f(c)exists, then f(c) = 0



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UProof of Fermats Theorem

The Fermats Theorem: Iffattains local maximum or local minimum at candif f

(c)exists, then f

(c) = 0

Proof: Suppose fattains local maximum at c.

=x (c1, c2) : f(x) f(c)

= f(x)

f(c)

x c 0 c1


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The Fermats Theorem

Remark: Local maximum and local minimum at c + existence off (c)=

f (c) = 0. But(

=)is NOT true.

Ex: Let f(x) =x3. We have f (x) = 3x2 and f (0) = 0 but f does NOTattain local max. or local min. at 0.

x

y



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Local minimum and local maximum

Ex: Function f(x) = |x| attains local minimum at x= 0 but it is NOTdifferentiable at x= 0.

x

y

0



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Critical numbers of a function

Corollary: If fhas a local minimum or local maximum at c then c is a

critical number of f

Proof: Fermats Theorem:

f (c) = 0

f

(c) does NOT exist = c is a critical number of f

fhas local min. or local max. at c = There are 2 cases:

f (c) does not exist=

c is a critical number

f (c) existsFermats Th.

= f (c) = 0= c is a critical number



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Finding absolute minimum and absolute maximum

The closed interval method: Let f : [a, b] R be a continuous function. Tofind the absolute max. and absolute min. of f, we follows the steps:

1 Find the values of fat critical numbers of f in (a, b);

2 Find the values f(a)and f(b);

3 The largest number in steps 1 and 2 is the absolute maximum value of f,and the smallest number in steps 1 and 2 is the absolute minimum of f.

Ex: Find abs. max. and abs. min. of f(x) = 2x3 9x2 + 12x+ 2 in[0, 3]Ans: f(x) = 6x2 18x+ 12 = 6(x2 3x+ 2)

1 f(x) = 0 x= 1 or x= 2, and f(1) = 7, f(2) = 62 f(0) = 2and f(3) = 113 Comparing the 4 values in Steps 1 and 2, we conclude

max[0,3]

f =f(3) = 11 and min[0,3]

f =f(0) = 2


Th l d i l h d


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The closed interval method

x

yy = 2x3 9x2 + 12x+ 5

0

2

7

1

6

2

11

3


E i


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Exercises

4.1:

310, 1520, 2728, 2936, 4755


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R ll Th


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Rolles Theorem

Rolles Theorem: Let f : [a, b] R be a function satisfying1 f is continuous in [a, b]

2 f is differentiable in (a, b)

3 f(a) =f(b). Then there exists a c (a, b)such that f(c) = 0

Proof:

The case: f(x) =cx [a, b]. then clearly f(c) = 0 c (a, b).The case: x (a, b) s.t. f(x)> f(a) =f(b).Since f is continuous in [a, b]

ExtremeValueTh.= c (a, b) such that

f(c) = max[a,b]

f

= f has a local max. at c + f(c) exists (f is differentiable in (a, b))Fermats Th.

= f(c) = 0The case: x (a, b) s.t. f(x)< f(a) =f(b). (Similar as the secondcase)


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R ll Th


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Rolles Theorem

Ex: Prove that equation x3 + x2 + 3x+ 1 = 0 has exactly one real root.

Ans: Denote f(x) =x3 +x2 + 3x+ 1.

f is a polynomial= fis differentiable (and thus continuous) in (,).f(1) = 2 and f(0) = 1. Since f is continuous in [1, 0] and sincef(

1)< 0 < f(0), the Intermediate value Theorem=

c

(

1, 0) s.t.

f(c) = 0= the equation has one root c (1, 0)Suppose that d=cis another root of the equation.

Ifc 0 for all x R= ContradictionIfc>d, similar argument= Contradiction

The equation has exactly one root c (1, 0).


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R ll Th


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Rolle s Theorem

x

y

y = x3 +x2 + 3x+ 1

0

1

c1

2


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Recall the Rolles Theorem


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Recall the Rolle s Theorem

x

y

(a, f(a)) (b, f(b))L

t(c, f(c))

(a, f(a))

(b, f(b))

(c, f(c))t

The slope of secant connecting (a, f(a)) and (b, f(b)) is f(b)

f(a)

b a = 0Rolles Theorem:c (a, b) s.t. f(c) = 0

= f(c) = f(b) f(a)b

a

t//L


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The Mean Value Theorem


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The Mean Value Theorem: Let f : [a, b] R be a function satisfying1 f is continuous in [a, b]

2 f is differentiable in (a, b)

Then there exists a c (a, b)such that

f(c) = f(b) f(a)

b aroof: Denote h(x) =f(x) f(b)f(a)

ba (x a) f(a). Then:h is continuous in [a, b] and h is differentiable in (a, b)

h(a)= f(a)

f(b)f(a)

b

a

(a

a)

f(a)= 0

h(b)= f(a) f(b)f(a)ba (b a) f(a)= 0

= h(a) =h(b)Rolles Theorem: there is a c (a, b)such that h(c) = 0=

f(c) = f(b) =f(a)

b aDuong T. PHAM - EEIT2014 October 22, 2014 23 / 60

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Theorem: Letf : (a, b) R be a function which is differentiable in (a, b)and f

(x) = 0 for all x (a, b). Then f is a constant function.Ans: Let x1, x2 (a, b]. Then f is continuous in [x1, x2] and differentiablein (x1, x2). By the Mean Value Theorem, there is a c (x1, x2) such that

f (c) = f

(x

2) f(x

1)x2 x1 .

Since f (x) = 0 for all x (x1, x2), we havef(x2)

f(x1)

x2 x1 = 0.It means that f(x1) =f(x2) and this is true for any x1, x2 (a, b). Hence,f is a constant function in (a, b).


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Exercises


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Exercises

4.2:

15, 7, 1114, 151720, 2231, 36


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Derivative and shape of a graph


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Derivative and shape of a graph

Increasing and decreasing Test: Let f : (a, b) R be a differentiablefunction.

1 If f (x)>0,x (a, b), then f is increasing in (a, b).2 If f (x)


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Increasing and decreasing Test

Ex: Determine when the functionf(x) = 3x4

4x3

12x2 + 5is increas-

ing and decreasing.

Ans: f (x) = 12x3 12x2 24x= 12x(x 2)(x+ 1)= f (x) = 0 x= 1 x= 0 x= 2

x 1 0 2 x 2 | | 0 +x | 0 + | +

x+ 1 0 + | + | +f (x) 0 + 0 0 +f(x) 0 5 27


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x

y

12

y = 3x4

4x3

12x2

+ 5

0


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The first derivative Test


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The First derivative Test: Suppose that c is a critical number of a continuousfunction f.

(a) If f changes from positive to negative at , then has a local maximum at c.

(b) Iff changes from negative to positive at c, then has a local minimum at c.

(c) If f does not change sign at c(for example, if f is positive on both sidesofcor negative on both sides), then fhas no local maximum or minimum

at c.

y

xc

f < 0 f > 0

local minimum

y

xc

f > 0 f < 0

local maximum

y

xc

f > 0

f > 0

no local min. or max.

y

xc

f < 0

f < 0

no local min. or max.


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Ex: Find the local minimum and maximum values of the function f(x) =3x4

4x3

12x2 + 5

Ans: f (x) = 12x3 12x2 24x= 12x(x 2)(x+ 1)= f (x) = 0 x= 1 x= 0 x= 2

x 1 0 2 x 2 | | 0 +x | 0 + | +

x+ 1 0 + | + | +f (x)

0 + 0

0 +

f(x) 0 5 27

fattains local minimum at1 and 2; and attains local maximum at0.


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x

y

12

y = 3x

4

4x

3

12x

2

+ 5

0


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Concave functions


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Def: Let f :I R.If the graph of f lies above all of its tangent lines, then f is said to

be upward concave on IIf the graph of f lies below all of its tangent lines, then f is said tobe downward concave on I

y

x

upwad concave

y

x

downward concave


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Concavity Test


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y

Concavity Test: Let f :I R. Then(a) If f(x)> 0 for all x

I, then the graph of f is concave upward on I.

(b) If f(x)< 0 for all x I, then the graph of f is concave downward on I.

Discussion:

f(x)> 0 for all x I = f(x) is increasing on If(x


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The second derivative Test


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The second derivative Test: Let f be a function such that f is continuousnear c. Then

(a) If f(c) = 0and f(c)> 0, then fhas a local minimum at c,(b) If f(c) = 0and f(c)< 0, then fhas a local maximum at c,


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Ex: Discuss the concavity, inflection points, local maxima and local minima ofthe curve y=x4 4x3.

Ans: Denote f(x) =x4 4x3. Thenf(x) = 4x3 12x2 = 4x2(x 3)f(x) = 12x2 24x= 12x(x 2)

f(x

) = 0x

= 0 x

= 3 and f(x

) = 0x

= 0 x

= 2

x 0 2 3x2 + 0 + | + | +

x 3 | | 0 +x

2

| 0 +

| +

f(x) 0 | 0 +f(x) + 0 0 + | +f(x) 0 16 27

up.con. infl.p. down.con. infl.p. up.con. local min. up.con.


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x

y

2

inflection point16

3

local min.27

0inflection point


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Exercises


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4.3:

12, 56, 912, 1921, 3340


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Indeterminate forms


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-VGUIndeterminate forms: consider limx

a

f(x)

g(x). Then

(a) If f(x) 0and g(x) 0when x a, then the limit is called anindeterminate form of type 00

(b) If f(x) (or) and g(x) (or) when x a, then the limitis called an indeterminate form of type

Ex: Evaluate limx1

x3 xx 1

Ans: limx1

x3 xx

1

= limx1

x(x2 1)x

1

= limx1

x(x 1)(x+ 1)x

1

= limx1

[x(x+ 1)] = 1(1 + 1) = 2.

Ex: limx1

ln x

x 1 = we need a tool to evaluate this limit.


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LHopitals Rule


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except possibly at number x=a (b, c). Suppose further that g(x) = 0for allx (b, c)\{a}. Iflimxa

f(x) = limxa

g(x) = 0

or limxa

f(x) = and limxa

g(x) = ,then

limxa

f(x)

g(x) = lim

xaf(x)

g(x)

provided the limit on the right hand side exists (or isor)

Ex: limx1

ln x

x 1LHopital.

= limx1

(ln x)

(x 1) = limx1

1

x1

= 1.


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LHopitals Rule


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Ex: limx

ex

x2LHopital.

= limx

(ex)

(x2)= lim

xex

2x

LHopital.= lim

x(ex)

(2x)= lim

xex

2 = .

Ex: limx0

sin x

x

LHopital.= lim

x0(sin x)

(x)= lim

x0cos x

1 = 1.


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Indeterminate products


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Intermediate products: The limit limxa[f(x)g(x)] in which f(x) 0 andg(x) (or) as x a is called an indeterminate form of type 0

Ex: Evaluate limx0+

xln x

Ans: We have

limx0+

xln x= limx0+

ln x1x

LHopital.= lim

x0+(ln x)

1x

= lim

x0+

1x

1x2

= limx0+

(x) = 0.



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Indeterminate powers


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-VGUThe limit limxa[f(x)]g(x) when f and g satisfy

(a) limxa

f(x) = limxa

g(x) = 0 type 00

(b) limxa

f(x) = and limxa

g(x) = 0 type0

(c) limx

af(x) = 1 and lim

x

ag(x) = type 1 are indeterminate powers.

Remark: To evaluate the above limits we can either

1 write y= [f(x)]g(x) and thus ln y=g(x) ln f(x), and then evaluate limxa

ln y

first; after that, we deduce limxa

y.

2 We can write [f(x)]g(x) =eg(x) ln f(x).

3 By these ways, we transfer it to the problem of finding limit of the type0 .


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Indeterminate powers


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-VGUEx: Evaluate lim

x0+

(1 + 4 sin 4x)cot x

Ans: Denote y= (1 + 4 sin 4x)cot x = ln y= cot xln(1 + 4 sin 4x)Then

limx0

+ln y= lim

x0

+cot x ln(1 + 4 sin 4x) = lim

x0

+

ln(1 + 4 sin 4x)

tan xLHopital.

= limx0+

16cos4x1+4 sin 4x

1cos2 x

= limx0+

16cos4x cos2 x1 + 4 sin 4x

= 16

Hence, limx0+

y= limx0+

eln y =elimx0+ ln y =e16.


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Exercises


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4.4:

Each student should do a third of exercises from 564.

6970,

7576



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Optimization problems


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Find max(xy)?2x+y= 50

Ans: We have A= xyand since 2x+y= 50, we replace y= 50 2x in A toobtain

A(x) =x(50 2x).Our task is now to find: max A(x), where 0 x 25. We have A = 50 4x,and

A= 0 50 4x= 0 x= 15.x 0 15 25A 50 + 0 50A 0 300 0

We can deduce from the table that

maxA= 300m2 when x= 15m


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Optimization problems


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20m

15m

15m

20m

Area = 300m2


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Applications in business and economics


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-VGUThe cost function C(x)= the cost of producing xunits of a certainproduct.

The marginal cost (=C(x)) is the change ofC(x)w.r.t. x

The demand function (or price function), denoted by p(x), is the priceper unit that the company can charge if it sells x units

If the company sells xunits and the price per unit is p(x), then the total

revenue is denoted by the revenue function,

R(x) =xp(x)

The derivative R is the marginal revenue function

Ifxunits are sold, the total profit

P(x) =R(x) C(x)and P is called the profit function.

P is called the marginal profit function


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Ex: A store sells 200 DVD burners a week at $350 each.

Survey: if each $10 rebate is offered, 20 more units will be sold every week.

Q.: Find demand and revenue functions. How large a rebate should be tomaximize its revenue?

Ans: Denote by xthe number of units sold every week= weekly increase insales is x

200.

If the price per unit decreases by $10, more 20 units are sold.

= the price per unit so that the weekly sale is xunits, is

p(x)= 350 x 20020

10 =450 x2.

The revenue function is R(x)= xp(x) =450xx2

2

Our task: Find the absolute maximum ofR(x) = 450xx2

2


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M-VGUOur task: Find the absolute maximum ofR(x) = 450x

x2

2R(x) = 450 x, and R(x) = 0 x= 450.consider the table

x 350 450

R(x) 100 + 0 R(x) 96250 101250

The revenue has an absolute maximum (101250) when the number ofweekly sold units is 450.

The price per unit is then p(450) = 450 4502 = 225The rebate should then be offered as350 225 = 125


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Exercises


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M-VGU

4.7:

26, 11, 1719, 35


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Newtons method


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M-VGU

Some examples:Solve 2x+ 5 = 0 x= 52Solve x2 5x+ 4 = 0.

= (5)2 4 1 4 = 9The solutions

x1 = 5

2 =

x2 = 5+

2

x1 = 5

9

2

x2 = 5+

9

2

x1 = 1

x2 = 4

Solve cos x x= 0?


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Newtons method


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M-VGU

x

y

y = f(x)

r

x1

f(x1)

t1

t1 : y = f (x1)(x x1) +f(x1)

x2

f(x2)

t2

t2 : y = f (x2)(x x2) +f(x2)

x3

Step 1: Choose some x1

x2 = x1 f(x1)

f

(x1)

x3 = x2 f(x2)

f (x2)

Step 2: xn = xn1 f(xn1)

f (xn1)


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Newtons method

E Fi d i d i l l h f h i


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D.T.

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Ex: Find, correct to six decimal places, the root of the equation cos x=x

Ans: The equation is equivalent tocos x

x= 0.

Denote f(x) = cos x x. Then f(x) = sin x 1Newtons method:

xn+1 =xn f(xn)f(xn)

=xn+cos xn xn

sin xn+ 1

Choose x1 = 0. Then

x2 = 0 +cos0 0sin 0 + 1

= 1; x3 = 1 +cos1 1sin 1 + 1

= 0.750363868

x4 = 0.737151911; x5 = 0.739446670

x6 = 0.739018516; x7 = 0.739097442

x8 = 0.739082860; x9 = 0.739085553

x10 = 0.739085055; x11 = 0.739085147


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Exercises


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GU

4.8:

58, 11, 12, 1722


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Antiderivatives


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GUDef: A function F is called an antiderivative of f on an interval I ifF

(x) =f(x) for all x in I.

Ex: F(x) = x3

3 is an antiderivative of f(x) =x2 for any x R.

Indeed, F(x)= x33

=x2 =f(x)

Theorem: IfF is an antiderivative of f on an interval I, then the mostgeneral antiderivative of fon the interval I is

F(x) +C,

where C is an arbitrary constant.



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Exercises


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4.9:

120, 2328, 48


Documents

Fundamental Engineering Maths 4