Cos x

DU? c. 5 - riesenie

1.

Vysetrite

priebeh

funkcie

f (x)

=

cos x cos 2x

a

nakreslite

jej

graf.

D(f )

=

R

\

{

4

+

k 2

,

k

Z}

Funkcia je periodicka? s peri?odou 2, preto stac?i jej priebeh vysetrit' na intervale

4

,

9 4

.

Py = [0; 1]

Px1

=

[

2

;

0]

Px2

=

[

3 2

;

0]

lim

x

4

+

cos x cos 2x

=

-

lim

x

3 4

-

cos x cos 2x

=

lim

x

3 4

+

cos x cos 2x

=

-

lim

x

5 4

-

cos x cos 2x

=

-

lim

x

5 4

+

cos x cos 2x

=

lim

x

5 4

-

cos x cos 2x

=

-

lim

x

5 4

+

cos x cos 2x

=

lim

x

5 4

-

cos x cos 2x

=

f

(x)

=

sin x(1+2 cos2 x)

cos2 2x

x

4

;

3 4

f (x)

+

x

3 4

;

+

x

;

5 4

-

x

5 4

;

7 4

-

x

7 4

;

2

-

x

2;

9 4

+

loka?lne maximum: [; -1]

loka?lne minimum: [2; 1]

f

(

2

)

=

-1

dotycnica

ku

grafu

funkcie

v

bode

[

2

,

0]

zviera

s

kladny?m

smerom

osi

x

uhol 135

f

(

3 2

)

=

-1

dotycnica

ku

grafu

funkcie

v

bode

[

3 2

,

0]

zviera

s

kladny?m

smerom

osi

x

uhol 135

f

(x)

=

( ) cos x 4 sin4 x+7 sin2 x+cos2 x+12 sin2 x cos2 x

cos3(2x)

x

4

;

2

f (x)

-

x

2

;

3 4

+

x

3 4

;

5 4

-

x

5 4

;

3 2

+

x

3 2

;

7 4

-

x

7 4

;

9 4

+

inflexn?e body:

2

;

0

,

3 2

;

0

1

Obra?zek

1:

f (x)

=

cos x cos 2x

2. Vysetrite priebeh funkcie f (x) = (x - 2 arctg (x - 5)) ? sgn x a nakreslite jej graf.

2 arctg (x - 5) - x pre x < 0,

f (x) = 0

pre x = 0,

x - 2 arctg (x - 5) pre x > 0.

D(f ) = R

Px = [a; 0], kde a -2,9

Py = [0; 0]

lim f (x) = lim (2 arctg (x - 5) - x) = 2 arctg(-5) -2,7

x0-

x0-

lim f (x) = lim x - 2 arctg (x - 5) = -2 arctg(-5) 2,7

x0+

x0+

k

=

lim

x

f (x) x

=

lim

x

1

-

2 arctg(x-5) x

=1

q = lim (f (x) - kx) = lim (-2 arctg(x - 5)) = -

x

x

k

=

lim

x-

f (x) x

=

lim

x-

2 arctg(x-5) x

-

1

= -1

q = lim (f (x) - k x) = lim (2 arctg(x - 5)) = -

x-

x-

asymptoty: y = x - , y = -x -

f (x) =

- x2-10x+24 x2-10x+26 x2-10x+24

x2-10x+26

pre x < 0, pre x > 0.

2

Obra?zek 2: f (x) = (x - 2 arctg (x - 5)) ? sgn x

lim

x0+

x2-10x+24 x2-10x+26

=

12 13

= arctg

12 13

42,7 (vid' graf)

lim

x0-

-

x2-10x+24 x2-10x+26

=

-

12 13

132,7

x (-; 0) x (0; 4) x (4; 6) x (6; )

f (x)

-

+

-

+

loka?lne

maximum:

[4;

4

+

2

]

loka?lne

minimum:

[6;

6

-

2

]

f (x) =

- 4x-20 (x2-10x+26)2

4x-20 (x2-10x+26)2

pre x < 0, pre x > 0.

x (-; 0) x (0; 5) x (5; )

f (x)

+

-

+

inflexny? bod: [5; 5]

3. Pomocou Taylorovho polyn?omu vypoc?itajte pribliznu? hodnotu 3 30 s chybou mensou ako

10-5.

Zostroj?ime

Taylorov

polyno?m

funkcie

f (x)

=

3x

v

bode

x0

=

27.

Potom

3 30

=

f (30)

T3(30).

Plat?i,

ze

f (n+1)(x)

=

(-1)n

?

x . 2?5?...?(3n-1) 3n+1

-

3n+2 3

Preto

2 ? 5 ? . . . ? (3n - 1) ? 3n+1 2 ? 5 ? . . . ? (3n - 1)

|Rn+1(30)| =

3n+1

3n+2 3

(n

+

1)!

=

3n+2 3

(n

+

1)!

,

3

kde 27 < < 30. Plat?i

2 ? 5 ? . . . ? (3n - 1) 2 ? 5 ? . . . ? (3n - 1) 2 ? 5 ? . . . ? (3n - 1)

|Rn+1(30)| =

3n+2 3

(n

+

1)!

<

27

3n+2 3

(n

+

1)!

<

33n+2(n + 1)! .

Hl'ada?me

tak?e n, ze

2?5?...?(3n-1) 33n+2(n+1)!

<

10-5.

To

plat?i pre

n

=

4,

preto

mus?ime

zostrojit'

Taylorov

polyno?m stvrt?eho stupna:

T4(x)

=

3

+

1 27

(x

-

27)

-

1 2187

(x

-

27)2

+

5 531441

(x

-

27)3

-

10 43046721

(x

-

27)4

.

Potom plat?i

3 30

T4(30)

=

3

+

1 9

-

1 243

+

5 19683

-

10 .

531441

4. Pomocou Taylorovho polyno?mu vypoc?itajte pribliznu? hodnotu e s chybou mensou ako 10-5.

Zostroj?ime Taylorov polyno?m funkcie f (x) = ex v bode x0 = 0. Potom e = f (1) T (1).

Plat?i,

ze

f (n+1)(x)

=

ex.

Preto

Rn+1(1)

=

e (n+1)!

,

kde

0

<

<

1.

Plat?i

e

e

3

|Rn+1(1)|

=

(n

+ 1)!

<

(n

+

1)!

<

(n

+

. 1)!

Hl'ada?me

tak?e

n,

ze

3 (n+1)!

<

10-5.

To

plat?i pre

n

=

8,

preto

zostroj?ime

Taylorov

polyn?om

o^smeho stupna:

x2 x3 x4 x5 x6

x7

x8

T8(x) = 1 + x +

2

+

6

++ + + +

.

24 120 720 5040 40320

Potom plat?i

11 1 1 1 1

1

e

T8(1)

=

2

+

2

+

6

+

24

+

120

+

720

+

5040

+

. 40320

5.

1

1

lim

-

= lim ex - 1 - sin x L='H lim

ex - cos x

L='H

x0 sin x ex - 1 x0 (ex - 1) sin x x0 (ex - 1) cos x + ex sin x

ex + sin x

1

= lim

=

x0 sin x + 2ex cos x 2

6.

1

lim (cos x) x2

ln cos x

= lim e x2

= e lim x0

ln cos x x2

(=1)

e-

1 2

=

1

x0

x0

e

ln cos x lim

L='H

lim

- sin x

= lim

-

1

sin x 1 =-

(1)

x0 x2

x0 2x cos x x0 2 cos x x

2

4

7. Pomocou L'Hospitalovho pravidla:

1 + tg x - 1 + sin x 1 + tg x + 1 + sin x

lim

x0

x3

?

=

1 + tg x + 1 + sin x

tg x - sin x

= lim

=

x0 x3 1 + tg x + 1 + sin x

=

lim

x0

tg x

- sin x x3

?

lim

x0

1

+

tg x

1 +

1

+

sin x

(2=,3)

1 4

1

1

lim

=

(2)

x0 1 + tg x + 1 + sin x 2

tg x - sin x lim

L='H

lim

1 cos2 x

- cos x

=

lim

1 - cos3 x

L='H

x0

x3

x0

3x2

x0 3x2 cos2 x

= lim

3 cos2 x sin x

= lim

cos2 x sin x

L='H

(3)

x0 6x cos2 x - 6x2 cos x sin x x0 2x cos2 x - x2 sin 2x

-2 cos x sin2 x + cos3 x

1

=

lim

x0

2 cos2

x

-

4x cos x sin x

-

2x sin 2x - 2x2

cos 2x

=

2

Pomocou Taylorovych polyn?omov:

1 + tg x = 1 + x - x2 + 11x3 + o(x3) 2 8 48

1

+

sin x

=

1

+

x

-

x2

-

x3

+

o(x3)

2 8 48

1 + tg x - 1 + sin x

lim

x0

x3

=

= lim x0

= lim x0

1

+

x 2

-

x2 8

+

11x3 48

+

o(x3)

-

x3

1 o(x3) 1

+

=

4 x3

4

1

+

x 2

-

x2 8

-

x3 48

+

o(x3)

=

lim

x3 4

+

o(x3)

=

x0

x3

5

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