1. Fugg v eny ert ekek - University of Pannoniamath.uni-pannon.hu/~szalkai/F1.pdf1. Fugg v eny ert ekek 1 Sz am tsuk ki az al abbi fuggv enyek ert ek et a megadott helyeken! a) f (x)

1. Fuggv�eny �ert�ekek

1 Sz�am��tsuk ki az al�abbi fuggv�enyek �ert�ek�et a megadott helyeken!

a) f (x) = 3x3 � 2x2 + x � 15 x = 5, 10,�5 B Ib) f (x) = sin

�1x

�x = 2π ,

4π ,�

32π ,

310π I

c) f (x) =

�arcsin(x) ha � 1 � x � 0arctg(x) ha 0 < x < +∞ x = �1,�

12 , 0, 1,

p3 I

2 Keressuk az al�abbi fuggv�enyek z�erushelyeit!

a) f (x) = 4x � 2x+log2(3) + 2 B Ib) f (x) = log2

�2x1+x

�I

3 Hol vesz fel pozit��v �ert�eket?

a) f (x) = 2+ x � x2 B Ib) f (x) = 3

px2 � 1� 2 I

c) f (x) = 2�px2 � 1 I

Matematikai anal��zis (PE) Fuggv�enyek 2007. okt�ober 27. 1 / 58

2. �Ertelmez�esi tartom�any, �ert�ekk�eszlet

4 Hol �ertelmezhet}ok a kovetkez}o fuggv�enyek, �es mi az �ert�ekk�eszletuk?

a) f 2 R ! R x 7! 1p4�x2 B I

b) f 2 R ! R x 7!psin(2x) I

c) f 2 R ! R x 7! lg�x2�3x+2x+1

�B I

d) f 2 R ! R x 7! xpx2 � 2 I

5 Mely fuggv�enyek p�arosak, illetve p�aratlanok?

a) f 2 R ! R f (x) = 22x + 4�x B Ib) f 2 R ! R f (x) = lg

�1+x1�x

�I

6 Mely fuggv�enyek peri�odikusak?

a) f (x) = sin2(2x) B Ib) f (x) = sin(2x2) Ic) f (x) = x � int(x) B I


3. M}uveletek fuggv�enyekkel

7 Adjuk meg az f + g f � g fggf fuggv�enyeket!

a) f (x) = x2 � 5x + 4 �es g(x) = x � 1 B Ib) f (x) =

px � 3� 1 �es g(x) =

p6� x � 2 I

c) f (x) = jx � 1j �es g(x) = j�x � 1j I8 V�azoljuk az al�abbi fuggv�enyek H halmazra lesz}uk��t�es�enek gra�konj�at!

a) f (x) = sin(x) H =��π2 ;�

π2

�I

b) f (x) = cos(x) H = [0;π] Ic) f (x) = tg(x) H =]� π2 ;�

π2 [ I

d) f (x) = ctg(x) H =]0;π[ I


4. Inverz fuggv�eny

9 Adjuk meg az f fuggv�eny inverz�et, ha l�etezik!

a) f (x) = 1+px � 2 B I

b) f (x) = 2x1+x2 I

c) f (x) = 2�x1+x Id) f (x) = ln(x2 � 1) x 2]�∞;�2[ I

10 Adjuk meg az f fuggv�eny olyan lesz}uk��t�es�et, ha szuks�eges, amelyinvert�alhat�o! �Abr�azoljuk a fuggv�enyt �es inverz�et!

a) f (x) = 2x � x2 Ib) f (x) = 1

1+x2 Ic) f (x) =

p1� x2 � I

d) f (x) = 32�px�2 � I

e) f (x) = 32+px�2 � I


5. Osszetett fuggv�eny

11 Adjuk meg az f � g g � f f � f g � g fuggv�enyeket!a) f (x) = x2 �es g(x) = x � 1 B Ib) f (x) =

px �es g(x) = x3 + 1 I

c) f (x) = xp1�x2 �es g(x) = sin(x) I

12 Adjuk meg az f � g fuggv�enyt!a) f (x) = exp

�4x2 � 1

�x 2 [�1; 3] �es g(x) = 1� 2x x 2 [0; 2] I

b) f (x) = 1� x2 x 2 [0; 1] �es g(x) = tgj]� π2 ; π2 [ (x) Ic) f (x) = cos(x) �es g(x) = arcsin(x) Id) f (x) = sin(x) �es g(x) = arccos(x) I


1.a) �Utmutat�as

Haszn�aljuk a Horner elrendez�est:

3x3 � 2x2 + x � 15 = ((3 � x � 2) � x + 1) � x � 15

x I



Vegyuk �eszre:

4x = (2x )2 2x+log2(3) = 3 � (2x )x I



Negat��v f}oegyutthat�oj�u m�asodfok�u fuggv�eny menete!

-3 -2 -1 1 2 3

-3

-2

-1

1

2

3

x I



A Df �ertelmez�esi tartom�anyhoz keressuk azokat az x 2 R sz�amokat, ahol anevez}o nem 0, �es a gyok alatti kifejez�es nem negat��v!Az Rf �ert�ekk�eszlethez keressuk azokat az y 2 R sz�amokat, melyekre vanmegold�asa az

f (x) = y x 2 Dfegyenletnek! x I


4.c) �Utmutat�as

Vizsg�aljuk ax2 � 3x + 2x + 1

tort el}ojel�et! x I



Ellen}or��zzukx 2 Df ) �x 2 Df

Vizsg�aljukf (�x) = �f (x)

teljesul�es�et!

x I



Keressunk olyan p 2 R+ sz�amot, amivel

x 2 Df ) x + p 2 Df �es f (x + p) = f (x)

x I


6.c) �Utmutat�as

Az eg�eszr�esz fuggv�enyre

intj[n,n+1[(x) = n n 2 Z

x I



Vizsg�aljuk, hol �ertelmezhet}o a fuggv�enyek

ossszege: Df \Dgszorzata: Df \Dgfg h�anyadosa: Df \Dg r fx 2 Dg j g(x) = 0g

x I



Ellen}or��zzuk, hogy teljesul-e:

x1,x2 2 Df : f (x1) = f (x2)) x1 = x2 (*)

y 2 Rf eset�en keressuk x 2 Df , amivel

f (x) = y (**)

teljesul.

Megjegyz�es: A (*) felt�etel ekvivalens azzal, hogy minden y 2 Rf eset�enpontosan egy megold�asa van a (**) egyenletnek. x I



Az f � g fuggv�enyhezkeressuk Df �g = fx 2 Dg j g(x) 2 Df g!adjuk meg f � g(x) = f (g (x))!

x I


1.a) Megold�as

f (5) = ((3 � 5� 2) � 5+ 1) � 5� 15 = 315

f (10) = ((3 � 10� 2) � 10+ 1) � 10� 15 = 2795f (�5) = ((3 � (�10)� 2) � (�10) + 1) � (�10)� 15 = �3225 x


1.a) Megold�as

f (5) = ((3 � 5� 2) � 5+ 1) � 5� 15 = 315f (10) = ((3 � 10� 2) � 10+ 1) � 10� 15 = 2795

f (�5) = ((3 � (�10)� 2) � (�10) + 1) � (�10)� 15 = �3225 x


1.a) Megold�as

f (5) = ((3 � 5� 2) � 5+ 1) � 5� 15 = 315f (10) = ((3 � 10� 2) � 10+ 1) � 10� 15 = 2795f (�5) = ((3 � (�10)� 2) � (�10) + 1) � (�10)� 15 = �3225 x


1.b) Megold�as

sin

�12π

�= sin

�π2

�1 = 1

sin

�14π

�= sin

�π4

�=

p22

sin

�1� 32π

�= sin

�� 2π3

�= �

p32

sin

�1310π

�= sin

�10π3

�= �

p32 x


1.c) Megold�as

arcsin(�1) = �π2 mert sin(�π2 ) = �1

arcsin(� 12 ) = �π6 mert sin(�

π6 ) = �

12

arcsin(0) = 0 mert sin(0) = 0

arctg(1) = π4 mert tg�

π4

�= 1

arctg(p3) = π3 mert tg

�π3

�=p3 x


2.a) Megold�as

0 = (2x )2 � 3 � (2x ) + 2

(2x )1,2 =3�

p32 � 4 � 22

% 2x = 2! x = 1& 2x = 1! x = 0

x


2.b) Megold�as

0 = log2

�2x

1+ x

�

20 =2x

1+ x

1+ x = 2x ! x = 1x


3.a) Megold�as

0 < 2+ x � x2 () �1 < x < 2 vagy x 2]� 1; 2[

x


3.b) Megold�as

0 <3px2 � 1� 2

2 <3px2 � 1

8 < x2 � 1! x 2]�∞;�3[[]3;+∞[

x


3.c) Megold�as

0 < 2�px2 � 1 �x2 � 1�p

x2 � 1 < 2x2 � 1 < 4! x 2]�

p3;p3[r]� 1; 1[=]�

p3;�1] [ [1;

p3[

x


4.a) Megold�as

p4� x2 6= 0, x 6= �2 �es 4� x2 � 0, �2 � x � 2

Teh�at Df =]� 2; 2[.

1p4� x2

= y x 2]� 2; 2[

Ha y � 0, nincs megold�as, ha pedig y > 0,

4� x2 = 1y2) x2 = 4� 1

y2.

Itt y2 < 4 esetben nincs megold�as, �es 2 � y eset�en x12 = �q4� 1

y2

megold�as, mert x12 2]� 2; 2[.Teh�at Rf = [2;+∞[. x


4.a) Megold�as

p4� x2 6= 0, x 6= �2 �es 4� x2 � 0, �2 � x � 2

Teh�at Df =]� 2; 2[.

1p4� x2

= y x 2]� 2; 2[


4� x2 = 1y2) x2 = 4� 1

y2.


y2

megold�as, mert x12 2]� 2; 2[.

Teh�at Rf = [2;+∞[. x


4.a) Megold�as

p4� x2 6= 0, x 6= �2 �es 4� x2 � 0, �2 � x � 2

Teh�at Df =]� 2; 2[.

1p4� x2

= y x 2]� 2; 2[


4� x2 = 1y2) x2 = 4� 1

y2.


y2

megold�as, mert x12 2]� 2; 2[.Teh�at Rf = [2;+∞[. x


4.b) Megold�as

sin(2x) � 0, k � 2π � 2x � k � 2π + π k 2 Z

k � π � x � k � π + π2

k 2 Z

Teh�at Df = [k2Z[k � π; k � π + π2 ].qsin(2x) = y x 2 Df

Ha y < 0, nincs megold�as, ha pedig y � 0,

sin(2x) = y2

egyenletnek mind��g van megold�asa, pl. x = 12 arcsin(y2) 2 [0; π4 ] 2 Df .

Teh�at Rf = R+0 . x


4.b) Megold�as

sin(2x) � 0, k � 2π � 2x � k � 2π + π k 2 Z

k � π � x � k � π + π2

k 2 Z

Teh�at Df = [k2Z[k � π; k � π + π2 ].

qsin(2x) = y x 2 Df


sin(2x) = y2




4.b) Megold�as

sin(2x) � 0, k � 2π � 2x � k � 2π + π k 2 Z

k � π � x � k � π + π2

k 2 Z

Teh�at Df = [k2Z[k � π; k � π + π2 ].qsin(2x) = y x 2 Df


sin(2x) = y2




4.c) Megold�as

x2 � 3x + 2 > 0, x < 1 vagy 2 < x �es x + 1 > 0, x > �1vagy x2 � 3x + 2 < 0, 1 < x < 2 �es x + 1 < 0, x < �1Teh�at Df =]� 1; 1[[]2;+∞[.

lg

�x2 � 3x + 2x + 1

�= y x 2 Df

x2 � 3x + 2x + 1

= 10y , x2 � (3+ 10y ) x + 2� 10y = 0

Ez ut�obbi egyenletnek akkor van gyoke, ha diszkrimin�ansa:0 � (3+ 10y )2 � 4 (2� 10y ) = 102y + 10 � 10y � 8 � 0,vagyis

p33� 5 � 10y , lg

�p33� 5

�� y .

Teh�at Rf = [ lg�p

33� 5�;+∞[. x


4.d) Megold�as

x2 � 2 � 0, x � �p2 vagy

p2 � x

Teh�at Df =]�∞;�p2[[]

p2;+∞[.

xpx2 � 2 = y x 2 Df

Ha y = 0) x = �p2 2 Df , ha y > 0 csak x 2]

p2;+∞[ intervallumban

lehet megold�as, ��gy az

x2�x2 � 2

�= y2 x 2]

p2;+∞[

z2 � 2z � y2 = 0 z = x2

egyenlet megoldhat�os�ag�at kell vizsg�alni. 0 � 4+ 4y2 mid��g teljesul, �esz1 =

2+2p1+y2

2 > 2) x =pz1 2]

p2;+∞[. Hasonl�oan kapunk a

]�∞;�p2[ intervallumban megold�ast y < 0 esetben.

Teh�at Rf = R. x


5.a) Megold�as

Df = R

f (�x) = 22(�x) + 4�(�x) = 4�x + 4x = f (x)Teh�at f p�aros fuggv�eny.

x


5.b) Megold�as

Df =]� 1; 1[

f (�x) = lg�1+ (�x)1� (�x)

�= lg

�1� x1+ x

�= lg

�1+ x

1� x

��1!=

= � lg�1+ x

1� x

�= �f (x)

Teh�at f p�aratlan fuggv�eny.x


6.a) Megold�as

x 2 Df = R ) f�x +

π

2

�= sin2

�2�x +

π

2

��=

= [sin(2x) � cos(π) + cos(2x) � sin(π)]2 == sin2(2x) = f (x)

Teh�at f peri�odikus π2 peri�odussal. x


6.b) Megold�as

x 2 Df = R ) f (x + p) = sin�2 (x + p)2

�= sin

�2x2 + 4px + 2p2

�= sin

�2x2�

ami csak �ugy lehet, ha minden x 2 R megold�asa valamely n 2 Z eset�en a

2x2 + 4px + 2p2 = 2x2 + 2nπ vagy 2x2 + 4px + 2p2 = π � 2x2 + 2nπ

egyenletek valamelyik�enek. Ezek megold�asainak halmaza azonban megsz�aml�alhat�ohalmaz, ez�ert f nem lehet peri�odikus. x


6.c) Megold�as

Df = R, �es ha x 2 [n; n+ 1[ az n eg�esz sz�am eset�en

f (x + 1) = (x + 1)� intj[n+1,n+2[(x + 1) = x + 1� (n+ 1) = x � nf (x) = x � intj[n,n+1[(x) = x � n

Teh�at f peri�odikus 1 peri�odussal. x


7.a) Megold�as

Df = Dg = Rh(x) = f (x) + g(x) = x2 � 4x + 3 x 2 R

h(x) = f (x) � g(x) = x3 � 6x2 + 9x � 4 x 2 Rh(x) = f (x)g (x) =

x2�5x+4x�1 x 2 Rr f1g

h(x) = g (x)f (x) =x�1

x2�5x+4 =1x�4 x 2 Rr f1, 4g x


7.a) Megold�as

Df = Dg = Rh(x) = f (x) + g(x) = x2 � 4x + 3 x 2 Rh(x) = f (x) � g(x) = x3 � 6x2 + 9x � 4 x 2 R

h(x) = f (x)g (x) =x2�5x+4x�1 x 2 Rr f1g

h(x) = g (x)f (x) =x�1

x2�5x+4 =1x�4 x 2 Rr f1, 4g x


7.a) Megold�as

Df = Dg = Rh(x) = f (x) + g(x) = x2 � 4x + 3 x 2 Rh(x) = f (x) � g(x) = x3 � 6x2 + 9x � 4 x 2 Rh(x) = f (x)g (x) =

x2�5x+4x�1 x 2 Rr f1g

h(x) = g (x)f (x) =x�1

x2�5x+4 =1x�4 x 2 Rr f1, 4g x


7.b) Megold�as

Df = [3;+∞[ Dg =]�∞; 6] Df \Dg = [3; 6]h(x) = f (x) + g(x) =

px � 3+

p6� x � 3 x 2 [3; 6]

h(x) = f (x) � g(x) ==p(x � 3) (6� x)� 2

px � 3�

p6� x + 2 x 2 [3; 6]

h(x) = f (x)g (x) =px�3�1p6�x�2 x 2 [3; 6]

h(x) = g (x)f (x) =p6�x�2px�3�1 x 2 [3; 4[[]4; 6] x


7.c) Megold�as

Df = Dg = R

h(x) = f (x) + g(x) =

8

8.a) Megold�as

-1.5 -1.0 -0.5 0.5 1.0 1.5

-1.0

-0.5

0.5

1.0

x


8.b) Megold�as

0.0 0.5 1.0 1.5 2.0 2.5 3.0

-1.0

-0.5

0.5

1.0

x


Megold�as

-1.5 -1.0 -0.5 0.5 1.0 1.5

-1.5

-1.0

-0.5

0.5

1.0

1.5

x


8.d) Megold�as

0.0 1.0 2.0 3.0

-2.0-1.5-1.0-0.5

0.51.01.52.0

-3.0

-2.0-1.5-1.0-0.5

0.51.01.52.0

3.0

x


9.a) Megold�as

Df = [2;+∞[ x1, x2 � 2 : 1+px1 � 2 = 1+

px1 � 2) x1 = x2

1+px � 2 = y x � 2

px � 2 = y � 1

Ha y < 1, nincs megold�as, ha y � 1

x = (y � 1)2 + 2 2 Df

Teh�at Rf = [1;+∞[, �es

f �1 : [1;+∞[! [2;+∞[ y 7! (y � 1)2 + 2

vagyf �1(x) = (x � 1)2 + 2 x 2 [1;+∞[

x


9.b) Megold�as

Df = R

2x

1+ x2= y

yx2 � 2x + y = 0 ha y 6= 0, D = 4(1� y2)

Ha y 6= 0, �es (1� y2) > 0, akkor x12 =1�p1�y2y 2 Df , teh�at k�et megold�as

van ahol f (x1) = f (x2), ez�ert f nem invert�alhat�o. x


9.c) Megold�as

Df = Rr f�1g x1, x2 6= �1 : 2�x11+x1 =2�x21+x2

) x1 = x2

2� x1+ x

= y x 6= �1

2� y = x(y � 1)

Ha y = 1, akkor nincs megold�as;Ha y 6= 1, akkor x = 2�yy�1 6= �1;

Teh�at Rf = Rr f1g, �es

f �1 : Rr f1g ! Rr f�1g y 7! 2� yy � 1

vagy

f �1(x) =2� xx � 1 x 6= 1

x


9.c) Megold�as

Df = Rr f�1g x1, x2 6= �1 : 2�x11+x1 =2�x21+x2

) x1 = x2

2� x1+ x

= y x 6= �1

2� y = x(y � 1)

Ha y = 1, akkor nincs megold�as;

Ha y 6= 1, akkor x = 2�yy�1 6= �1;


f �1 : Rr f1g ! Rr f�1g y 7! 2� yy � 1

vagy

f �1(x) =2� xx � 1 x 6= 1

x


9.c) Megold�as

Df = Rr f�1g x1, x2 6= �1 : 2�x11+x1 =2�x21+x2

) x1 = x2

2� x1+ x

= y x 6= �1

2� y = x(y � 1)

Ha y = 1, akkor nincs megold�as;Ha y 6= 1, akkor x = 2�yy�1 6= �1;


f �1 : Rr f1g ! Rr f�1g y 7! 2� yy � 1

vagy

f �1(x) =2� xx � 1 x 6= 1

x


9.d) Megold�as

Df =]�∞;�2] x1, x2 � �2 : ln(x21 � 1) = ln(x22 � 1)) jx1j = jx2j )x1 = x2

ln(x2 � 1) = y x � �2x2 � 1 = ey

x1 =pey + 1 �2 x2 = �

pey + 1 � �2 ha y � 0

Teh�at Rf = [0;+∞[, �es

f �1 : R+0 ! R y 7! �pey + 1

vagyf �1(x) = �

pex + 1 x 2 R+0

x


10.a) Megold�as

Df = R, 2x � x2 = y ! x2 � 2x + y = 0 D = 4(1� y)Ha y = 1, akkor x = 1, ha y < 1, akkor a k�et megold�as x12 = 1�

p1� y ? 1.

Teh�at f �1j]1;+∞[(x) = 1+p1� x /x < 1/.

-1 1 2 3

-11

23

xMatematikai anal��zis (PE) Fuggv�enyek 2007. okt�ober 27. 45 / 58

10.b) Megold�as

Df = R, 11+x2 = y ! yx2 = 1� y

Ha 0 < y � 1 akkor a k�et megold�as x12 = �q1�yy ? 0. Teh�at

f �1jR+(x) =q1�xx x 2]0; 1].

-11

23

-1 1 2 3

x


10.c) Megold�as

Df = [�1; 1],p1� x2 = y � 1 � x � 1

Ha y < 0 akkor nincs megold�as, ha 0 � y � 1, akkorx2 = 1� y2 ! x12 = �

p1� y2

Teh�at

f �1j[0;1](x) =p1� x2 x 2 [0; 1].

0.25 0.50 0.75 1.00-0.25 0.25 0.50 0.75 1.00-0.25

0.25

0.50

0.75

1.00


10.d) Megold�as

Df = [2; 6[[[6;+∞[, f �1(x) =�2� 3x

�2+ 2 x 2]�∞; 0[[[ 32 ;+∞[

-10 -8 -6 -4 -2 2 4 6 8 10

-10

-8

-6

-4

-2

2

4

6

8

10


10.e) Megold�as

Df = [2;+∞[, f �1(x) =�2� 3x

�2+ 2 x 2]0; 32 ]

-1 2 4 6 8 10-1

2

4

6

8

10


11.a) Megold�as

Df = Dg = Rx 2 R �es g(x) 2 R ,x 2 Rf (g (x)) = f (x � 1) = (x � 1)2 = x2 � 2x + 1

f � g : R ! R f � g(x) = x2 � 2x + 1

x 2 R �es g(x) 2 R ,x 2 R g (f (x)) = g�x2�= x2 � 1

g � f : R ! R g � f (x) = x2 � 1

x 2 R �es g(x) 2 R ,x 2 R f (f (x)) = f�x2�= x4

f � f : R ! R f � f (x) = x4

x 2 R �es g(x) 2 R ,x 2 R g (g (x)) = g (x � 1) = x � 2

g � g : R ! R g � g(x) = x � 2

x


11.a) Megold�as

Df = Dg = Rx 2 R �es g(x) 2 R ,x 2 Rf (g (x)) = f (x � 1) = (x � 1)2 = x2 � 2x + 1

f � g : R ! R f � g(x) = x2 � 2x + 1

x 2 R �es g(x) 2 R ,x 2 R g (f (x)) = g�x2�= x2 � 1

g � f : R ! R g � f (x) = x2 � 1

x 2 R �es g(x) 2 R ,x 2 R f (f (x)) = f�x2�= x4

f � f : R ! R f � f (x) = x4

x 2 R �es g(x) 2 R ,x 2 R g (g (x)) = g (x � 1) = x � 2

g � g : R ! R g � g(x) = x � 2


11.b) Megold�as

Df = R+0 Dg = Rx 2 R �es g(x) = x3 + 1 � 0() x � �1f (g (x)) = f

�x3 + 1

�=px3 + 1

f � g : [�1;+∞[! R f � g(x) =px3 + 1

x 2 R+0 �es f (x) 2 R , x 2 R+0

g (f (x)) = g�px�=px3 + 1

g � f : R ! R g � f (x) =px3 + 1

x 2 R+0 �es f (x) 2 R+0 , x 2 R

+0 f (f (x)) = f

�px�= 4px

f � f : R ! R f � f (x) = 4px

x 2 R �es g(x) 2 R , x 2 Rg (g (x)) = g

�x3 + 1

�=�x3 + 1

�3+ 1 = x9 + 3x6 + 3x3 + 2

g � g : R ! R g � g(x) = x9 + 3x6 + 3x3 + 2


11.c) Megold�as

Df =]� 1; 1[ Dg = Rx 2 R �es �1 < sin(x) < 1() x 6= n � π n 2 Z

f � g : Rr fn � π j n 2 Zg! R f � g(x) = sin(x)jcos(x)j

g � f :]� 1; 1[! R g � f (x) = sin�

xp1� x2

�x 2]� 1; 1[ �es xp

1�x2 2]� 1; 1[, x 2]�12 ;12 [

f � f :]� 12;1

2[! R f � f (x) = xp

1� 2x2

g � g : R ! R g � g(x) = sin (sin(x))x


11.c) Megold�as



g � f :]� 1; 1[! R g � f (x) = sin�

xp1� x2

�

x 2]� 1; 1[ �es xp1�x2 2]� 1; 1[, x 2]�

12 ;12 [

f � f :]� 12;1

2[! R f � f (x) = xp

1� 2x2



11.c) Megold�as



g � f :]� 1; 1[! R g � f (x) = sin�

xp1� x2

�x 2]� 1; 1[ �es xp

1�x2 2]� 1; 1[, x 2]�12 ;12 [

f � f :]� 12;1

2[! R f � f (x) = xp

1� 2x2



12.a) Megold�as

Df = [�1; 3] Dg = [0; 2]

x 2 [0; 2] �es �1 � 1� 2x � 3() x 2 [0; 1]

f � g : [0; 1]! R f � g(x) = exp�16x2 � 16x + 3

�x


12.a) Megold�as

Df = [�1; 3] Dg = [0; 2]x 2 [0; 2] �es �1 � 1� 2x � 3() x 2 [0; 1]

f � g : [0; 1]! R f � g(x) = exp�16x2 � 16x + 3

�x


12.b) Megold�as

Df = [0; 1] Dg =]� π2 ;π2 [

x 2]� π2 ;π2 [ �es 0 � tg(x) � 1() x 2 [0;

π4 ]

f � g : [0; π4]! R f � g(x) = 1� tg2(x)

x


12.c) Megold�as

Df = R Dg = [�1; 1]

x 2 [�1; 1] akkor arcsin(x) = y 2 [�π2 ;�π2 ] 2 R, �es sin(y) = x ,

0 � cos(y) =p1� x2

f � g : [�1; 1]! R f � g(x) =p1� x2

x


12.c) Megold�as

Df = R Dg = [�1; 1]x 2 [�1; 1] akkor arcsin(x) = y 2 [�π2 ;�

π2 ] 2 R, �es sin(y) = x ,

0 � cos(y) =p1� x2

f � g : [�1; 1]! R f � g(x) =p1� x2

x


12.d) Megold�as

Df = R Dg = [�1; 1]

x 2 [�1; 1] akkor arccos(x) = y 2 [0;π] � R, �es cos(y) = x ,0 � sin(y) =

p1� x2

f � g : [�1; 1]! R f � g(x) =p1� x2

x


12.d) Megold�as

Df = R Dg = [�1; 1]x 2 [�1; 1] akkor arccos(x) = y 2 [0;π] � R, �es cos(y) = x ,0 � sin(y) =

p1� x2

f � g : [�1; 1]! R f � g(x) =p1� x2

x


10.c) Megjegyz�es

Vegyuk �eszre, hogy

f �1j[0;1] = fj[0;1] illeteve f�1j[�1;0] = �fj[�1;0]

x


10.d-e) Megjegyz�es

Vegyuk �eszre, hogy a 10. d), e) feladat inverz fuggv�enyeinek k�eplete(hozz�arendel�esi utas��t�asa) azonos, de �ertelmez�esi tartom�anyaik kulonboznek!

x


FüggvényekFüggvény értékekÉrtelmezési tartomány, értékkészletMuveletek függvényekkelInverz függvényÖsszetett függvény

ÚtmutatásokMegoldásokMegjegyzések

Documents

1. Fugg v eny ert ekek - University of Pannoniamath.uni-pannon.hu/~szalkai/F1.pdf1. Fugg v eny ert ekek 1 Sz am tsuk ki az al abbi fuggv enyek ert ek et a megadott helyeken! a) f (x)