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AYT Matematik Ders İşleyiş Modülü Limit ve Türev

Published by Nesibe Aydın Eğitim Kurumları, 2019-08-22 09:23:23

Description: AYT Matematik Ders İşleyiş Modülü Limit ve Türev

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5ÑSFW\"MNB,VSBMMBS TEST - 4 1. f ( x ) = ax3 + bx2 +DY+ d , f ' ( 1 ) = 1 ve f '' ( 1 ) = 2 5. f ( x ) = ( x - 1 ) ( x - 2 ) ( x - 3 ) … ( x - 12 ) PMEVôVOBHÌSF B-DGBSLOOFöJUJLBÀUS PMEVôVOBHÌSF Gh JGBEFTJOJOEFôFSJLBÀUS A) - # $ % & A) - # - $ - 4! % & 6. f^ x h = x PMEVóVOBHËSF x+3 2. f ( x ) = 5 + 2 x lim f ^ 4 + h h- f^ 4 h h\"0 h PMEVôVOBHÌSF Gh JGBEFTJOJOFöJUJLBÀUS MJNJUJOJOFöJUJLBÀUS A) - 1 # - 1 $ 1 % 1 1 A) 4 # 2 $ 1 % 4 E) 1 E) 5 5 5 25 25 3 12 16 12 6 3. f ( x ) = ( x - 2 ) 20 . ( x + 1 ) 10 7. f^ x h = 1 PMEVóVOBHËSF PMEVôVOBHÌSF Ghh + f' ( - UPQMBNOOFöJUJ 3 x2 - 2x + 5 LBÀUS lim f^ 3 + h h- f^ 3 - h h A) - # - $ % & h\"0 h MJNJUJOJOFöJUJLBÀUS A) - 1 # - 1 $ % 2 E) 4 6 12 3 3 4. f ( x ) = ( 3x + 1 ) 20 8. f^ x h = x + 1 fonksiyonu veriliyor. PMEVôVOBHÌSF Ghh JGBEFTJOJOFöJUJLBÀUS x+9 \" # $ Buna göre Gh LBÀUS A) 16 # 8 $ 8 % 4 % & E) 4 27 81 27 81 27 1. C 2. D 3. C 4. \" 49 5. \" 6. D 7. \" 8. &

TEST - 5 5ÑSFW\"MNB,VSBMMBS 1. y = P ( x ) polinomu 5. f^ x h = x + 2 fonksiyonu veriliyor. P' ( x ) + P ( x ) = x2+ 3x + x2 - 4 FõJUMJóJOJTBóMZPS #VOBHÌSF 1 LBÀUS Buna HÌSF Gh LBÀUS A) - # $ - 1 % 1 E) 0 44 \" # $ % & 2. f^ x h = x2 3 x2 - 1 + x. x2 + 16 6. f^ x h = x2 - 2x + 1 fonksiyonu veriliyor. fonksiyonu veriliyor. x2 + 4x + k #VOBHÌSF Gh LBÀUS f' ( 0 ) = -PMEVóVOBHËSF QP[JUJGLHFSÀFLTBZT LBÀUS \" # $ % & \" # $ % & 7. P ( x ) = 4 ( x - 1 ) + Y- 1 )2 + 4 ( x - 1 )3+ ( x - 1)4 3. f^ x h = x - 1 fonksiyonu veriliyor. polinomu veriliyor. x+1 Buna göre, lim P^ 2 + 5h h - P^ 2 h LBÀUS d2 f^ x h h\"0 2h Buna göre, dx2 EFôFSJLBÀUS \" # $ % & x=0 A) - # - $ - % 1 E) 4 4. f ( x ) = ( x2 - 1 ) ( x + 1) ( x2 - 9 ) fonksiyonu verili- 8. fn( x ) = n . n x , n ` { 2, ..., 100}GPOLTJZPOMBSWF- yor. riliyor. Buna göre, f' ( - LBÀUS Buna göre, f'2^ 1 h + f3' ^ 1 h + . . . + f1' 00^ 1 hUPQMB- A) - # - $ % & NLBÀUS \" # $ % & 1. & 2. & 3. \" 4. C 50 5. C 6. B 7. D 8. D

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 6. MODÜL ·/÷7&34÷5&:&)\";*3-*, 1\"3¦\"-*'0/,4÷:0/-\"3*/5·3&7÷ 1BSÀBM'POLTJZPOMBSO5ÑSFWJ ÖRNEK 3 %m/*m f^ x h = * x2 + 1 x # 2 fonksiyonu veriliyor. f^ x h = * g1 ^ x h x≤a 4x x > 2 g2 ^ x h x>a Buna göre, f' ( 2+ ) ve f' ( 2– EFôFSMFSJWBSTBLBÀUS y = f ( x ) fonksiyonunda x = a apsisli nokta kri- lim f^ x h ≠ lim– f^ x h f, x = 2 için süreksiz, UJLOPLUBES + x\"2 x\"2 lim f^ x h- f^ 2 h = lim 4x - 5 r f fonksiyonu x = a için süreksiz ise f'(a) + x-2 + x+2 x\"2 x\"2 yoktur. MJNJUJHFSÀFMTBZEFôJMEJSGh +) yoktur. f^ x h- f^ 2 h x 2 +1-5 x-2 r f fonksiyonu x = a için sürekli ise lim– x-2 = lim– = lim– ^x+2h= 4 x\"2 x\"2 x\"2 i) g1' (a) áH2'(a) ise f'(a) yoktur. f'(2-) = 4 ii) g '(a) = g '(a) ise f'(a) = g '(a) = g '(a) 1 2 1 2 olur. ÖRNEK 1 8<$5, f^ x h = * - x, x < 0 ise 4ÐSFLMJPMNBZBOOPLUBMBSEBTBóEBOWFTPMEBOUÐSFW x, x ≥ 0 ise BMOSLFOUÐSFWJOSFTNJUBONLVMMBOMNBMES'POL- TJZPO TÐSFLMJ PMEVóVOEB UÐSFW BMNB LVSBM LVMMBO- Gh Gh Gh m EFôFSMFSJOJIFTBQMBZO[ labilir. lim+ f^ x h = lim– f^ x h = f^ 0 h sürekli x\"0 x\"0 -1 x < 0 ÖRNEK 4 f'^ x h = ( 1 x>0 Z x2 - 2x + 1 x≤0 ]] f'(0+) = 1, f'(0-) = -PMEVôVOEBOGh ZPLUVSGh = 1 ve f^ x h = [ 3x + 1 0 < x ≤ 1 foOLTJZPOVUBONMBOZPS f'(-2) = -1 olur. ]] \\ x3 - 2 x >1 Buna göre, f ' ( 0– ) , f ' ( 0+ ) , f ' ( 1– ), f ' ( 1+ ) EFôFSMFSJ varsa bulunuz. ÖRNEK 2 Z lim f^ x h = lim– f^ x h = f^ 0 h sürekli ]] 3x + 5, x11 + x\"0 1 # x # 3 fonksiyonu veriliyor. x\"0 f^ x h = [ 4x + 1, x23 lim f^ x h ≠ lim– f^ x h = , x = 1 için süreksiz ]] \\ 2x - 3, + x\"1 x\"1 Buna göre, y =Gh Y GPOLTJZPOVOVZB[O[ x < 0 f' ( x ) = 2x - 2, f'(0) = -2 f^ x h ≠ lim– f^ x h X > 0 f' ( X ) = 3, f' ( 0+ ) = 3 lim+ x\"1 süreksiz, f'(1) yoktur. f^ x h- f^ 1 h x3 - 2 - 4 x-1 x\"1 f'^ 1+ h = lim+ = lim+ lim f^ x h ≠ lim f^ x h süreksiz, f'(3) yoktur. x\"1 x-1 x\"1 x\"3 ++ MJNJUHFSÀFMTBZEFôJMG'(1+) yoktur. x\"3 Z 3 , x<1 f'(1-) = f^ x h- f^ 1 h 3x + 1 - 4 ]] =3 f'^ x h = [ 4 , 1 < x < 3 lim = lim– ]] x-1 x-1 \\ + x\"1 x\"1 2 , x>3 f'(1-) = 3 3, x < 112351 3. f'(2+) yoktur, f'(2–) = 4 4. f'(0-) = –2, f'(0+) = 3, f'(1–) = 3, f(1+) yoktur. 1. f'(0) yoktur, f'(1) = 1, f'(–2) = –1 2. f(x) = 4, 1 < x < 3 2, x > 3

·/÷7&34÷5&:&)\";*3-*, 6. MODÜL -÷.÷57&5·3&7 www.aydinyayinlari.com.tr ÖRNEK 5 ÖRNEK 8 ôFLJMEFZ=G Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõUJS f^ x h = * 2x2 + mx + 4 , x 1 1 fonksiyPOVUBONMBOZPS y nx2 + 2x + 1 , x $ 1 y = f(x) x =BQTJTMJOPLUBTOEBZ= f ( x ) fonksiyonu türevli ol- EVóVOBHËSF NWFOHFSÀFMTBZMBSOCVMVOV[ 2 1 –5 –3 x lim– f^ x h = lim+ f^ x h = f^ 1 h O2 –1 x\"1 x\"1 2 +N+ 4 = n + 2 + 1 jN- n = -3 f ( 1+) = f ( 1-) y = f ( x ) fonksiyonu ( - BSBMôOEBLJ LBÀ UBOF 2n + 2 = 4 +NjN- 2n = -2 UBNTBZEFôFSJJÀJOUÑSFWMJEJS m-n =-3 x = -TÑSFLTJ[PMEVôVJÀJO Y= -5 ve x =TBôTPMUÑ- 4n = - 1, m = - 4 SFWGBSLMPMEVôVJÀJOUÑSFWTJ[EJS m - 2n = - 2 11 -3 =UBOFUBNTBZEFôFSJJÀJOUÑSFWMJEJS .VUMBL%FôFS'POLTJZPOVOVO5ÑSFWJ ÖRNEK 6 TANIM f : R Z R ve f^ x h = * 2x2 + n, x # 1 f^ x h = g^ x h = * -g^ x h g^ x h < 0 g^ x h g^ x h $ 0 mx + 3, x 2 1 fonksiyonu x =BQTJTMJOPLUBTOEBUÑSFWMJPMEVôV- f'^ x h = * - g'^ x h g^ x h < 0 OBHÌSF NOÀBSQNLBÀUS g'^ x h g^ x h > 0 lim f^ x h = f^ 1 hPMNBM g ( x ) = EFOLMFNJOJTBóMBZBOOPLUBMBSEBTBó- dan ve soldan türeve bakmak gerekir. x\"1 lim+ f^ x h = m + 3 _ bb x\"1 `m+3=2+n jN- n = -1 lim– f^ x h = 2 + n bb (FOFMPMBSBLCVLËLMFSJO¿PLLBUMPMNBTEVSV- a NVOEBUÐSFWMFSJWBSESWFTGSBFõJUUJS x\"1 f'^ x h = ( 4x x<1 m x > 1 x =EFFöJUMJLLVMMBOBCJMNFLJÀJO f'(1+) = f'(1-) f'(1+) =N= 4 = f'(1- N= 4, n = NO= 20 ÖRNEK 7 ÖRNEK 9 mx + n , x # 2 | |f ( x ) = x + 2 PMEVóVOBHËSF f^ x h = * x2 + nx + 2 , fonksiyonu veriliyor. x22 f' ( 0 ), f' ( -3 ), f' ( -1 ), f' ( -2 ) EFôFSMFSJOJIFTBQMBZO[ #VGPOLTJZPOVOUÑNSFFMTBZlardBUÑSFWMJPMBCJMNFTJ f^ x h = ( x+2 x $-2 JÀJO N O JLJMJTJOFPMNBMES -x - 2 x <-2 lim+ f^ x h = lim– f^ x h = f^ 2 h 1 x >-2 f'^ x h = ( - 1 x <-2 x\"2 x\"2 4 + 2n + 2 = 2m + n & _ 2m - n = 6 x = -2 de türevsizdir. = 2x + n 44 + n = m & bb m-n=4 x>2 f'^ x h = ` m = 2, n = f'(0) = f'(-1) = 1 j f'(-3) = -1 x<2 f'^ x h m bb a - 2 5. 8 6. 20 7. (2, –2) 52 8. Nm Om 9. f'(0) = 1, f'(–1) = 1, f'(–3) = –1, f'(–2) = yoktur.

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 6. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 10 ÖRNEK 13 f ( x ) = | x2 - 2x | | | | |f ( x ) = x . x3 - 1 - 2x 1 - 3x2 PMEVôVOBHÌSF Gh Gh Gh Gh EFôFSMF PMEVôVOBHÌSF G LBÀUS SJOJIFTBQMBZO[ x = 2 için x3-JGBEFTJQP[JUJG - 3x2 ifadesi negatiftir. Z x2 - 2x x<0 f(x) = x ( x3 - 1) - 2x ( 3x2 - 1) ] 0#x<2 f(x) = x4 - x - 6x3 + 2x ]] f'(x) = 4x3 - 18x2 + 1 f^ x h = [ - 2 + 2x x$2 f''(x) = 12x2 - 36x j f''(2) = -24 ] x x<0 ÖRNEK 14 0<x<2 ]] x2 - 2x | |f ( x ) = x2 + ax + 4 \\ x>2 ifadesi r x `3JÀJOUÑSFWMJPMEVôVOBHÌSF BHFSÀFM Z TBZMBSOOBMBCJMFDFôJEFôFSBSBMôOFEJS ]] 2x - 2 f'^ x h = [ - 2x + 2 f(x) = |x2 + ax + 4|GPOLTJZPOVUÑSFWMJPMNBTJÀJO ]] \\ 2x - 2 x2 + ax + 4 =EFOLMFNJOJOLÌLÑZPLUVSZBEBLÌLMFS ÀJGULBUPMNBMES f'(0) ve f'(2) yoktur. D # 0 a2 - 16 # 0 j -4 # a # 4 j a ` [-4, 4] f'(1) = -2 + 2= 0 j f'(3) = 6 - 2 = 4 ÖRNEK 15 ÖRNEK 11 | |f ( x ) = ( x - 1) ( x + 2 ) ( x - 3 ) ( x + 3 )2 | |f ( x ) = x3 + x2 GPOLTJZPOVOVO UÑSFWTJ[ PMEVôV LBÀ GBSLM Y EFôFSJ PMEVôVOBHÌSF Gh Gh -1 ) , f ' ( 2 ) , f ' ( - EFôFS WBSES MFSJOJIFTBQMBZO[ y = f(x) fonksiyonu x = 1 ve x = 3 için türevsizdir. f^ x h = * -x3 - x2 x <-1 x = -2 ve x = - ÀPL LBUM LÌLMFS PMEVôV JÀJO UÑSFWMFSJ x3 + x2 x $-1 WBSESWFTGSES5ÑSFWTJ[UBOFYEFôFSJWBSES f'^ x h = * - 3x2 - 2x x <-1 x >-1 2 + 2x 3x f'(-1) yoktur. f'(0) = 0 ( 0 çift kat) f'(2) = 12 + 4 = 16 , f'(-2) = -12 + 4 = -8 ÖRNEK 12 ÖRNEK 16 | |f ( x ) = x . x B CWFDTGSEBOGBSLMUBNTBZMBSPMNBLÐ[FSF PMEVóVOBHËSF Gh Gh EFôFSMFSJOJIFTBQMBZO[ | |f ( x ) = ( x + 1 ) a. xb . ( x + 2 )D fonksiyonu veriliyor. 2 x$0 2x x $ 0 f fonksiyonu r x `3JÀJOUÑSFWMJPMEVôVOBHÌSF f'^ x h = ( a + b +DUPQMBNen azLBÀUS f^ x h = * x 2 - 2x x < 0 f, r x `3JÀJOUÑSFWMJJTFB C Dä - x<0 2+3+4=9 x f'(0+) = f'(0-) =PMEVôVOdan f'(0) = 0 f'(2) = 4 10. f'(0), f'(2) yoktur, f'(1) = 0, f'(3) = 4 11. f'(–1) yoktur, 53 13. –24 14. [–4, 4] 15. 2 16. 9 f'(0) = 0, f'(2) = 16, f'(–2) = –8 12. f'(2) = 4, f'(0) = 0

·/÷7&34÷5&:&)\";*3-*, 6. MODÜL -÷.÷57&5·3&7 www.aydinyayinlari.com.tr ;JODJS,VSBMWF#JMFöLF'POLTJZPOVO5ÑSFWJ ÖRNEK 19 TANIM fl^ 2 h = 3 _ b gl^ 5 h = - 4 ` PMEVôVOBHÌSe, HPG h LBÀUS GWFHUÐSFWMFOFCJMJSGPOLTJZPOMBSJ¿JOZ= fog ( x ) b a GPOLTJZPOVOVOUÐSFWJOJIFTBQMBZBMN f^ 2 h = 5 fog^ x + h h - fog^ x h g' ( f ( 2 ) ) .f' ( 2 ) = g' ( 5 ) .f' ( 2 ) = -4.3 = -12 lim = ^ fog h'^ x h h\"0 h PMEVóVOVUÐSFWUBONOEBOCJMJZPSV[ fog^ x + h h - fog^ x h g^ x + h h - g^ x h lim · h \" 0 g^ x + h h- g^ x h h ÖRNEK 20 g ( x ) = u için lim g (x + h) = lim u + k olur. f : R Z R ve f ( 3x + 1 ) = x3 +Y2 + 2x - 7 h\"0 k\"0 PMEVôVOBHÌSF Gh LBÀUS f^ u + k h- f^ u h g^ x + h h- g^ x h f'(3x + 1) . 3 = 3x2+ 10x + 2 lim · lim 3f'(4) = 15 j f'(4) = 5 1k4\"404 4 4 2 4k4 4 4 4 3 h \" 0 h f'^ u h f' ( u ) . g' ( x ) = f' ( g ( x ) ) . g' ( x ) ( fog )' ( x ) = f' ( g ( x ) . g' ( x ) olur. :VLBSEBV=H Y EËOÐõÐNÐZBQMEóOEB y = f^ u h 4 FMEFFEJMNJõUJ ÖRNEK 21 u = g^ v h f : R Z R, 5ÐSFWBMOEóOEBJTF f ( x ) = ( 3x - 1 )2 PMEVôVOB HÌSF GPG h LBÀUS dy = dy · du elde edilir. f' ( x ) = 2 ( 3x - 1 ) . 3 dx du dx ( fof )' ( 0 ) = f' ( f ( 0 ) ) . f' ( 0 ) = f'(1) . f'(0) = 12. (-6) = -72 &MEFFUUJóJNJ[CVLVSBMB[JODJSLVSBMBEWFSJMJS ÖRNEK 22 ÖRNEK 17 f ( x ) = x2 + 3x - 1 ve g ( x ) = x2 - 1 f ( x ) = x2 +Y H Y = 3x GPOLTJZPOMBSJÀJO GPH h+Gh H LBÀUS PMEVôVOBHÌSF Z= GPH h Y JOFöJUJOJCVMVOV[ ( fog ) ( 3 ) )' = 0 f' ( g ( 3 ) ) = f'^ 8 h = 2 8 + 3 = 4 2 + 3 (fog)' (x) = f'(g(x)) . g'(x) f'^ x h = 2x + 5 ^ 2^ 3x h + 5 h.3 4 g'^ x h = 3 18x + 15 ÖRNEK 18 ÖRNEK 23 fl^ 7 h = 3 _ g ( x ) = x. f 2 ( 2x + 1 ) ve f ( 3 ) = 1, f ' ( 3 ) = 2 b gl^ 5 h = 4 ` PMEVôuna göre, ( fog ) ' ( 5 ) kaçUS PMEVôVOBHÌSF Hh LBÀUS b g^ 5 h = 7 a g'(x) = 1.f2(2x + 1) + x.2.f(2x + 1) . f'(2x + 1).2 g'(1) = f2(3) + 4f(3).f'(3) = 1 + 8 = 9 f' ( g ( 5 ) ) . g' ( 5 ) = f' ( 7 ) . g' ( 5 ) = 3.4 = 12 17. 18x + 15 18. 12 54 19. –12 20. 5 21. –72 22. 4 2 + 3 23. 9

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 6. MODÜL ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 24 ÖRNEK 27 f ( x ) + f f 1 p + f^ x2 - x + 1 h = x3 + 2x + 1 y = t2 - 2t _ xx bb FöJUMJôJOJTBôMBZBOZ= f ( x ) fonksiyonu için f' ( 1 ) kaç- t = u3 + 1 ` US bb u= x a f'^ x h - 1 f'd 1 n + ^ 2x - 1 h f'^ x2 - x + 1 h = 2x - 1 dy ifadesininJOFöJUJOJCVMVOV[ PMEVôVOBHÌSF dx 2x 2 xx f' ( 1 ) - f' ( 1 ) + f' ( 1 ) = 1j f' ( 1 ) = 1 dy · dt · du = ^ 2t - 2 h.3u2· 1 dt du dx 2x = 2 x x ·3x 1 = 2 3x 2x ÖRNEK 28 y = t2 + 2t _ bb ÖRNEK 25 t=3 u ` f^ x h- f^ 2 h u = 8x bb lim = 10 a x\"4 x-4 FöJUMJôJOJTBôMBZBOZ= f ( x ) için fh LBÀUS dy PMEVôVOBHÌSF dx JGBEFTJOJOFöJUJLBÀUS x=1 y = f^ x h için, dy = f'^ x h· 1 dx 2 x dy dt du = ^ 2t + 2 h. 1 –2/3 · · 3 u .8 dt du dx f'^ 2 h. 1 = 10 & f'^ 2 h = 40 4 1 x = 1, u = 8, t = 2 j 6· ·8 = 4 3.4 ÖRNEK 26 ÖRNEK 29 y = 1 , x = u , u = t2 G Y QPMJOPN GPOLTJZPOV JÀJO EFSG Y ä PMEVôV- x na göre, dy dy I. y = f ( x ) çift fonksiyon ise y = f' ( x ) fonksiyonu tek fonksiyondur. PMEVóVOBHöre, dt JGBEFTJOJOEFôFSJJMF du II. y = f ( x ) tek fonksiyon ise y = f' ( x ) fonksiyonu çift t=1 u=4 fonksiyondur. ifadesiniOEFôFSJGBSLOONVUMBLEFôFSJLBÀUS III. y = f ( x ) tek fonksiyon ise f'' (- x ) = Ghh Y FõJUMJóJ TBóMBOBCJMJS dy dx du 11 · · =- · ·2t JGBEFMFSJOEFOLBÀUBOFTJEPôSVEVS dx du dt 2 2u *G -x) = f ( x ) j -f'(-x) =Gh Y EPôSVEVS x **G -x ) = -f ( x ) j -f'(-x) = -Gh Y EPôSVEVS ***G Y =NYJÀJOGh Y =NWFGhh Y = 0 olur. 1 t = 1, u = 1, x = 1 j - 1· ·2 = - 1 f'' ( -x ) = f''(x) =FöJUMJôJTBôMBOS * **WF***EPôSVEVS 2 dy dx 11 · =- · dx du 2 2u x 11 1 1 15 u = 4, x = 2 j - · = - j - 1 + = 44 16 61 16 24. 1 25. 40 15 55 27. 3x2 28. 4 29. * **WF*** 26. 16

TEST - 6 1BSÀBM'POLTJZPOMBSO5ÑSFWJ 1. f^ x h = * ax2 + 1 x $ 1 5. f^ x h = * x2 – 1 x ≤ 2 x +1 4x + b x 1 1 x>2 fonksiyonu türevlenebilir CJSGPOLTJZPOPMEVôV- fonksiyonu veriliyor. OBHÌSF BCÀBSQNLBÀUS Buna göre, f ' ( 2+ ) - f ' ( 2- LBÀUS A) - # - $ % & A) - # - $ % & 2x – 1 x ≤ 1 6. f^ x h = * x2 + 3x fonksiyonu veriliyor. x >1 2. f^ x h = * ax2 + 4 x 1 1 Buna göre BöBôEBLJ TFÀFOFLMFSEFO IBOHJTJ ZBOMöUS 3x4 + b x $ 1 A) f' ( 1+ ) =UJS fonksiyonu türevlenebilir CJSGPOLTJZPOPMEVôV- # y = f ( x ) fonksiyonu x = 0 için süreklidir. na göre, f'( - JGBEFTJOJOEFôFSJLBÀUS $ y = f ( x ) fonksiyonu x = 1 için türevsizdir. \" # $ - % -12 E) -24 % f ' ( 1- ) = 2 dir. E) f' ( 2+ ) = 7 dir. 3. \"öBôEBLJGPOLTJZPOMBSOIBOHJTJOJOY= 1 nok- 7. f : R Z R UBT OEBUÑSFWJWBSES 2x - 3 x < 1 f^ x h = * x3 + 4x x $ 1 A) f (x) = ' x + 1 x # 1 # f (x) = * x2 + 1 x # 1 2x - 2 x 2 1 f^ x h- f^ 1 h x+1 x21 fonksiyonu için lim MJNitinin so- x \" 1+ 3x x # 1 x-1 nucVLBÀUS $ f (x) = * 0 x = 1 % f (x) = ' x + 1 x 1 1 3x - 1 x $ 1 x+1 x21 E) f (x) = * x2 + 1 x 1 1 A) -1 # $ 2x x $ 1 % & ZPLUVS Z ax2 Z x2 - 4 x<2 ] ] x = 2 fonksiyonu veriliyor. + x12 ] 3 8. f^ x h = [ 0 x>2 4. f^ x h = [ bx - 5 x=2 ]] x 2 + x ] \\ \\ 8x + c x 2 2 Buna göre, f' ( 2+ ) vaSTBLBÀUS fonksiyonu türevlenebilir bir fonksiyon olduôV- na göre, a + b +DUPQMBNLBÀUS \" # $ \" # $ % & % & :PLUVS 1. \" 2. & 3. & 4. C 56 5. \" 6. \" 7. D 8. &

1BSÀBM'POLTJZPOMBSO5ÑSFWJ TEST - 7 1. ôFLJMEFZ=G Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõUJS 4. f ( x ) = | x2 - 4 | + | x2| y y = f(x) GPOLTJZPOVLBÀYSFFMTBZTJÀJOUÑSFWTJ[EJS 5 \" # $ % & –7/2 O 5 x 6 –1 –1 Buna göre, f' ( -1- ) + f' ( 0 ) + f' ( 7+ UPQMBNOO 5. | | |f ( x ) = x2 - 1 + x - 2 | EFôFSJLBÀUS GPOLTJZPOVLBÀYSFFMTBZTJÀJOUÑSFWTJ[EJS \" # $ % & \" # $ % & Z x2 - 3 x#1 ] 1<x<2 ] 2. f^ x h = [ - 2x x$2 ]] \\ x2 - 6x fonksiyonu veriliyor. Buna göre, 6. f ( x ) = | x | + | x - 1 | + | x - 2 | I. f fonksiyonu r x ` R için süreklidir. II. f' ( 2 ) = -2 dir. GPOLTJZPOVOVOUÑSFWTJ[PMEVôVYEFôFSMFSJUPQ- III. f' ( 1+ ) = -2 dir. MBNLBÀUS IV. f' ( 3 ) =ES \" # $ % & JGBEFMFSJOEFOLBÀUBOFTJEPôSVEVS \" # $ % & f^ x h - 2x f^ x h > g^ x h | |7. f ( x ) = x3 - 8 - x2 f^ x h # g^ x h 3. h^ x h = * g^ x h.x2 fonksiyonu için f ' ( -1 ) +Gh UPQMBNOOFöJ- UJLBÀUS GPOLTJZPOVUBONMBOZPS A) - # - $ % & Buna göre, g ( x ) = 2x + 3 ve f ( x ) = x2 fonk- TJZPOMBS için lim h^ x h- h^ -1 h ifadesinin + x+1 x \" –1 TPOVDVWBSTBLBÀUS \" :PLUVS # - $ % & 1. \" 2. C 3. C 57 4. C 5. D 6. D 7. D

TEST - 8 1BSÀBM'POLTJZPOMBSO5ÑSFWJ | |1. f ( x ) = x3 - x2 5. f ( x ) = | x | . x3 + PMEVôVOBHÌSF Gh +Gh EFôFSJWBSTBLBÀ- fonksiyonu için f '' ( 2 ) + f '' ( -1 ) ifadesinin de- US ôFSJLBÀUS \" :PLUVS # $ \" # $ % & % & 2. ôFLJMEFZ=G Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõUJS 6. y =G Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõUJS y y y = f(x) 2 y = f(x) O x 1 5 –2 O –6 –1 2 3 x –2 | |Buna göre, y = f ( x ) fonksiyonunun türevsiz –3 PMEVôVOPLUBMBSOBQTJTMFSJOJOUPQMBNLBÀUS \" # $ - % -4 E) -1 | |Buna göre, g ( x ) = f ( x ) fonksiyonu ile ilgili olarak; I. g' ( 0 ) yoktur. II. ZHh Y GPOLTJZPOVOPLUBEBUBONT[ES III. y = g ( x ) fonksiyonu r x ` R için süreklidir. | |3. f ( x ) = x2 + 4x - k fonksiyonunun r x ` R için tü- JGBEFMFSJOEFOIBOHJMFSJLFTJOMJLMFEPôSVEVS SFWMJPMEVóVCJMJOJZPS \" :BMO[* # **WF*** $ :BMO[*** #VOB HÌSF L HFSÀFM TBZTOO BMBCJMFDFôJ EF- % *WF** & *WF*** ôFSMFS LÑNFTJOEFLJ en büyük JLJ UBN TBZOO UPQMBNLBÀUS A) - # - $ - % -12 E) -13 | |4. f ( x ) = x2 + bx +D fonksiyonunun türevsiz oldu- 7. I. y = f ( x ) fonksiyonu r x ` R için türevliyse óVOPLUBMBSOBQTJTMFSJOJO¿BSQNES | |y = f ( x ) fonksiyonu da türevlidir. #VOB HÌSF C HFSÀFL TBZTOO BMBCJMFDFôJ EF- II. y = f ( x ) fonksiyonu r x ` R için türevliyse ôFSLÑNFTJBöBôEBLJMFSEFOIBOHJTJEJS | |y = f ( x ) fonksiyonu da türevlidir. A) 6- 2 6, 2 6 @ # ^ - 3, - 2 6 h , ^ 2 6, 3 h $ [ Þ % -Þ III. y = f ( x ) fonksiyonu x = a için türevsiz ise | |y = f ( x ) fonksiyonu da türevsizdir. | |IV. y = f ( x ) fonksiyonu rx ` R için türevli ise G N G O FõJUTJ[MJóJOJTBóMBZBONWFOHFS- ¿FLTBZMBSZPLUVS JGBEFMFSJOEFOLBÀUBOFTJLFTJOMJLMFEPôSVEVS E) ( -Þ \" # $ % & 1. D 2. & 3. B 4. B 58 5. \" 6. B 7. \"

;JODJS,VSBMWF#JMFöLF'POLTJZPOVO5ÑSFWJ TEST - 9 1. x f ( x ) f ' ( x ) g ( x ) g ' ( x ) 5. f ( 3x + 1 ) = 2x - 1, ( gof ) ( x ) = x2 - 2x ve g ' ( a ) = 9 134 2 PMEVôVOBHÌSF BHFSÀFMTBZTLBÀUS 3 –3 7 2 \" # $ % & :VLBSEBLJUBCMPZBHÌSF HPG h EFôFSJkaç- US A) - # - $ - % & 6. f ( x ) = x2 + 2x + 1, g ( x ) = 2x - 1 ve ( fog ) ' ( a ) = ( gof ) ' ( 2 ) PMEVôVOBHÌSF BHFSÀFMTBZTLBÀUS 2. f ( x2 ) = 2 g ( 3 - 4x ), g ' ( 1 ) = 3 A) 4 # 3 $ 7 % & 3 24 PMEVôVOBHÌSF f' c 1 mEFôFSJLBÀUS 4 A) - # - $ - % - & -24 y = 2t3 - 3t - 10 _ b 7. t = u ` b u = 3x + 1 a 3. g ( x ) = x + 2 ve f ( x + 1 ) = x2. g ( 2x ) oldVôVOBHöre, dy JGBEFTJOJOFöJUJLBÀUS PMEVôVOBHÌSF Gh EFôFSJLBÀUS dx x = 1 A) - 63 # - $ - 15 4 4 \" # $ % & % 15 E) 63 4 4 4. f^ x h = x2 + 2x + 1 8. f : R Z R, f ( x ) = 3 x g^ x h = x2 + 1 fonksiyonu veriliyor. PMEVôVOBHÌSF ( fog ) ' ( 2 ) deôFSJLBÀUS ( gof ) ( x ) = x \" # $ % & PMEVôVOBHÌSF Hh EFôFSJLBÀUS A) 1 # 1 $ 1 % & 27 9 3 1. D 2. & 3. & 4. D 59 5. \" 6. B 7. & 8. &

TEST - 10 ;JODJS,VSBMWF#JMFöLF'POLTJZPOVO5ÑSFWJ 1. y =G Y EPóSVTBMGPOLTJZPOVJ¿JO fog^ x h r ( fof )' ( 2 ) = 4 4. h^ x h = r ( fof ) ( x ) = mx + 9 r f ( 1903 ) > f ( 1907 ) g^ x h bilgileri veriliyor. FõJUMJóJOJTBóMBZBOGWFHUÐSFWMFOFCJMJSGPOLTJZPOMBS #VOBHÌSF G N LBÀUS için g ( 0 ) = g'( 0 ) = 1 ve h' ( 0 ) = 1 dir. A) - # - $ % & Buna göre, y = f ( 2x ) . g ( 1 - 2x ) fonksiyonunun x = 1 noktaTOEBLJUÑSFWJLBÀUS 2 A) - # - $ 1 % & 2 2. f^ x h = x + 1 , g ( x ) = x2 - 2x ve h ( x ) = x4 fonk- 5. f^ x h = x4 + x2 + 1 fonksiyonu veriliyor. siyonlaSWFSJMJZor. x4 - x2 + 1 Buna göre, d 6hofog^ x h@JGBEFTJOJOFöJUJBöB- Buna göre, f'^ 1 h f^ 1 h dx LBÀUS - JGBEFTJOJO FöJUJ ôEBLJMFSEFOIBOHJTJEJS f'^ - 1 h f^ - 1 h A) 4 ( x - 1 )3 # Y+ 1)3 $ Y+ % Y- 1 ) A) - # - $ % & E) 4 ( x2- 1 ) 2x 6. k = {1, 2, ..., n} olmak üzere, 3. y =G Y GPOLTJZPOVOVOHSBGJóJõFLJMEFLJHJCJEJS fk( x ) = kx + k2 - k +GPOLTJZPOMBSUBONMBOZPS y y = f(x) d a f1of2of3o. . .ofn^ x h k dx O1 x –4 2 1 ifadesinin x =J¿JOEFóeri 72PMEVóVOBHËSF Buna göre, y = f ( -x2 + 2x - 1 ) fonksiyonu kaç ( n + 1 ) fn ( x ) - f1 ( x ) ifadesinin x =JÀJOFöJUJ OPLUBEBUÑSFWTJ[EJS LBÀUS \" 3 # 3 $ 3 - 1 \" # $ % & % 3 - & 3 + 1 1. \" 2. \" 3. B 60 4. & 5. \" 6. \"

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 .0%·- ·/÷7&34÷5&:&)\";*3-*, 5·3&75&03&.-&3÷ 0SUBMBNB%FôFS5FPSFNJ 3PMMF5FPSFNJ TANIM TANIM y = f ( x ) fonksiyonu [ a, b ]BSBMóOEBTÐSFLMJWF y = f ( x ) fonksiyonu [ a, b ]BSBMóOEBTÐSFLMJWF B C BSBMóOEBUÐSFWMJPMTVO B C BSBMóOEBUÐSFWMJPMTVO f'^ c h = f^ b h - f^ a h b-a f ( a ) =G C JTFGh D =FõJUMJóJOJTBóMBZBOFO õBSUOTBóMBZBOFOB[CJSD` B C HFS¿FMTB- B[CJSD` B C HFS¿FMTBZTWBSES ZTWBSES yy y y = f(x) f(a) = f(b) f(b) f(a) = f(b) O a c b x O a c1 c2 c3 b x f(a) b x ÖRNEK 3 O ac f : [ 0, 2 ] Z [-1, 0], f ( x ) = x2 - 2xGPOLTJZPOVOVO y y = f(x) 3PMMFUFPSFNJOJTBôMBZBOD` HFSÀFMTBZTOO f(b) FöJUJOJCVMVOV[ G =G =PMEVôVOEBOD` Gh D =PMVS D-=jD= f(a) c2 c3 b x O a c1 ÖRNEK 1 ÖRNEK 4 f : [ 0, 3] Z [-1, 3] f : [0, 1] Z R f ( x ) = x2 - 2xGPOLTJZPOVOVOUBONLÑNFTJOEFPS- f ( x ) = x3 - 6x2 +YGPOLTJZPOVWFSJMJZPS UBMBNBEFôFSUFPSFNJOJTBôMBZBOHFSÀFMTBZLBÀUS #VOBHÌSF Z=G Y GPOLTJZPOVJÀJO3PMMFUFPSFNJOJ TBôMBZBOD` HFSÀFMTBZTLBÀUS G Y =Y2JÀJOGh Y =Y- G =G =PMEVôVOEBOsD` Gh D =PMVS f^ 3 h - f^ 0 h = f'^ c hj 3 - 0 = 2c - 2 D2-D+5=j c 1, 2 12 ± 144 - 60 3-0 3-0 = 6 3 6 - 21 =D-jD= D1= 3 2 ÖRNEK 2 ÖRNEK 5 f : [ 1, 2 ] Z [ 1, 4 ],f ( x ) = x2GPOLTJZPOVWFSJMJZPS f ( x ) = x3 - x GPOLTJZPOVOVO[- ]BSBMôOEB3PM- #VOBHÌSF CVGPOLTJZPOVOPSUBMBNBEFôFSUFPSFNJ- MFUFPSFNJOJTBôMBZBODHFSÀFMTBZEFôFSMFSJOJOÀBS- OJTBôMBZBOD` HFSÀFMTBZTLBÀUS QNLBÀUS G Y =Y2JÀJOGh Y =Y G - =G =jsD`[- ]Gh D = f'^ c h = f^ 2 h - f^ 1 h & 4 - 1 & c = 3 D2-=jD1= 1 D2= - 1 33 2-1 2-1 2 11 1 D1D2= ·- =- 3 33 33 61 1 6 - 21 - 1 33 22

TEST - 11 5ÑSFW5FPSFNMFSJ 1. f^ x h = x3 + x2 + x + 1 5. -2 < a <PMNBLÐ[FSF 3 f ( x ) = x3 -YGPOLTJZPOVWFZ=BEPóSVTVOVO LFTJNOPLUBMBSY1, x2WFY3UÐS GPOLTJZPOVOVO[ ]BSBMôOEBPSUBMBNBEFôFS UFPSFNJOJTBôMBZBODHFSÀFMTBZTLBÀUS A) 7 - 1 B) 7 + 1 C) 6 - 1 Y1<Y2<Y3PMEVôVOBHÌSF GPOLTJZPOVOVO D) 1 + 6 E) 2 [Y1 Y] BSBMôOEB SPMMF UFPSFNJOJ TBôMBZBO D HFSÀFM TBZ EFôFSMFSJ BöBôEBLJMFSEFO IBOHJTJ- EJS A) - 3 ve 3 B) - 2 ve 2 2. f ( x ) = x2 +YGPOLTJZPOVWFSJMJZPS C) - 5 ve 5 D) -WF #VOBHÌSF \" WF# OPLUBMBSOEBOHF- & 4BEFDF ÀFO EPôSVZB QBSBMFM WF Z = G Y GPOLTJZPOVOB UFôFUPMBOEPôSVOVOUFNBTOPLUBTOOPSEJOBU 6. BWFCUBNTBZPMNBLÐ[FSF LBÀUS y = ( x2 - 4x + 3 )2 fonksiyonunun [a, b]BSBMóOEB A) 1 B) 2 C) 3 D) 4 E) 5 3PMMFUFPSFNJOJTBóMBZBOGBSLMDHFS¿FLTBZEF- óFSJWBSES 3. f^ x h = x - 1 #VOBHÌSF B+CUPQMBNLBÀUS A) 2 B) 4 C) 5 D) 3 E) 6 x+1 7. y = x - 2 x GPOLTJZPOVOVO[ ]BSBMôOEBPSUBMBNBEFôFS GPOLTJZPOVOVO[ ]BSBMôOEB3PMMFUFPSFNJOJ UFPSFNJOJTBôMBZBODHFSÀFMTBZTLBÀUS TBôMBZBODHFSÀFMTBZTJÀJOG D LBÀUS A) -1 B) - 2 C) - 3 D) -2 E) -15 A) 1 B) 3 - 1 C) 2 - 1 2 3 2 D) E) 4 2 4. y = x3 - x2 8. f1( x ) = sinx, f2( x ) =DPTY 32 f3( x ) = x2 -ÕY G4( x ) = x- π GPOLTJZPOVOVOÐ[FSJOEFLJ\" B O WF# C N OPL- 2 UBMBSOEBOHF¿FOEPóSVOVOFóJNJEJS f5( x ) =Fx + 1 #V EPôSVOVO GPOLTJZPOV [B C] BSBMôOEB GPOLTJZPOMBSOEBOLBÀUBOFTJOJO[ Ö]BSBMôO- OPLUBEBLFTUJôJCJMJOEJôJOFHÌSF B C BSBMôO- EBFOB[CJSUBOFDHFSÀFMTBZTJÀJOUÑSFWJOJO EBPSUBMBNBEFôFSUFPSFNJOJTBôMBZBODHFSÀFM PMEVôVOV3PMMFUFPSFNJOJLVMMBOBSBLHÌTUFSFCJ- TBZMBSOOGBSLOOQP[JUJGEFôFSJLBÀUS MJSJ[ A) 1 B) 2 C) 3 D) 4 E) 5 A) 4 B) 1 C) 3 D) 0 E) 2 A C C C 62 % B A &

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 .0%·- ·/÷7&34÷5&:&)\";*3-*, 5·3&7÷/'÷;÷,4&-:036.6 5ÑSFWJO'J[JLTFM:PSVNV ÖRNEK 3 TANIM ,ÑSFöFLMJOEFLJCJSCBMPOVOZBSÀBQDNEFODN ZFÀLBSMSTBIBDNJOEFLJEFôJöJNJOZBSÀBQOBHÌSF y y = f(x) x PSUBMBNBEFôJöJNI[LBÀUS y1 x1 Dy Dx 4 3 4 3 4 y0 Dx x0 Dt DV ·4 π - ·2 π π ·56 V= = 33 3 112π = = 2 x t ort Dr 4-2 23 x1 t1 cm O x0 O t0 ôFLJM* ôFLJM** 0SUBMBNBEFóJõJNI[CJSOJDFMJóJOEFóJõJNJOJO CBõLB CJS OJDFMJLUF EFóJõJNF LZBTMB PSUBMBNB ÖRNEK 4 OFLBEBSPMBDBóOHËTUFSFOPSBOES #JSIBSFLFUMJOJOLPOVN-[BNBOEFOLMFNJY=U2 +UEJS ôFLJM*EFHËSÐMFDFóJÐ[FSFCVPSBO #VOBHÌSF CVIBSFLFUMJOJOU1=WFU2=TBOJZF- MFSJBSBTOEBLJPSUBMBNBEFôJöJNJI[LBÀNTOEJS m = y1 - y0 = Dy JMFJGBEFFEJMJS x1 - x0 Dx #VLBWSBNEBIBË[FMMFõUJSNFLJTUFSTFLCJSIB- SFLFUMJOJO LPOVNVOEBLJ EFóJõJNJO [BNBOEBLJ a 2 + 6.5 k - a 2 2 + 6.2 k 55 - 16 DX 5 V= = = ort Dr 5-2 3 EFóJõJNF PSBOO PSUBMBNB I[ PMBSBL BEMBOE- SBCJMJSJ[ Vort = Dx JMFJGBEFFEFCJMJS ôFLJM** 39 Dt = = 13 m/s 3 ÖRNEK 1 ÖRNEK 5 #JSIBSFLFUMJTBBUUFLNI[MBTBBU TBBUUFLNI[- ôFLJMEFLJHSBGJLUFCJSIBSFLFUMJOJOTBBUCPZVODBI[OO MBTBBUJMFSMFNJõUJS [BNBOBHËSFEFóJõJNJHËTUFSJMNJõUJS #VOBHÌSF CVIBSFLFUMJOJOZPMCPZVODBPSUBMBNBI- 7 LNTBBU [LBÀLNTBBUUJS 100 Dx 40.3 + 30.2 180 V= = = = 36 km/sa 50 ort Dt 3+2 5 30 34 6 U TBBU O1 ÖRNEK 2 #VOB HÌSF m TBBU BSBTOEBLJ WF m TBBUMFSJ BSB- TOEBLJI[OOPSUBMBNBEFôJöJNMFSJOJ JWNF IFTBQ- 0DBLEFPMBOCJSJMJOIBWBTDBLMó0DBL MBZO[ EF-PMEVóVOBHËSF CVJMJOIBWBTDBLMôHÑO- EFPSUBMBNBLBÀEFSFDFEÑöNÑöUÑS DT - 4 - 8 12 a = DV = 30 - 0 = 5 2 V= = = - = - 3 derece/gün ort Dt 6-0 km/sa ort Dt 20 - 16 4 a = DV = 100 - 50 = 50 2 ort Dt 4-3 km/sa LNTBmEFSFDFHÑO 63 112π DN2NTO5.LNTB LNTB2 3

·/÷7&34÷5&:&)\";*3-*, .0%·- -÷.÷57&5·3&7 www.aydinyayinlari.com.tr TANIM ÖRNEK 7 #JSOJDFMJLUFLJEFóJõJNJOCBõLBCJSOJDFMJLUFEF- :FSEFOZVLBSZBEPóSVGSMBUMBOCJSUPQVO[BNBOBCBóM óJõJNFHËSFBOMLOFLBEBSEFóJõUJóJOJHËTUFSFO ZPMEFOLMFNJx = -U2 +UEJS orana BOMLI[EFOJS0SUBMBNBEFóJõJNI[O- 5PQVOLBÀODTBOJZFEFLJI[NTOEJS EBO GBSLM PMBSBL CVSBEB EFóJõJNMFS ¿PL LÐ¿ÐL PMNBMES 4GSB ¿PLZBLO dx V = = - 10t + 80 j-U+=jU= x dt x2 Dx ÖRNEK 8 x1 Dt t x0 t1 t2 :FSEFO EJL PMBSBL ZVLBS GSMBUMBO CJS UBõO U TBOJZFEF- O t0 LJZFSEFOZÐLTFLMJóJ I U = -U2 +U NFUSF GPSNÐMÐZ- MFWFSJMNJõUJS U1J¿JOBOMLI[IFTBQMBOSLFOZVLBSEBLJHSB- #VOBHÌSF UBöOZFSFÀBSQNBI[LBÀNTOEJS GJLUFO EF BOMBõMBDBó Ð[FSF IFSIBOHJ CJS TB- I U =JÀJO-U2+U=jU= U= CJUU2 EFóFSJJ¿JOPSUBMBNBI[IFTBQMBNBLCJ[F dh ZBOMõTPOV¿WFSFDFLUJS V = = - 2t + 20 V = x2 - x1 dt t2 - t1 7=-+=-NTO U1 =U2BMONBTEVSVNVOEBEBIFTBQMBNBZB- QMBNB[ ÖRNEK 9 #V EVSVNEB UÐSFW LPOVTVOEBO ZBSEN BMBSBL s =G U =U2 +U- 1 GPOLTJZPOVJMFIBSFLFUFEFOCJS IBSFLFUMJOJOU=WFU=BOOEBLJBOMLI[WFBO CVTPSVOV¿Ë[FDFóJ[ MLJWNFTJOJCVMVOV[ UTBOJZF TNFUSFJMFËM¿ÐMÐZPS V = lim x-x = dx ds V = = 6t + 5 1 dt anlık t\"t t - t1 dt U=JÀJO7=NTO U=JÀJO7=NTO 1 a = dV = 6 jU=WFU=JÀJONT2 ÖRNEK 6 dt 4BBUUFLNI[MBZVLBSZBEPóSVBUMBOCJSDJTNJOZFS- ÖRNEK 10 EFOZÐLTFLMJóJOJWFSFOEFOLMFNI U =U-U2JMFJGB- EFFEJMNJõUJS ;BNBOBCBôMZPMEFOLMFNJY=-U2+U-PMBO IBSFLFUMJOJOI[OOTGSPMEVôVBOEBIBSFL FUFCBö- #VOB HÌSF U = TO BOOEB CV DJTNJO I[ LBÀ NTO MBEôOPLUBZBPMBOV[BLMôLBÀNFUSFE JS PMVS dx dh V = = 0 & - 12t + 60 = 0 jU= V = = 40 - 10t j-=NTO dt dt U=JÀJOY=- U=JÀJOY=-+-= 6[BLMLNFUSF NTO mNTONTO NTO NTO2 NTO2 64 N

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 .0%·- ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 11 ÖRNEK 15 #JSLBSFOJOIFSCJSLFOBSDNTOPSBOZMBBSUNBLUBES N V[VOMVóVOEB CJS NFSEJWFO EVWBSB EBZBM PMBSBL ,BSFOJO CJS LFOBS DN PMEVôVOEB LBSFOJO BMBO EVSNBLUBES IBOHJPSBOEBBSUBS 13 V = dA = 2a· da ==DN2TO dt dt EVWBS ÖRNEK 12 ZFS #JSEBJSFOJOBMBO 3 ÕDN2TOPSBOZMBBSUNBLUBES .FSEJWFOJOEVWBSBEFóFOBZBóEVWBSEBO- 1 m/snI- 2 2 %BJSFOJO BMBO Ö DN2 PMEVôVOEB ZBSÀBQ IBOHJ [ZMBLBZBSLFO NFSEJWFOJOZFSFEFóFOBZBóEVWBSEBn V[BLMBõNBLUBES PSBOMBBSUBS .FSEJWFOJO ÑTU VDV ZFSEFO N ZÑLTFLMJLUFZLFO ÖS2=ÖjS= NFSEJWFOJO BMU VDV EVWBSEBO LBÀ NTO I[MB V[BL- MBöS dA dr V = = 2πr· dt dt dr 3 dr 1 %VWBSBV[BLMôY ZFSEFOZÑLTFLMJôJIPMTVO 2 π ·3· = π j = DNTO dt 2 dt 4 Y2+I2=jI=DN Y= dx dh 2x· + 2h· = 0 dt dt ÖRNEK 13 5· dx + 12d - 1 n = 0 j dx = 6 NTO dt 2 dt 5 ,ÐSFõFLMJOEFCJSCBMPOHFOJõMFNFLUFEJS ÖRNEK 16 ,ÑSFOJOZBSÀBQDNTOPSBOZMBBSUUôOBHÌSF LÑ- SFOJOZBSÀBQDNPMEVôVOEBIBDNJOEFLJBOMLEF- N ôJöJNPSBOLBÀUS V= dH = 4 2 dr =Ö2=DN3TO ,POJ õFLMJOEFLJ CJS TV EFQPTV- OVOUBCBOZBS¿BQNWFZÐL- dt 3 ·3π .r · dt TFLMJóJ N EJS 4V EFQPTVna ÖRNEK 14 N 3 N3ELI[ZMBTVQPNQBMBO- #JSEJLEËSUHFOJOHFOJõMJóJTBOJZFEFDNB[BMSLFO CPZV 2 TBOJZFEFDNB[BMNBLUBES NBLUBES #PZVO DN HFOJöMJôJ DN PMEVôV BOEB LÌöFHFO %FQPEB UBN N TFWJZFTJOEF TV PMEVôV BOEB TV V[VOMVôVOVOBOMLEFôJöJNPSBOOFPMVS ZÑLTFLMJôJOJOBSUöI[OFEJS (FOJöMJôJB CPZVCLÌöFHFOJLPMTVO r h6 h B2+C2=L2 h = & =r da db dk r2 3 2a· + 2b· = 2k· )BDJN= 1 ÖS2I= 1 ÖI3 dt dt dt 3 27 2.6.^ - 2 h + 2.8.^ - 4 h = 2.10· dk = - 4, 4 cm/sn dH 1 ÖI2p dh dt V= = dt 27 dt 1 dh 3 dh 27 & π . 3 .4· = j = m/dk 27 dt 2 dt 8π DN2TO 1 DNTODN3TOm DNTO 65 6 27 4 NTO NEL 5 8π

TEST - 12 5ÑSFWJO'J[JLTFM:PSVNV 1. :FSEFO EJL PMBSBL ZVLBS BUMBO CJS IBSFLFUMJOJO U 4. #JS LÑSFOJO ZÑ[FZ BMBOOO Ö DN2TO WF ZBS- TBOJZFEFLJZFSEFOZÐLTFLMJóJ ÀBQOO DNTO I[ZMB BSUUô BOEB LÑSFOJO IBDNJIBOHJI[MBEFôJöJS I U =U-U2GPSNÐMÐZMFWFSJMNJõUJS A) 5π 4π 16π D) 5π & Õ #VOBHÌSF UBöOJMLTOJÀFSJTJOEFLJPSUBMBNB 4 B) C) I[LBÀNTEJS 5 52 A) 40 B) 10 C) 20 D) 15 E) 5 2. Y LN 55 50 30 U TBBU 5. %Ð[CJSZPMEBCJS\"OPLUBTOEBOZÐSÐNFZFCBõMB- 20 17 ZBO CJS IBSFLFUMJOJO [BNBO J¿FSJTJOEF \" OPLUBTO- EBOOFLBEBSV[BLUBCVMVOEVóVOVHËTUFSFOEFOL- 0 t1 t2 t3 t4 t5 t6 t7 MFNY=U3 -U2 +UJMFWFSJMNJõUJS :VLBSEBLJLPOVN-[BNBOHSBGJLMFSJOEF #VOBHÌSF CVIBSFLFUMJTOJÀFSJTJOEF\"OPL- UBTOEBOFOÀPLLBÀNFUSFV[BLMBöNöUS i `;WFâJâJ¿JOUiTBZMBSBSEõLTBZMBSES )[OFLTJ¿LUóBSBMLUBIBSFLFUJOUFSTZËOEFZB- \" # $ % & QMEóO CJMFO öMLFS PSUBMBNB I[MBS IFTBQMBSLFO IBSFLFU ZËOÐOÐ JINBM FEJQ TPOV¿MBS QP[JUJG CVMV- ZPS #VOB HÌSF BöBôEBLJ [BNBO BSBMLMBSO IBOHJ- TJOEFPSUBMBNBI[FOB[CVMVS \" mU1 # U3 -U4 $ U6 -U & U1 -U2 % U4 -U5 3. V0JMLI[ZMBBõBóEPóSVBUMBOCJSDJTNJOU[BNBO 6. 4JMJOEJSõFLMJOEFLJCJSTVEFQPTVOVOUBCBOZBS¿B- J¿JOEFBMEóZPMEFOLMFNJ QNZÐLTFLMJóJNEJS%FQPZB ÕN3ELI[- S = V0t + 1 gt2 EJS(g: yer çekimi ivmesi) MBTVBLUBOCJSNVTMVLUBLMZPS 2 #VOBHÌSF TVTFWJZFTJOJOZFSEFONZÑLTFL- MJLUF PMEVôV BOEB TVZVO ZÑLTFLMJôJOJO BOML EFôJöJNI[LBÀNELEJS H = NTO2 PMNBL Ñ[FSF 7 = NTO I[- \" # $ % & MBBöBôBUMBOCVDJTNJOTBOJZFEFLJI[LBÀ NTOEJS \" # $ % & B C & 66 A & %

5ÑSFWJO'J[JLTFM:PSVNV TEST - 13 1. ,FOBSV[VOMVLMBSDNWFDNPMBOEJLEËSUHFOJO 4. #JSPCKFY2 + y2 =¿FNCFSJ¿FWSFTJOEFTBBUZË- LTB LFOBS DNTO I[MB BSUBSLFO V[VO LFOBS 22 DNTOI[MBBSUUSMZPS OÐOEFIBSFLFUFUNFLUFEJS0CKFf , p nok- #VOB HÌSF EJLEÌSUHFOJO V[VO LFOBSOO DN 22 PMEVôV BOEB BMBOOO BOML EFôJöJN I[ LBÀ UBTOEBO HF¿FSLFO PCKFOJO QP[JTZPOVOV JGBEF DN2TOEJS FEFO Y Z OPLUBTOO Y LPPSEJOBUOO TBOJZFEF CS PSBOZMB BSUUó CJMJOEJóJOF HËSF Z LPPSEJOBU \" # $ % & IBOHJPSBOEBEFôJöJS 2. ôFLJMEF CJS BZSUOO V[VOMVóV NFUSF PMBO LBSF \" CSTOBSUBS # CSTOBSUBS CJ¿JNJOEF CFZB[ CJS QFSEF WBSES %Ð[FOFóF ZBS- $ CSTOBSUBS % CSTOB[BMS ¿BQNFUSFPMBOTBZEBNPMNBZBOCJSLÐSFZFSMFõ- UJSJMNJõUJS,ÐSFOJONFSLF[JJMFQFSEFOJOBóSMLNFS- & 4BCJULBMS LF[JBSBTOEBLJV[BLMLNFUSFEJS N O1 N 5. :BS¿BQV[VOMVóVDN ZÐLTFLMJóJDNPMBOCJSLP- N N OJOJOZBS¿BQZÐLTFLMJLPSBOTBCJULBMBDBLõFLJMEF #VLÐSFOJOWFLBSFOJOBóSMLNFSLF[JJMFBZOEPó- ZBS¿BQ DNTOI[MBBSUUSMZPS SVMUV Ð[FSJOEF LBMBDBL õFLJMEF LÐSFOJO NFSLF[JO- #VOBHÌSF ZBSÀBQOODNPMEVôVBOEBIBD- EFONFUSFV[BLMóBOPLUBTBMCJSõLLBZOBóLPO- NJOJOEFôJöJNJOJOBOMLI[LBÀDN3TOPMVS NVõUVS %BIB TPOSB õL LBZOBó LÐSFZF NTO \" Õ # Õ $ Õ % Õ & Õ I[MBZBLMBõUSMZPS 6. #JSBZSUOOV[VOMVóVDNPMBOCJSLÐQÐ BZSUMBS- #VOBHÌSF TBOJZFEFQFSEFÑ[FSJOEFPMVöBO OO[BNBOBHËSFEFóJõJNEFOLMFNJ a =U2 +U+ 1 EBJSF öFLMJOEFLJ HÌMHFOJO BMBOOO BOML EFôJ- EFOLMFNJJMFWFSJMNJõUJS öJNI[LBÀN2TOPMVS #VOB HÌSF CV LÑQÑO CJS BZSUOO DN PMEV- ôVBOEBIBDNJOJOEFôJöJNJOJOBOMLJWNFTJLBÀ A) 30 3 π # Õ $ 6 3 π DN3TO2PMVS \" # $ % & % Õ & Õ 3. U>PMNBLÐ[FSF Y U =U2 +U+ 10 EFOLMFNJJMFTBCJUJWNFMJCJSIBSFLFUMJOJOLPOVN–[B- NBOEFOLMFNJWFSJMNJõUJS #VOB HÌSF CV IBSFLFUMJOJO JWNFTJ LBÀ NT2 EJS \" # $ % & A % % 67 % A A

·/÷7&34÷5&:&)\";*3-*, .0%·- -÷.÷57&5·3&7 www.aydinyayinlari.com.tr 5·3&7÷/(&0.&53÷,:036.6 TANIM ÖRNEK 1 y dt f ( x ) = x3 - 2x + 1 GPOLTJZPOVOB\" OPLUBTO B(x1, y1) EBOÀJ[JMFOUFôFUJOEFOLMFNJOJWFOPSNBMJOEFOLMF- a y = f(x) NJOJCVMVOV[ O A(x0, f(x0)) G =PMEVôVOEBO Gh Y =Y2- dn Gh =-==NU x Z-= Y- Z=Y-UFôFUEFOLMFNJ :VLBSEBLJ HSBGJLUF Z = f ( x ) fonksiyonu, bu N5N/=-jN/=-jN/=-1 Z-=- Y- Z=-Y+OPSNBMEFOLMFNJ GPOLTJZPOB\"OPLUBTOEBUFóFUPMBOEUEPóSVTV WF\"OPLUBTOEBLJOPSNBMJPMBOEn EPóSVTVWF- ÖRNEK 2 SJMNJõUJS/PSNBMEPóSVTV UFóFUEPóSVTVOBEJL- f : R Z R , G Y =|-Y2|GPOLTJZPOVOVOBQTJTJ UJS PMBO OPLUBTOEBLJ UFôFUJOJO EFOLMFNJOJ WF OPSNBMJ- EUUFóFUEPóSVTVOVOFóJNJ OJOEFOLMFNJOJCVMVOV[ r mAB = f^ x0 h - y1 \"OBMJUJLHFPNFUSJ \"QTJTJPMBOOPLUBOOPSEJOBUG =-= x0 - x1 `[- ]PMEVôVOEBO r UBOa 5SJHPOPNFUSJ G Y =-Y2jGh Y =-Y f^ x h- f^ x h Gh =-UFôFUJOFôJNJ Z-=- Y- jZ=-Y+UFôFUEPôSVTV r f'^ x h = lim 0 5ÐSFW x\"x x - x0 0 GPSNÐMMFSJZMFCVMVOVS 5ÐSFWGPSNÐMÐJ¿JOLVMMBOMBO\"OPLUBTOOFóSJ- N5N/=-jN5=-j m = 1 Y- ZJ TBóMBEóOB EJLLBU FEJMNFMJEJS 5FóFU EPóSV- TVÐ[FSJOEFFóSJZJTBóMBNBZBOOPLUBMBSJ¿JOUÐ- / 2 SFW GPSNÐMÐ LVMMBOMNB[ ±ODF UFNBT OPLUBT IFTBQMBOS 15 Z-=Y-j y = x + OPSNBMEPôSVTV 22 EnOPSNBMEPóSVTVOVOFóJNJJTFEJLEPóSVMBSO ÖRNEK 3 FóJNMFSJOEFOCVMVOVS Z= Y- 2FôSJTJOF\" OPLUBTOEBOÀJ[JMFOUF- NTN/ = -1 ôFUMFSJOEFOLMFNMFSJOJCVMVOV[ 5FóFUWFOPSNBMEPóSVMBSOOEFOLMFNMFSJBOB- MJUJLHFPNFUSJEFOZBSBSMBOMBSBLCVMVOVS Gh Y = Y- PMVS\"ODBL\" OPLUBTFôSJZJTBôMB- NBEôOEBOFôSJÑ[FSJOEF# O L OPLUBTBMOS ET : y - f (x ) =NT Y- x) m = k - 0 = f'^ n hPMVS T n-3 0 0 y - f ( x0 ) =Gh Y0 ) ( x - x0 ) k - 0 = 2^ n - 2 h & ^ n - 2 h2 = 2^ n - 2 h n-3 n-3 E/: y - f ( x0 ) =N/ Y- x0 ) O=WFZBO-= O- y - f(x ) = - 1 ^x - x0h O-=-+OjO=CVMVOVS#VEVSVNEB OPL- f'^ x0 h UBTOEBOHFÀFOUFôFUJOFôJNJPMVSLFO OPLUBTO- 0 EBOHFÀFOUFôFUJOFôJNJPMVSZ-= Y- jZ= Z-= Y- jZ=Y-UFôFUMFSJCVMVOVS 68 1.ZYmUFôFU ZmYOPSNBM2.ZmYUFôFU x+5 Z OPSNBMZ ZYm 2

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 .0%·- ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 4 ÖRNEK 7 Z=Y2æ-Y+FôSJTJOJOZ=Y+EPôSVTVOBQB- f ( x ) = x3 - ( n - 2 ) x2 + 12x + 1 SBMFMUFôFUJOJOEFOLMFNJOFEJS GPOLTJZPOVOVO Y FLTFOJOF QBSBMFM UFôFUJ CVMVONB- EôOB HÌSF O UBN TBZMBSOO BMBCJMFDFôJ LBÀ GBSL- Z=Y+EPôSVTVOBQBSBMFMEPôSVOVOFôJNJUÑS MEFôFSWBSES Gh Y =Y-PMVS Y-= Y=UFôFUOPLUBTOOBQTJTJ Gh Y =EFOLMFNJOJOÀÌ[ÑNLÑNFTJCPöLÑNFPMNBMES G =-+=PSEJOBU Gh Y =Y2- O- Y+= Z-= Y- Z=Y- D< 4^ n - 2 h2 - 4 .3.12 < 0 O- 2-< O+ O- < ÖRNEK 5 –4 8 y = f ( x ) = x3 - 5x2 + 6x - 1 + –+ FôSJTJOJOY+Z=EPôSVTVOBEJLUFôFUMFSJOJOEFô- NFOPLUBMBSOOBQTJTMFSJÀBSQNLBÀUS - BSBTOEBUBNTBZWBSES 2 ÖRNEK 8 Y+Z=EPôSVTVOVOFôJNJ - f ( x ) = x4 + 3x2 + bx +DGPOLTJZPOVOVOHSBGJôJY=- 33 BQTJTMJOPLUBTOEBYFLTFOJOFUFôFUPMEVô VOBHÌSF EJLEPôSVOVOFôJNJJTF PMNBMES C+DUPQMBNLBÀUS 2 Y=-EFYFLTFOJOFUFôFUJTFG - =Gh - = Gh Y =Y2-Y+ Gh Y =Y3+Y+C Y2-Y+= 3 --+C=jC= G - =+-+D=jD= 2 C+D=+= 29 3x - 10x + = 0 2 3 x .x = 12 2 ÖRNEK 6 ÖRNEK 9 f ( x ) = x2 - 3x + 2 f ( x ) = x2 +YGPOLTJZPOVOVOCBõMBOH¿OPLUBTOEBO¿J- [JMFOUFóFUJÐ[FSJOEFCJS\" O L OPLUBTBMOZPS GPOLTJZPOVÐ[FSJOEFLJ\" O L OPLUBTOEBO¿J[JMFOUFóFU YFLTFOJJMFQP[JUJGZËOMÐB¿ZBQNBLUBES | |OA = 4 5 CSPMEVóVOBHËSF LHFSÀFMTBZTOOBMB- #VOBHÌSF O+LUPQMBNLBÀUS CJMFDFôJQP[JUJGEFôFSLBÀUS y Gh Y =Y+ A(n,k) L=O2-O+ UBO=Gh O 4 5 2a N=UBOa=Gh = =O- L=-+ Oa x 5 a = 4 5 jB= a –2 =O L= L=B= O= jO+L= 3 69 118 ZYm 2 2

·/÷7&34÷5&:&)\";*3-*, .0%·- -÷.÷57&5·3&7 www.aydinyayinlari.com.tr ÖRNEK 10 ÖRNEK 12 f ( x ) = x3 + kx2 - 3x + 5 y ôFLJMEFZ= f ( x ) fonksiyo- GPOLTJZPOVÑ[FSJOEF\" O WF# - I OPLUBMBSO- y = f(x) nunun HSBGJóJWF\"OPLUB- EBOÀJ[JMFOUFôFUMFSCJSCJSJOFQBSBMFMPMEVôVOBHÌSF A TOEBLJUFóFUJWFSJMNJõUJS LHFSÀFMTBZTLBÀUS O 13 x Gh Y =Y2+LY- –2 Gh =Gh - 3 + 2k - 3 = 12 - 4k - 3 #VOBHÌSF H Y =Y2G Y FôSJTJOFÑ[FSJOEFLJ L= Y = BQTJTMJ OPLUBTOEBO ÀJ[JMFO UFôFUJO EFOLMFNJ 3 OFEJS L= 2 y y = f(x) Hh Y =YG Y +Y2Gh Y 2;f^ 3 h + 3<f'^ 3 h = 14 A x 42 24 2 13 H =G = Z-= Y- –2 Z=Y- ÖRNEK 11 ÖRNEK 13 y =G Y GPOLTJZPOVOVOHSBGJóJWF\"OPLUBTOEBLJUFóFUJ y ôFLJMEFZ= f ( x ) fonksi- õFLJMEFLJHSBGJLUFWFSJMNJõUJS ZPOVOVO HSBGJóJ WF \" y = f(x) 3 y = f(x) OPLUBTOEBLJ UFóFUJ WF- y x SJMNJõUJS A(n, 4) –3 –2 O 45° x O A –2 d1 g^ 2x - 4 h = f^ 3x h PMEVô VOBHÌSF Hh - LBÀUS g^ x h = x2 FöJUMJôJOJTBôMBZBOQP[JUJGHFSÀFMTBZ- x2 - 2 f^ x h MBSEBUBONMZ=H Y GPOLTJZPOVJÀJOHh O LBÀUS 4-^-2h y = 3 x + 3 UFôFUEPôSVTV f^ - 3 h = - 9 + 3 = - 3 ; = 1jO= 2 22 n-0 f'^ - 3 h = 3 G = Gh =UBO= 2 f'^ 3x h.3^ x2 - 2 h - 2x.f^ 3x h g'^ x h = 2x.f^ x h - x2f'^ x h = 12.4 - 36.1 = 12 = 3 g'^ 2x - 4 h.2 = ^ x2 - 2 h2 f2^ x h 16 16 4 3f'^ - 3 h.^ - 1 h + 2f^ - 3 h g'^ - 6 h.2 = ^ - 1 h2 96 15 Hh - = - - jHh - = - 22 4 33 15 ZYm - 4 24

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 .0%·- ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 14 ÖRNEK 17 ôFLJMEF Z = G Y WF Z = H Y GPOLTJZPOMBSOO HSBGJLMF- f ( x ) = x3 + x2 - 4x +GPOLTJZPOVÐ[FSJOEFLJ\" OPLUBTOEBO¿J[JMFOUFóFUFóSJZJCBõLBCJS# O L OPLUB- SJWFSJMN JõUJS TOEBLFTJZPS y y #VOBHÌSF OHFSÀFMTBZTLBÀUS y = f(x) y = g(x) 5 2 –1 –3 –1 O Gh Y =Y2+Y-jGh = O2 x x 32 k-2 =1 & n + n - 4n + 4 - 2 =1 n-1 n-1 #VOBHÌSF Z= GPH Y FôSJTJOFÑ[FSJOEFLJY=- n3 - 1 + n2 - 4n + 3 & =1 BQTJTMJ OPLUBTOEB ÀJ[JMFO UFôFUJO EFOLMFNJOJ CVMV- OV[ n-1 O2+O++O-=jO2+O-= O+ O- =jO=-3 G = H - = 5 Hh - = Gh = 3 GPH - =G = G2 5 5 Gh H Y Hh Y =Gh g^ - 1 h Hh - = ·1 = 33 ÖRNEK 18 5 y d Z-= Y- y = f(x) A 3 B x Z-=Y+jZ-Y-= O ÖRNEK 15 f ( x ) = x2 - 2x GPOLTJZPOVOVOLÌLMFSJOEFOÀJ[JMFOUF- ôFUMFSJOJOBSBTOEBLJEBSBÀOOUBOKBOULBÀUS Gh Y =Y- :VLBSEBLJõFLJMEFHSBGJóJWFSJMFOZ=G Y JLJODJEFSFDF- EFOGPOLTJZPOVJMFJMHJMJPMBSBL Gh =-WFGh = r EEPóSVTVGPOLTJZPOB\"OPLUBTOEBUFóFUPMVQFóJ- tan a = m -m + 2-^-2h =- 4 NJUÐS 12 r [ AB ]0Y 1 + m .m 1 + ^ - 2 h^ 2 h 3 r 'POLTJZPOVOVOBMBCJMFDFóJFOLÐ¿ÐLEFóFS-EJS 12 r f ( -1 ) = f ( 5 ) ÖRNEK 16 CJMHJMFSJWFSJMJZPS f ( x ) = x2 + 4x + 1QBSBCPMÑOÑOZ=Y-EPôSVTV- #VOBHÌSF G - LBÀUS OBFOZBLOPMEVôVOPLUBOOBQTJTJLBÀUS G - =G JTFG =G PMVS Y2+Y+=Y- NE=JTF-Gh =Gh =PMVS Y2-Y+=D< Gh Y =BY+C FO ZBLO OPLUB CV EPôSVZB QBSBMFM PMBO UFôFU ZBSENZ- Gh =-jC=-4 MBCVMVOVS Gh =jB-=jB= Y+=jY= G =-j-+D=-jD= Y=TJNFUSJFLTFOJ G Y =Y2-Y+ G - = 4 71 –38 ZmYm - 1 3

TEST - 14 5ÑSFWJO(FPNFUSJL:PSVNV 1. f ( x ) = x2 - 3x + 1 5. f ( x ) = x3 - 2x2 - 3x +FóSJTJWFSJMJZPS GPOLTJZPOVOBY=OPLUBTOEBÀJ[JMFOUFôFUJO Z=G Y GPOLTJZPOVOVOBQTJTJPMBOOPLUBTO- FôJNJLBÀUS EBLJOPSNBMJOJOEFOLMFNJBöBôEBLJMFSEFOIBO- HJTJEJS A) -3 B) -1 C) 0 D) 1 E) 3 A) y = x B) y = x - 1 C) y = 2x + 1 D) y = -x E) y = 3 - 2x 2. f(x) = x3 - x + 2 GPOLTJZPOVOBÑ[FSJOEFLJY=BQTJTMJOPLUBTO- 6. f ( x ) = x2 - 2ax + b EBO ÀJ[JMFO UFôFUJO EFOLMFNJ BöBôEBLJMFSEFO IBOHJTJEJS FôSJTJOJOY=-BQTJTMJOPLUBTOEBLJUFôFUJ Z=Y-EPôSVTVPMEVôVOBHÌSF,C-BLBÀUS A) y = 26x - # Z= 26x - 52 A) -3 B) -2 C) 0 D) 1 E) 3 C) y = 26x - 26 D) y = 26x - 12 E) y = 26x + 26 7. f^ x h = x2 + 2x + 6 3. f ( x ) = | x2 - 3x + 2 | + 2x2 - 3 GPOLTJZPOVOBY=BQTJTMJOPLUBT OEBOÀJ[JMFO UFôFUJOEFOLMFNJBöBôEBLJMFSEFOIBOHJTJEJS FôSJTJOFY=-BQTJTMJOPLUBTOEBOÀJ[JMFOOPS- NBMJOFôJNJLBÀUS A) 2x - 3y += 0 B) 2x + 3y + 1 = 0 A) - 1 B) - 1 C) 1 D) 1 E) 1 C) 2x - 3y - 11 = 0 D) 2x + 3y += 0 9 8 8 9 7 E) 2x - 3y + 11 = 0 4. f ( x ) = x2 +NY+OWFH Y = -x2 +UY 8. y = -3x2 - kx - 2 FôSJMFSJOJO \" OPLUBTOEB CJSCJSMFSJOF UF- QBSBCPMÑOFY=-BQTJTMJOPLUBTOEBOÀJ[JMFO ôFUPMEVôVCJMJOEJôJOFHÌSF N+O-UJGBEFTJOJO UFôFUY-Z+B=EPôSVTVOBQBSBMFMPMEV- FöJUJLBÀUS ôVOBHÌSF LHFSÀFMTBZTLBÀUS A) -4 B) -2 C) 0 D) 2 E) 4 A) 6 16 D) 14 13 B) C) 5 E) 3 33 B B % C 72 % & A B

5ÑSFWJO(FPNFUSJL:PSVNV TEST - 15 1. y = x2 - 4x + 12 5. f ( x ) = x3 - 3x2 + 4x + 3 FôSJTJOJOZ=YEPôSVTVOBFOZBLOOPLUBT- FôSJTJOJOZ=Y+EPôSVT VOBQBSBMFMUFôFUMF- OOPSEJOBULBÀUS SJBSBTOEBLJV[BLMLLBÀCJSJNE JS A) - 3 B) 0 C) 3 D) 6 E) 9 A) 4 B) 2 2 C) 2 3 D) 4 E) 2 5 17 2. y = 2x2 QBSBCPMÑOÑOZ=Y+BEPôSVTVOBUF- 6. f ( x ) = x3 - 3x2 + ax + C GPOLTJZPOVOVO HSBGJóJ ôFUPMEVôVOBHÌSF BHFSÀFMTBZTOOFöJUJLBÀ- y = 2x +EPóSVTVOBY=BQTJTMJOPLUBEBUFóFU US UJS A) - # - $ % & #VOBHÌSF CHFSÀFMTBZTLBÀUS 3. f^ x h = 1 x3 + ax2 - 6x + 1 A) -2 B) -1 C) 0 D) 2 E) 3 32 GPOLTJZPOVOVO Y FLTFOJOF QBSBMFM UFôFUMFSJOJO 7. y = x2 - ax + EFôNFOPLUBMBSOOBQTJTMFSJOJOUPQMBN-PM- QBSBCPMÑOÑOPSJKJOEFOHFÀFOUFôFUMFSJCJSCJSMFSJ- EVôVOBHÌSF BHFSÀFMTBZTOOFöJUJLBÀUS OFEJLPMEVôVOBHÌSF BHFSÀFMTBZTOOOFHBUJG A) 0 B) 1 C) 2 D) 3 E) 4 EFôFSJLBÀUS 4. y = x3 +QY2 + 4x + 1 A) – 6 B) - 3 3 C) - 2 3 FôSJTJOJOYFLTFOJOFQBSBMFMUFôFUJOJOPMNBNB- D) - 2 2 E) - 2 TJÀJOQHFSÀFMTBZTOOBMBCJMFDFôJEFôFSBSB- MôBöBôEBLJMFSEFOIBOHJTJEJS 8. y = 4 FôSJTJJMFZ=Y2QBSBCPMÑOÑOLFTJöUJL- A) -12 <Q< 12 B) -4 <Q< 4 x C) - 2 3 1 p 1 2 3 MFSJOPLUBEBOFôSJMFSFÀJ[JMFOUFôFUMFSBSBTOEB- D)Q< -ZBEBQ> 2 E) p 2 2 3 ZBEB p 1 - 2 3 LJEBSBÀOOUBOKBOULBÀUS A) 12 B) 5 C) 8 D) 1 E) 10 31 6 9 9 & A C C 73 A & B A

TEST - 16 5ÑSFWJO(FPNFUSJL:PSVNV 1. y = ( x + 1 ) 2 5. y = x3 FôSJTJOJO \" OPLUBTOEBO HFÀFO UFôFU- x MFSEFO CJSJOJO EFOLMFNJ BöBôEBLJMFSEFO IBO HJTJEJS FóSJTJOF OPLUBTOEBO¿J[JMFOUFóFUFóriyi bir A OPLUBTOEBLFTJZPS A) y =Y+ 16 B) y =Y+ $ Z=Y- #VOBHÌSF \"OPLUBTOOBQTJTJLBÀUS D) y = 4x - 4 E) y = 5x - 5 A) 1 + 2 B) 1 C) -1 + 2 D) -1 - 2 E) -3 2. y = x2 QBSBCPMÑOF - OPLUBTOEBOÀJ[JMFO 6. f ( x ) = x3 - 3x2 + bx +DGPOLTJZPOVOVOHSBGJóJ UFô FUMFSJO EFôNF OPLUBMBS BSBTOEBLJ V[BLML x =BQTJTMJOPLUBEBY–FLTFOJOFUFóFUPMEVóVOB LBÀCJSJNEJS HËSF DHFSÀFMTBZTLBÀUS \" # $ % & A) - # $ % & 3. y = x2 - 3x +Q 7. y = 4 x FôSJTJJMFZ=Y2+BFôSJTJ\" Y1 Z QBSBCPMÑOFLÌLMFSJOEFOÀJ[JMFOUFôFUMFSJOFôJN- OPLUBTOEBCJSCJSMFSJOFUFôFUPMEVôVOBHÌSF MFSJÀBSQN-PMEVôVOBHÌSF QHFSÀFMTBZT BHFSÀFMTBZTLBÀUS LBÀUS A) 2 B) 3 C) 4 D) 5 E) 6 A) -2 B) -3 C) -4 D) -5 E) -6 8. y = -x2 - 4x - 3 y = x2 - 4x + 3 4. y = x3 -FóSJTJOFÐ[FSJOEFLJY= -BQTJTMJOPL FôSJMFSJOJOPSUBLUFôFUMFSJOEFOCJSJOJOFôJNJBöB- ôEBLJMFSEFOIBOHJTJEJS UBTOEBO¿J[JMFOUFóFUFóSJZJCJS\"OPLUBTOEBLFTJ ZPS A) - 12 3 B) -12 C) - #VOBHÌSF \"OPLUBTOOPSEJOBULBÀUS D) - 2 3 + 4 E) 2 3 - 4 \" # $ % & C % % % 74 % % B &

5ÑSFWJO(FPNFUSJL:PSVNV TEST - 17 1. y ôFLJMEFZ= f ( x ) 4. 2 x A fonksiyonunun grafi- y y = óJ WF \" OPLUBT OEB 2 UFóFUJWFSJMNJõUJS –1 O 2 x A y = f(x) x g ( x ) =YG2( 3x - 1 ) O PMEVôVOBHÌSF Hh LBÀUS :VLBSEBLJõFLJMEF y = 2 FóSJTJWFFóSJZF\"OPL- \" # $ % & x UBTOEBO¿J[JMFOUFóFUJOHSBGKóJWFSJMNJõUJS #VOBHÌSF UFôFUJOFLTFOMFSMFPMVöUVSEVôVÑÀ- HFOJOBMBOLBÀCJSJNLBSFEJS \" # $ % & y y 2. A b 5. d y = f(x) A(a, b) y = f(x) –a O 2a x d –3 x O E EPóSVTV Z = G Y GPOLTJZPOVOB \" OPLUBTOEB :VLBSEBLJ õFLJMEF \" B C OPLUBTOEB Z = f ( x ) UFóFUUJS GPOLTJZPO VOBUFóFUEEPóSVTV¿J[ JMNJõUJS g ( x ) = x2G Y WFHh - a ) = -BPMEVôVOBHÌ- f ( x ) = ( 2x + Gh Y PMEVôVOBHÌSF BHFSÀFM TBZTLBÀUS SF G -B JGBEFTJOJOFöJUJLBÀUS A) -2 B) -1 C) 0,25 D) -1,5 E) -2,5 A) 3 5 3 D) -2 E) -3 B) C) 22 6. y y = g(x) y 3. y A(2, 3) y = f(x) x B(3, 2) y = g(x) y = f(x) O O6 x –3 A 60° x :VLBSEBLJHSBGJLMFSZ=H Y WFZ= f ( x ) fonksiyon- –2 –1 O MBSOBTSBTZMB\"WF#OPLUBMBSOEBO¿J[JMFOUFóFU- MFSWFSJMNJõUJS ôFLJMEFZ=G Y WFZ=H Y GPOLTJZPOMBS\"OPLUB TOEBUFóFUUJS I Y = GPH Y PMEVóVOBHËSF Z=I Y GPOLTJZP OVOBÑ[FSJOEFLJY=BQTJTMJOPLUBTOEBOÀJ[J I Y =G Y H Y GPOLTJZ POVOVOY=-BQ- MFOUFôFUJOEFOLMFNJOFEJS TJTMJOPLUBTOEBLJUFôFUJOJOFôJNJLBÀUS A) y = 2x + 5 B) y = -2x + 10 A) 3 B) 2 3 C) 5 3 C) y = -2x + 9 D) y = -2x + 6 D) 7 3 E) 9 3 E) y = -2x + 4 C A & 75 C B %

·/÷7&34÷5&:&)\";*3-*, .0%·- -÷.÷57&5·3&7 www.aydinyayinlari.com.tr \"35\"/\";\"-\"/'0/,4÷:0/-\"3 TANIM TEOREM f : [a, b] Z3PMNBLÐ[FSF f : [a, b] Z3UBONMCJSGGPOLTJZPOV[a, b] ara- r x1, x2 ` [a, b]WFY1 < x2J¿JO MóOEBTÐSFLMJ B C BSBMóOEBUÐSFWMJPMTVO f ( x1 ) < f ( x2 ) oluyorsa f fonksiyonu [a, b]BSBM- r x ` B C J¿JO óOEBBSUBOES Gh Y > 0 oluyorsa f, [a, b]BSBMóOEBBSUBO f ( x1 ) > f ( x2 ) oluyorsa f fonksiyonu [a, b]BSBM- Gh Y < 0 oluyorsa f, [a, b]BSBMóOEBB[BMBO Gh Y = 0 oluyorsa f, [a, b]BSBMóOEBTBCJUGPOL- óOEBB[BMBOES TJZPOEVS f ( x1 ) = f ( x2 ) oluyorsa f fonksiyonu [a, b]BSBM- y d NE =UBOi =Gh Y > 0 0<i< π óOEBTBCJUUJS 2 yy i 88 Oa x b x –2 O 3 x y π < 0 <Õ 2 f(b) 2 2 x –2 NE =UBOi =Gh Y < 0 O –8 –8 f(x) = –x3 f(a) x i f : [-2, 2] Z [- ] f : [-2, 2] Z [- ] O ax b [-2, 2]BSBMóOEBB[BMBO [-2, 2]BSBMóOEBBSUBO d ÖRNEK 1 ÖRNEK 3 y = 4 - x2 GPOLTJZPOVOVO BSUBO WF B[BMBO PMEVôV f ( x ) = x2 + 4 GPOLTJZPOVOVOBSUBOWFB[BMBOPMEVôV BSBMLMBSZB[O[ BSBMLMBSUÑSFWZBSENZMBCVMVOV[ Gh Y =Y f' 0 –+ y -ß ] GPOLTJZPO BS- -Þ ]B[BMBO 4 f UBO [ Þ BSUBO [ ß GPOLTJZPO B[B- –2 2 O MBO x ÖRNEK 2 ÖRNEK 4 | | | |f : R Z R, f ( x ) = x + x - 2 GPOLTJZPOVOVOBSUBO f(x) = 1 x B[BMBOWFTBCJUPMEVôVBSBMLMBSCVMVOV[ GPOLTJZPOVOVOBSUBOWFB[BMBOPMEVôVBSBMLMBSCV- MVOV[ 1 0 Gh Y = - f' –– -ß ]GPOLTJZPOBSUBO f y [ ]GPOLTJZPOTBCJU 2 2 y = f(x) [ Þ GPOLTJZPOBSUBO O2 x x -ß B[BMBO ß B[BMBO #JSMFöJNLÑNFTJB[BMBOEFôJMEJS G =>-=G - mß >BSUBO < ß B[BMBO 76 mÞ >B[BMBO< Þ BSUBO mß >BSUBO < >TBCJU < Þ BSUBO mß B[BMBO ß B[BMBO

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 .0%·- ·/÷7&34÷5&:&)\";*3-*, UYARI ÖRNEK 8 #JS GPOLTJZPOVO B[BMBO WFZB BSUBO PMEVóV BSBML- f ( x ) = x3 GPOLTJZPOVOVOBSUBOPMEVôVBSBMôCVMV- MBSO CJSMFõJN LÐNFTJOEF GPOLTJZPO B[BMBO WFZB OV[ BSUBO PMNBZBCJMJS±SOFLÐJODFMFZJOJ[ ÖRNEK 5 y Gh = PMNBTOB SBô- y = x3 NFOZ=G Y GPOLTJZPOV f ( x ) = 2x3 - 9x2 - 24x + 5 BSUöOBEFWBNFEJZPS x O G GPOLTJZPOV 3 EF BS- UBOES GPOLTJZPOVOVO BSUBO WF B[BMBO PMEVôV BSBMLMBS JO- DFMFZJOJ[ Gh Y =Y2-Y- –1 4 +–+ -ß -1]BSUBO f' UYARI [- ]B[BMBO f #JS GPOLTJZPOVO BSUBO WFZB B[BMBO PMEVóV BSBML- [ ß BSUBO MBSEBUÐSFWJTOSMTBZEBOPLUBJ¿JOPMBCJMJS ÖRNEK 6 ÖRNEK 9 f_ x i = x2 - 6x + 5 f : R Z R, f ( x ) = x3 - 3x2 + ax GPOLTJZPOVOVOBSUBOWFB[BMBOPMEVôVBSBMLMBSZB- [O[ GPOLTJZPOV3EFBSUBOPMEVôVOBHÌSF BHFSÀFMTBZ- TOOFOLÑÀÑLUBNTBZEFôFSJLBÀUS f'^ x h = 2x - 6 1 35 Gh Y =Y2-Y+B 2 f + ––+ Dã -Bã 2 x - 6x + 5 ãB BFOB[PMVS -ß ]B[BMBO f' – – + + [ ß BSUBO f(x) UBONT[ ÖRNEK 7 n ` Z+PMNBLÐ[FSF ZG Y GPOLTJZPOV B C BSBMóO- EBOFHBUJGEFóFSMJB[BMBOCJSGPOLTJZPOEVS #VOBHÌSF Z=G Y Z=G Y y = 1 y = 1 ÖRNEK 10 f_ x i f2_ x i f : R – {-4} Z R GPOLTJZPOMBSOEBOLBÀUBOFTJBZOBSBMLUBBSUBOES f_ x i = ax - 3 GOFHBUJGEFôFSMJB[BMBOPMEVôVOBHÌSF x+4 G< Gh<ES GPOLTJZPOVOVO -ß, - WF - ß BSBMLMBSOEBBS- UBO GPOLTJZPO PMNBT JÀJO B HFSÀFM TBZTOO EFôFS d 2 ^ x h =G Y Gh Y >BSUBO BSBMôOCVMVOV[ dx f f'^ x h = 4a + 3 > 0 ^ x + 4 h2 d G3 Y =G2 Y Gh Y <B[BMBO 3 dx B+>jB> - d G-1 Y =-G-2 Y Gh Y >BSUBO 4 dx d G-2 Y =-G-3 Y Gh Y <B[BMBO dx UBOFTJBSUBOES mß m>BSUBO <m >B[BMBO < ß BSUBO 77 3d - 3 , 3 n mß >B[BMBO < ß BSUBO2 4

·/÷7&34÷5&:&)\";*3-*, .0%·- -÷.÷57&5·3&7 www.aydinyayinlari.com.tr ÖRNEK 11 ÖRNEK 14 y = f'(x) :BOEBLJõFLJMEF f ( x ) = x4 + 2x3 - 12x2 + 4x - 5 y GPOLTJZPOVWFSJMJZPS y =Gh Y GPOLTJZPOVOVO #VOB HÌSF Z = Gh Y GPOLTJZPOVOVO B[BMBO PMEVôV –3 O 3 HSBGJóJWFSJMNJõUJS BSBMôCVMVOV[ –1 5 x Gh Y =Y3+Y2-Y+ –2 1 #VOBHÌSF Ghh Y =Y2+Y- f'' + – + * ( 3, R BSBMóOEBGGPOLTJZPOVBSUBOCJSGPOLTJZPO- EVS [- ]GhB[BMBO f' ** BSBMóOEBGGPOLTJZPOVB[BMBOCJSGPOLTJZPO- ÖRNEK 12 EVS y *** Gh - ZPLUVS 2 1 *7 ( - BSBMóOEBGhh Y <ES 7 Ghh - ZPLUVS 7* Ghh Gh <ES ÌOFSNFMFSJOEFOLBÀUBOFTJEPôSVEVS –6 4 x –3 1 5 –1 3 –4 –2 O 23 f' – + –+ f' –1 y = f(x) f f'' + – + :VLBSEBLJZ=G Y GPOLTJZPOVOVOHSBGJóJOFHËSF [ ]B[BMBO Þ BSUBO *ZBOMö * [ 2, 4]BSBMóOEBGGPOLTJZPOVB[BMBOES Gh - WBSES***ZBOMö ** [ 0, 2 ]BSBMóOEBGGPOLTJZPOVBSUBOES Ghh Gh >PMEVôVOEBO7*ZBOMö *** (-Þ - BSBMóOEBGhQP[JUJGUJS ÖRNEK 15 :VLBSEBLJHSBGJL *7 (- BSBMóOEBGhOFHBUJGUJS y =Gh Y GPOLTJZPOV- 7 f ( - Gh < 0 y OBBJUUJS 7* Gh - G = 0 4 3 JGBEFMFSJOEFOLBÀUBOFTJEPôSVEVS x ²OFSNFMFSEFO TBEFDF 7 ZBOMöUS G - > Gh > ÀBSQNMBSQP[JUJGUJSUBOFTJEPôSVEVS –2 1 –1 O1 ÖRNEK 13 y = Gh Y GPOLTJZPOV- #VOBHÌSF OVOHSBGJóJWFSJMNJõUJS * ( -R, - BSBMóOEBGGPOLTJZPOVBSUBOES y ** ( 1, R BSBMóOEBGGPOLTJZPOVB[BMBOES 2 3x #VOB HÌSF Z = G Y *** Gh - ZPLUVS y = f'(x) GPOLTJZPOVOVO BSUBO *7 Gh - ZPLUVS –2 PMEVôVBSBMLOFEJS 7 ( - 2, - BSBMóOEBGhh Y >ES O JGBEFMFSJOEFOLBÀUBOFTJEPôSVEVS Gh>PMEVôVBSBML - UÑS -ß - BSBMôOEBB[BMBOES*ZBOMö Gh - =PMEVôVHSBGJLUFBÀLUS***ÌODÑMZBOMöUS [- ]BSBMôOEBGBSUBOES %JôFSÌODÑMMFSEPôSVEVS <m >5<m > 78 33

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 .0%·- ·/÷7&34÷5&:&)\";*3-*, :FSFM&LTUSFNVN/PLUBMBS ÖRNEK 16 TANIM f ( x ) = 3x - x2 GPOLTJZPOVOVOFLTUSFNVNOPLUBMB- SOCVMVOV[ :FUFSJODFLÐ¿ÐLCJSSQP[JUJGHFS¿FLTBZTJ¿JO r x ` D-S D+S BSBMóOEBG D $ f ( x ) G D # f ( x )) oluyorsa, y = f ( x ) fonksiyonunun Gh Y =-Y 3/2 x =DBQTJTMJOPLUBTOEBZFSFMNBLTJNVNV ZF- SFMNJOJNVNV WBSESG D EFóFSJZFSFMNBLTJ- fd 3 n = 9 - 9 =- 9 f' + – NVNEFóFSJ ZFSFMNJOJNVNEFóFSJ EJS 2 24 4 d 3 , - 9 nZFSFMNBLTJNVN f 24 yerel maks. f'(x3) yoktur. yerel maks. f' < 0 ÖRNEK 17 y = f(x) f'(x0) = 0 f ( x ) = 3x5 + 45 x4 + 10x3 + 5 f'(x2) = 0 f' > 0 f'(x4) yoktur. 4 f' < 0 f' > 0 f' < 0 f' > 0 GPOLTJZPOVOVOZFSFMFLTUSFNVNOPLUBMBSOOBQTJT- f' > 0 yerel min. MFSJUPQMBNLBÀUS f'(x1) = 0 f'(x5) yoktur. yerel min. a x0 x1 x2 x3 x4 x5 b TEOREM –2 –1 0 Gh Y =Y4+Y3+Y2 f' + – + + =Y2 Y2+Y+ x = D BQTJTMJ OPLUBEB TÐSFLMJ CJS G GPOLTJZPOV- f OVO CVOPLUBEBUÐSFWJZPLWFZBGh D =PMTVO Gh Y=DEFJõBSFUEFóJõUJSJZPSTB D G D OPLUB- -ZFSFMNBLTJNVN -ZFSFMNJOJNVNOPLUBTOOBQTJ- TZFSFMFLTUSFNVNPMVS TJ-+ - =- xc xc ÖRNEK 18 f' – + f' + – | |f ( x ) = x - 1 + 2UPQMBNOOZFSFMFLTUSFNVNEFôF- f f SJLBÀUS yerel minimum yerel maksimum f ( x ) = ( x+1 x $ 1 f' ( x ) = ( 1 x>1 -x + 3 x < 1 -1 x<1 NOT Y=JÀJOGhZPLUVSY=ZFSFMNJOJNVNBQTJTJ 4ÐSFLMJCJSGGPOLTJZPOVOVOY=DBQTJTMJOPLUB- ZFSFMNJOJNVNOPLUBT TOEBZFSFMFLTUSFNVNVWBSTBGh D ZPLUVSWF- ZBGh D =PMVS UYARI ÖRNEK 19 y = f(x) x f (x) = 6x x0 x1 k x2 x3 x2 + 1 GPOLTJZPOVOVO FLTUSFNVN OPLUBMBSOO PSEJOBUMBS UPQMBNLBÀUS 6^ 2 + 1 h - 2x.6x 6 - 6x2 &LTUSFNVNOPLUBMBSPMBOYi âJ< 4, i `; BQ- f' ( x) = x = TJTMJOPLUBMBSUÐSFWUFTUJZMFUFTQJUFEJMFNFZFCJMJS#V OPLUBMBS UBON ZBSENZMB CVMVOVS L BQTJTMJ OPLUB- ^ x2 + 1 h2 ^ x2 + 1 h2 EBGOJOFLTUSFNVNVZPLUVS4ÐSFLTJ[GPOLTJZPOMBS- EBUÐSFWUFTUJEPóSVTPOV¿WFSNFZFCJMJS –1 1 66 - + =0 f' – + – f 22 79 d 3 , - 9 nm 24

·/÷7&34÷5&:&)\";*3-*, .0%·- -÷.÷57&5·3&7 www.aydinyayinlari.com.tr ÖRNEK 20 ÖRNEK 24 f ( x ) = x3 + ax + b GPOLTJZPOVOVO OPLUBTOEB x3 - 3x2 + a = 0 EFOLMFNJOJOÑÀUBOFGBSLMSFFMLÌ- ZFSFMNJOJNVNVWBSTBB-CLBÀUS LÑOÑOPMBCJMNFTJJÀJOBIBOHJBSBMLUBPMNBMES Gh =WFG =PMNBMES &LTUSFNVNOPLUBMBS[UJöBSFUMJPSEJOBUBTBIJQPMNBMES Gh Y =Y2+Bj+B=jB=-3 G =-+C=jC= Gh Y =Y2-Y= 04 B-C=-7 jY=WY= +– + G G <jB B- < B` ÖRNEK 21 ÖRNEK 25 f ( x ) = ( a - 1 ) x3 - 3ax + 1 y y = f'(x) :BOEB Z = Gh Y fonksiyonunun gra- FôSJTJOJOZFSFMFLTUSFNVNOPLUBTOOPMNBNBTJÀJO –6 O 1 BHFSÀFMTBZTIBOHJBSBMLUBPMN BME S –4 –3 –2 x GJóJWFSJMNJõUJS 35 Gh Y = B- Y2-B 01 D#0jB B- # D+ – + #VOB HÌSF Z = G Y GPOLTJZPOVOVO ZFSFM NJOJNVN OPLUBMBSOOBQTJTMFSJÀBSQNLBÀUS B`[ ] ÖRNEK 22 5ÑSFWJOJöBSFUEFôJöUJSEJôJ-EFO+ZBHFÀUJôJZFSMFS- EFZFSFMNJOJNVNWBSES GGPOLTJZPOVOVOWFUÐSFWMFSJWBSTBWFY=BBQTJTMJ OPLUBTZFSFMNJOJNVNOPLUBTJTFGhh B >PMVS :FSFMNJONVNOPLUBMBSOOBQTJTMFSJ-WFUÑS f^ x h = x3 + ax2 + bx + 1 -WFÀBSQNMBS-EJS 3 GPOLTJZPOVOVO Y = BQTJTMJ OPLUBTOEB ZFSFM NJOJ- ÖRNEK 26 NVNVWBSTBBHFSÀFMTBZTIBOHJBSBMLUBPMNBMES y ôFLJMEFGhGPOLTJyonu- Gh =WFGhh > OVOHSBGJóJWFSJMNJõUJS Gh Y =Y2+BY+C (–3, 2) Ghh Y =Y+B y = f'(x) +B>jB>-1 B` - Þ 37x –5 O ÖRNEK 23 (5, –1) f ( x ) = x3 + 2x2 + ax + b GPOLTJZPOVOVOUFSTJOJOPMB- CJMNFTJJÀJOBHFSÀFMTBZTIBOHJBSBMLUBPMNBMES #VOBHÌSF * x = -UFGGPOLTJZPOVOVOZFSFMNJOJNVNVWBSES GGPOLTJZPOVIFQBSUBOZBEBB[BMBOPMNBMES ** x = -UFGGPOLTJZPOVOVOZFSFMNBLTJNVNVWBSES Gh Y =Y2+Y+B *** x =UFGGPOLTJZPOVOVOZFSFMNJOJNVNVWBSES *7 Ghh - 3 ) =ES 4 7 ( - R, - BSBMóOEBGhh Y >ES D#0j-B#j #B 7* x =UFGhGPOLTJZPOVOVOZFSFMNJOJNVNVWBSES 3 JGBEFMFSJOEFOLBÀUBOFTJEPôSVEVS Y=- GhJÀJOZFSFMNBLTJNVNEVS Y= GJÀJOZFSFMNBLTJNVNEVS %JôFSJGBEFMFSEPôSVEVS **WF***ZBOMöUS –7< > m Þ = 4 , 3 G –124 3

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 .0%·- ·/÷7&34÷5&:&)\";*3-*, .VUMBL&LTUSFNVN/PLUBMBS ÖRNEK 28 TANIM f:RZR f ( x ) = -2x2 -Y+ 5 y Mutlak y Mutlak maksimum maksimum GPOLTJZPOVOVO NVUMBL FLTUSFNVN OPLUBMBS WBSTB f(b) OFEJS a xa x b x1 b Gh Y =-Y- –2 f(a) G - = Mutlak Mutlak - NVUMBLFLTUSFNVNOPLUBT- f' + – minimum minimum ES f y Mutlak maksimum x1 x ab Mutlak minimum 4ÐSFLMJ CJS G GPOLTJZPOVOVO [ a, b ] BSBMóOEB ÖRNEK 29 BMBCJMFDFóJ FO CÐZÐL FO LÐ¿ÐL EFóFSF NVU- | |f : [-5, 3] Z R, f ( x ) = x2 - 3x + 2 MBL NBLTJNVN NVUMBL NJNJNVN EFOJS 4F[HJTFMPMBSBLPBSBMLUBHSBGJóJOFOZÐLTFL- GPOLTJZPOVOVONVUMBLFLTUSFNVNEFôFSMFSJOJCVMV- UFLJ FOBõBóEBLJ OPLUBTES OV[ Bir f fonksiyonunu [ a, b ] BSBMóOEBLJ NVUMBL x –5 1 2 3 FLTUSFNVN OPLUBMBSO CVMNBL J¿JO G GPOLTJ- f(x) x2– 3x + 2 –x2 + 3x – 2 x2– 3x + 2 ZPOVOVO B C BSBMóOEBLJ D1 D2 wDn ZFSFM FLTUSFNVNOPLUBMBSCVMVOVS f'(x) Ymæ Yæ Ymæ f'(x) – + {G B G D1 G D2 G Dn), f ( b ) } LÐNFTJ- f(x) +– OJOFOCÐZÐLFMFNBO FOLÐ¿ÐL NVUMBLNBL- 3/2 TJNVN NVUMBLNJOJNVN EFóFSJEJS yerel yerel yerel yerel yerel NBLT NJO NBLT NJO NBLT G - = G = fd 3 n = 1 G = G = 24 NVUMBLNBLTJNVNEFôFSJ NVUMBLNJOJNVNEFôFSJ ÖRNEK 27 f : [-1, 3] Z R ÖRNEK 30 f ( x ) = x3 + 3x2 + 5 GPOLTJZPOVOVOWBSTBNVUMBLFLT- f(x) = 1 USFNVNEFôFSMFSJOJCVMVOV[ x2 + 1 Gh Y =Y2+Y GPOLTJZPOVOVO WBSTB NVUMBL FLTUSFNVN OPLUBMBS- OCVMVOV[ Y=EBZFSFMNJOJNVN NVU- MBLNJOJNVN x –1 0 3 + Y = - WFZB Y = UF NVUMBL f' – 2x 0 NBLTJNVNWBSES - f Gh Y = - ^ x2 + 1 h2 f' + – f NVUMBLNBLTJNVNOPLUBT yerel NVUMBL NBLTJNVN EFôFSJ NVUMBL NJOJ- NBLTJNVN NVNEFôFSJ NVUMBLNJOJNVNEFôFSJ NVUMBLNBLTJNVNEFôFSJ 81 m NVUMBLNBLTJNVNEFôFSJ NVUMBLNJOJNVNEFôFSJ

TEST - 18 \"SUBO\"[BMBO'POLTJZPOMBS 1. f^ x h = x3 + x2 - 6x + 1 5. f^ x h = x - 1 32 x2 GPOLTJZPOV BöBôEBLJ BSBMLMBSO IBOHJTJOEF GPOLTJZPOVOVO BSUBO PMEVôV FO HFOJö BSBML B[BMBOES BöBôEBLJMFSEFOIBOHJTJEJS A) -3 # x # 2 B) -2 # x # 3 C) -1 # x # 6 \" mÞ > B) > 4 , 3 p $ > 3 D) -6 # x # 1 E) x $ 0 D) >1, 4 H 4 3 E) f 0 , H 3 2. y 6. f ( x ) =NY3 - 2x2 - 4x + 1 –2 –1 1 y = f(x) GPOLTJZPOVOVOEBJNBB[BMBOPMNBTJÀJONHFS- 3x ÀFMTBZTOOEFôFSBSBMôBöBôEBLJMFSEFOIBO- 3O HJTJEJS 2 y = g(x) A) m # - 1 # N< 0 $ N> 0 3 E) - 1 < m < 0 :VLBSEBLJ G WF H GPOLTJZPOMBS JMF UBONMB- D) m $ 1 3 OBO WF Ih Y = Gh Y H Y LPöVMVOV TBôMBZ BO 3 Z = I Y GPOLTJZPOV BöBôEBLJ BSBMLMBSEBO I BOHJTJOEFLFTJOMJLMFBSUBOES A) x < 1 B) -1 < x < 1 C) -1 < x < 3 7. f ( x ) = x3 + ax2 + bx + 3 D) x < -2 E) 1 < x < 3 GPOLTJZPOVOVO B[BMBO PMEVôV FO HFOJö BSBML PMEVôVOBHÌSF BCÀBSQNLBÀUS A) 0 B) -3 C) -9 D) - & - 3. f^ x h = x2 - 2x - 3 GPOLTJZPOVWFSJMJZPS Z = G Y GPOLTJZPOVOVO B[BMBO PMEVôV BSBML 8. y =G Y GPOLTJZPOV BSBMóOEBOFHBUJGEFóFS- BöBôEBLJMFSEFOIBOHJTJEJS MJBSUBOCJSGPOLTJZPOEVS A) ( -Þ > # -Þ -> $ < Þ #VOBHÌSF % < Þ & 3- ( -1, 3 ) r y = - 1 4. f ( x ) = x3 - 4x2 + 3x -GPOLTJZPOVWFSJMJZPS f2_ x i r y = f_ x2 i #VOBHÌSF GhGPOLTJZPOVOVOBSUBOPMEVôVBSB- MLBöBôEBLJMFSEFOIBOHJTJEJS r y = f_ x i A) >- 4 , 4 H # < R) C) > 4 , 3 p r y = f ( f ( x ) ) 33 3 r y = f3 ( x ) D) R E) ( -R > GPOLTJZPOMBSOEBOLBÀUBOFTJBZOBSBMLUBBSUBO GPOLTJZPOEVS A) 0 B) 1 C) 2 D) 3 E) 4 A % B C 82 % A & C

\"SUBO\"[BMBO'POLTJZPOMBS TEST - 19 1. f ( x ) = x3 - x2 + ax - 3 4. Z = G Y GPOLTJZPOV B C BSBMôOEB OFHB- GPOLTJZPOVOVOEBJNBBSUBOPMNBTJÀJOBHFS- UJG EFôFSMJ WF BSUBO CJS GPOLTJZPO PMEVôVOB HÌ- ÀFMTBZTOOEFôFSBSBMôBöBôEBLJMFSEFOIBO- SF BöBôEBLJMFSEFOIBOHJTJBZOBSBMLUBEBJNB HJTJEJS B[BMBOES \" <G Y >3 B) 5 + f ( x ) C) x3 + f ( x ) A) a > 1 B) a $ 1 C) a # 1 D) 3 E) 1 3 3 f^ x h f (x2) D) a < 0 E) a < -1 5. ôFLJMEFLJHSBGJL Z=G Y GPOLTJZPOVOBBJUUJS y y = f(x) 2. f^ x h = 3x - a –3 x –1 O x-1 Ih Y = ( x2 - x - Gh Y FöJUMJôJOJ TBôMBZBO I GPOLTJZPOVOVOY>JÀJOEBJNBB[BMBOPMNBT GPOLTJZPOVOVO B[BMBO PMEVôV BSBML BöBôEBLJ JÀJOBHFSÀFMTBZTOOEFôFSBSBMôBöBôEBLJ- TFÀFOFLMFSEFOIBOHJTJEJS MFSEFOIBOHJTJEJS A) a < 3 B) 1 < a < 3 C) a > 3 A) (-Þ -3] B) [ 0, 2] D) 3 < a < 4 E) a > 4 C) (-Þ -1] b [0, 2] D) (-Þ -3] b [1, 2] E) [-1, 2] 3. x a b c 6. (FS¿FMTBZMBSLÐNFTJOEFTÐSFLMJWFUÐSFWMFOFCJMJS f' + – – + y =G Y GPOLTJZPOVOVOBSUBOPMEVóVFOHFOJõBSB- ML[ 0, 6 ]ES (FS¿FMTBZMBSLÐNFTJOEFUÐSFWMFOFCJMJSCJSGGPOLTJ- #VOBHÌSF ZPOVOVOUÐSFWJOJOJõBSFUUBCMPTVZVLBSEBLJHJCJEJS #VOBHÌSF * y = f ( 2 -Y GPOLTJZPOV<- >BSBMóOEBBS- * <B D>BSBMóOEBGB[BMBOES UBOES ** (-Þ B>b<D Þ BSBMóOEBGBSUBOES *** <C Þ BSBMóOEBGhBSUBOES ** y = f ( 4 - Y GPOLTJZPOV <- > BSBMóOEB B[BMBOES *** y = fa x kGPOLTJZPOV<- >BSBMóOEBBSUBO- ES JGBEFMFSJOEFOIBOHJMFSJLFTJOMJLMFEPôSVEVS JGBEFMFSJOEFOIBOHJMFSJLFTJOMJLMFEPôSVEVS \" :BMO[* # *WF** $ *WF*** \" :BMO[* # *WF** $ *WF*** % **WF*** & * **WF*** % **WF*** & :BMO[** B A A 83 % B &

TEST - 20 \"SUBO\"[BMBO'POLTJZPOMBS 1. f ( x ) = x3 - 3x2 - 24x + 1 5. f ( x ) = | x2 - 4 | - 3 GPOLTJZPOVOVOZFSFMNJOJNVNEFôFSJLBÀUS GPOLTJZPOVOVONVUMBLNJOJNVNOPLUBMBSBSB- A) - # -69 C) -12 D) 0 E) 29 TOEBLJV[BLMLLBÀCJSJNEJS \" # $ % & 2. G3Z3PMNBLÑ[FSF 6. f fonksiyonunun x =LBQTJTMJOPLUBTOEBZFSFMNBL- f ( x ) = x3 - 3x + a TJNVNVWBSTBGhh L <ES f ( x ) = x3 + ax2 + bx +DGPOLTJZPOVOVO\" -1 ) GPOLTJZPOVOVO ZFSFM NBLTJNVN EFôFSJ - PM- EVôVOBHÌSF BHFSÀFMTBZTOOEFôFSJLBÀUS OPLUBTOEBZFSFMNBLTJNVNVWBSES A) -3 B) -2 C) 0 D) 2 E) 3 #VOB HÌSF B TBZTOO BMBCJMFDFôJ FO CÑZÑL UBNTBZEFôFSJJÀJODHFSÀFMTBZTLBÀUS A) 3 B) 2 C) 1 D) -2 E) -3 7. f ( x ) = ax3 + ( a - 1 ) x2 + 2x + 3 3. f : R Z3PMNBLÐ[FSF GPOLTJZPOVOVO FLTUSFNVNV PMNBNBT JÀJO B HFSÀFMTBZTOOBMBCJMFDFôJFOHFOJöEFôFSBSB- f ( x ) = ax3 + bx2 + 2 MôBöBôEBLJMFSEFOIBOHJTJEJS GPOLTJZPOVOVO\" - OPLUBTOEBZFSFMNJ- A) [ -3, -2 ] B) ( 0, 3 ) OJNVNVPMEVôVOBHÌSF BCÀBSQNLBÀUS C) ( -4, 4 ) D) 6 4 - 15, 4 + 15 @ E) 6 - 15 , 15 @ \" # $ % & 8. a >PMNBLÐ[FSF f ( x ) = ax3 + 2x2 + 3x - 5 4. f: R Z3PMNBLÐ[FSF GPOLTJZPOVOVOUFSTJOJOPMBCJMNFTJJÀJOBHFSÀFM TBZTOOBMBCJMFDFôJFOHFOJöEFôFSBSBMôBöB- f ( x ) = x3 - 2ax2 + 3x -GPOLTJZPOVWFSJMJZPS ôEBLJMFSEFOIBOHJTJEJS y =Gh Y GPOLTJZPOVOVOZFSFMNJOJNVNEFôFSJ -PMEVô VOBHÌSF BHFSÀFMTBZTOOQP[JUJGEF- A) ; 2 , 3 m B) [ 0, R) ôFSJLBÀUS 9 A) 1 B) 2 C) 3 D) 5 E) C) ;0 , 2 E D) ;0 , 4 E 9 9 E) ; 4 , 3 m 9 A A C C 84 % & % &

\"SUBO\"[BMBO'POLTJZPOMBS TEST - 21 1. f ( x ) = x3 - x2 + a 4. ôFLJMEFLJHSBGJLZ=G Y GPOLTJZPOVOBBJUUJS EFOLMFNJOJO GBSLM ÑÀ SFFM LÌLÑ PMEVôVOB HÌ- y y = f(x) SF BHFSÀFMTBZTOOBMBCJMFDFôJFOHFOJöEFôFS BSBMôBöBôEBLJMFSEFOIBOHJTJEJS O1 x A) ( 0, 5 ) B) ( 0, 3 ) C) f - 4 , 4 p –4 –3 –2 5 27 27 D) c 0, 4 m E) c 0, 1 m 27 3 2. ôFLJMEF Z = G Y GPOLTJZPOVOVO CJSJODJ UÐSFWJOJO #VOBHÌSF HSBGJóJWFSJMNJõUJS * x = -EFGGPOLTJZPOVOVOZFSFMNBLTJNVNV WBSES y ** x = EF G GPOLTJZPOVOVO ZFSFM NJOJNVNV WBSES *** -YEFGh Y ES *7 Gh -2 ) =ES 7 Gh - ES 7* Gh - G ES 7** x ` ( -3, R JTFGh Y ES 7*** Gh - ES –5 O 1 4x JGBEFMFSJOEFOLBÀUBOFTJEPôSVEVS –2 2 3 y = f'(x) \" # $ % & #VOBHÌSF Z=G Y GPOLTJZPOVOVOZFSFMNJOJ- 5. ôFLJMEFLJHSBGJLZ=G Y GPOLTJZPOVOBBJUUJS NVNOPLUBMBSOOBQTJTMFSJUPQMBNLBÀUS y y = f(x) A) -5 B) -3 C) -1 D) 1 E) 6 –5 –3 O 2 x –7 5 3. ôFLJMEF Z = G Y GPOLTJZPOVOVO CJSJODJ UÐSFWJOJO #VOBHÌSF HSBGJóJWFSJMNJõUJS * x = -UFGGPOLTJZPOVOVOZFSFMNJOJNVNVWBS- ES y ** x = - UF G GPOLTJZPOVOVO ZFSFM NBLTJNVNV WBSES –3 –2 O 2 x *** Gh =ES –5 4 6 89 *7 -YBSBMóOEBGh Y ES 7 x ` ( 2, R BSBMóOEBGh Y ES y = f'(x) 7* Gh - ES #VOBHÌSF Z=G Y GPOLTJZPOVOVOZFSFMNBL- 7** Gh -5 ) = f ( 5 ) TJNVNOPLUBMBSOOBQTJTMFSJUPQMBNLBÀUS JGBEFMFSJOEFOLBÀUBOFTJEPôSVE VS \" # $ % & \" # $ % & % B % 85 B %

TEST - 22 \"SUBO\"[BMBO'POLTJZPOMBS 1. ôFLJMEFLJHSBGJLZ=Gh Y GPOLTJZPOVOBBJUUJS 3. ôFLJMEFZ=Gh Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõUJS y y 5 y = f'(x) 2 –3 1 –3 O O 14 x y = f'(x) #VOBHÌSF Z=G Y GPOLTJZPOVOVOHSBGJôJBöBôEBLJMFSEFO * y = f ( x ) fonksiyonunun x = -OPLUBTOEB IBOHJTJPMBCJMJS ZFSFMNJOJNVNVWBSES ** y =Gh Y GPOLTJZPOVOVO OPLUBTOEBZF- A) y B) y SFMNBLTJNVNVWBSES –3 2 x –2 13 x *** ( -R, - BSBMóOEBZ= f ( x ) fonksiyonu ar- O1 O UBOES C) y D) y *7 Gh WFG ZPLUVS Ox 7 Gh - Ghh ES Ox 7* Ghh =ES –1 1 2 7** ( -3, R BSBMóOEBG Y GPOLTJZPOVBSUBOES JGBEFMFSJOEFOLBÀUBOFTJEPôSVEVS \" # $ % & E) y Ox 2. ôFLJMEFZ=Gh Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõUJS y y = f'(x) –7 –4 O2 6x 4. –2 x– a b c f' – + +– #VOBHÌSF :VLBSEBLJ JõBSFU UBCMPTV UÐN HFS¿FM TBZMBSEB * ( -R, - BSBMóOEBGGPOLTJZPOVB[BMBOES UÐSFWMFOFCJMJSZ=G Y GPOLTJZPOVOVOUÐSFWJOFBJU- ** (- BSBMóOEBGGPOLTJZPOVB[BMBOES UJS *** x = - OPLUBTOEB G GPOLTJZPOVOVO ZFSFM #VOBHÌSF NBLTJNVNVWBSES * Gh B Gh C =Gh D = 0 ** x =BOPLUBTOEBZ=G Y JOZFSFMNJOJNVNV *7 x =OPLUBTOEBGGPOLTJZPOVOVOZFSFMNJOJ- WBSES NVNVWBSE S *** x =COPLUBTOEBZ=G Y JOZFSFMNBLTJNV- 7 ( 2, R BSBMóOEBGhh Y ES NVWBSES 7* ( - BSBMóOEBGhh Y ES 7** Gh - Ghh - *7 DJTFG G UJS 7*** f ( -1 ) > f ( 3 ) 7 rx `3J¿JOG Y TÐSFLMJEJS JGBEFMFSJOEFOLBÀUBOFTJEPôSVEVS JGBEFMFSJOEFOLBÀUBOFTJEPôSVE VS \" # $ % & A) 1 B) 2 C) 3 D) 4 E) 5 C B 86 & C

\"SUBO\"[BMBO'POLTJZPOMBS TEST - 23 1. ôFLJMEFy =Gh Y GPOLTJZPOVOVOHSBGJóJWFSJMNJõUJS 3. ôFLJMEFLJHSBGJLZ=G Y GPOLTJZPOVOBBJUUJS y y y = f'(x) –3 –2 2 4 x O –5 –2 O 1 4x 5 #VOBHÌSF BöBôEBLJMFSEFOIBOHJTJZBOMöUS #VOBHÌSF ^ x - 3 h.f^ x h 1 0FöJUTJ[MJôJOJTBôMB A) x = -EFZ=G Y GPOLTJZPOVOVOZFSFMNJOJ- f'^ x h NVNVWBSES ZBOYEFôFSMFSJOJOBSBMôBöBôEBLJMFSEFOIBO B) x =EFZ=G Y GPOLTJZPOVOVOZFSFMNBLTJ- HJTJEJS NVNVWBSES A) ( -3, -2 ) b ( 4, R) C)Ghh -2 ) =ES B) ( -3, -2 ) b ( 2, 3 ) D) f ( -2 ) < f ( 1 ) C) ( -3, -2 ) b ( 3, 4 ) E) f ( - < f ( -5 ) D) ( 2, 4 ) E) (-2, 2 ) b ( 3, R) 2. y y = G Y EFSFDF- 4. y =G Y GPOLTJZPOVOVOUÐSFWJOJOHSBGJóJõFLJMEFWF- 3 y = f(x) EFOQPMJOPNVOHSB- –3 SJMNJõUJS GJóJWFSJMNJõUJS y O 1x y = f'(x) –2 1 x –1 O 2 Z=Gh Y GPOLTJZPOVOVOHSBGJôJBöBôEBLJMFSEFO #VOBHÌSF Z=G Y GPOLTJZPOVOVOHSBGJôJBöB- IBOHJTJEJS ôEBLJMFSEFOIBOHJTJPMBCJMJS A) y B) y –3 – 1 A) y B) y 3 Ox O x 1 1 1 –2 –1 O –1 O x 12 x C) y D) y –5 3 Ox O 2x C) y D) y 1 –2 –2 O2 x O2 –2 x E) y E) y O 3x 1 Ox & C 87 B %

TEST - 24 \"SUBO\"[BMBO'POLTJZPOMBS 1. ôFLJMEFLJHSBGJLPSJKJOFHËSFTJNFUSJLPMBOZ=Gh Y g_ x i , g_ x i $ h_ x i 3. f_ g_ x i, h_ x i i = * g_ x i < h_ x i GPOLTJZPOVOBBJUUJS , y h_ x i 2 y = f'(x) GPOLTJZPOVUBONMBOZPS #VOBHÌSF Z=G Y2 Y+ GPOLTJZPOVJMFJMHJ- MJPMBSBLWFSJMFO –3 x * .VUMBLNJOJNVNEFóFSJPMVS O 3 –2 ** .VUMBLNBLTJNVNEFóFSJZPLUVS *** :FSFMFLTUSFNVNEFóFSJZPLUVS GGPOLTJZPOV3EFTÑSFLMJPMEVôVOBHÌSF JGBEFMFSEFOIBOHJMFSJLFTJOMJLMFEPôSVEVS * y = f ( x ) fonksiyonunun x =BQTJTMJOPLUBTO- \" :BMO[* # *WF** $ :BMO[*** EBNVUMBLNBLTJNVNVWBSES % **WF*** & *WF*** ** y =G Y GPOLTJZPOVOVOUBOFZFSFMNJOJNVN OPLUBTWBSES *** f ( - LG FõJUTJ[MJóJOJTBóMBZBOLHFS¿FM TBZTJ¿JOG Y =LEFOLMFNJOJOUBOFGBSLMLË- LÐWBSES ZBSHMBSOEBOIBOHJMFSJLFTJOMJLMFEPôSVEVS 4. y =G Y GPOLTJZPOVJMFJMHJMJPMBSBL \" :BMO[* # **WF*** $ *WF*** r ¥JGUGPOLTJZPOEVS r G G G ES % *WF** & * **WF*** r(FS¿FM TBZMBS LÐNFTJOEF UÐSFWMFOFCJMJS CJS 2. (FS¿FLTBZMBSLÐNFTJOEFTÐSFLMJCJSGGPOLTJZPOV- GPOLTJZPOEVS OVOUÐSFWJOJOHSBGJóJBõBóEBLJõFLJMEFWFSJMNJõUJS #VOBHÌSF Z=G Y GPOLTJZPOVOVOFLTUSFNVN y OPLUBMBSOOTBZTFOB[LBÀUS A) 2 B) 3 C) 4 D) 5 E) 6 2 O 2x y = f'(x) –1 –2 #VOBHÌSF 5. f : R Z R * GGPOLTJZPOV < >BSBMóOEBBSUBOES ** G Y FõJUMJóJOJTBóMBZBOUBOFYHFS¿FLTB- x, 0#x<2 ZTWBSES f_ x i = * *** GGPOLTJZPOVOVOZFSFMFLTUSFNVNOPLUBMBSOO 4-x , 2#x<5 PSEJOBUMBSGBSLOONVUMBLEFóFSJPMVS fonksiyonu r x ` 3 J¿JO G Y = f (x + FõJUMJóJOJ TBóMZPS #VOBHÌSF Z=G Y GPOLTJZPOVOVOFLTUSFNVN OPLUBMBSOEBOLBÀUBOFTJ BSBMôOEBLB- MS JGBEFMFSJOEFOIBOHJMFSJLFTJOMJLMFEPôSVEVS \" :BMO[* # *WF** $ *WF*** A) 20 B) 25 C) 40 D) 41 E) 50 % **WF*** & * **WF*** B C 88 B % A

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 .0%·- ·/÷7&34÷5&:&)\";*3-*, .\",4÷.6..÷/÷.6.130#-&.-&3÷ TANIM ÖRNEK 3 #JSGPOLTJZPOVNBLTJNJ[FWFZBNJOJNJ[FFUNF ¦FWSFTJCSPMBOEJLEÌSUHFOMFSEFOLÌöFHFOJFOLÑ- QSPCMFNJOJO¿Ë[ÐNÐOÐ ¿Ë[ÐNJ¿JOJ[JOWFSJMFO ÀÑLPMBOOBMBOLBÀCS2EJS CJS LÐNF EBIJMJOEFLJ EFóFSMFSJOJ TJTUFNBUJL CJS õFLJMEF LVMMBOBSBL BSBNB JõMFNJOF PQUJNJ[BT- B2+C2=L2WFB+C=JÀJO 6 ZPOEFOJS'J[JL CJZPMPKJ NÐIFOEJTMJL FLPOPNJ –+ HJCJCJS¿PLBMBOEBFOJTUFOJMFOTPOVDVFMEFFU- ,ÌöFHFO V[VOMVôVOVO LBSFTJOJ NFLBNBDZMBLVMMBOMS WFSFOGPOLTJZPOL B PMTVO L B =B2+ -B 2 .BLTJNVN NJOJNVN QSPCMFNMFSJOEF UÐSFW ZBSENZMB FO JTUFOJMFO TPOV¿MBS FMEF FUNFZF Lh B =B+ -B - =B- ¿BMõBDBó[#VOVOJ¿JOJTUFOJMFOJGBEFZJUFLEF- óJõLFOMJGPOLTJZPOIºMJOFHFUJSFSFLCVGPOLTJZP- \"MBO== OVO CFMJSMFONJõ LÐNFEF BMBCJMFDFóJ FO CÐZÐL ÖRNEK 4 FOLÐ¿ÐL EFóFSMFSJIFTBQMBZBDBó[ \",ËõFMFSJ Ð¿HFOJO Ð[FSJOEF PMBO NBLTJNVN BMBOM EJL- ÖRNEK 1 EËSUHFOJO BMBO Ð¿HFOJO BMBOOO ZBSTES ÌOFSNFTJ- OJO EPôSV PMEVôVOV WFZB ZBOMö PMEVôVOV JTQBUMB- 'BSLMBS PMBO JLJ TBZOO ÀBSQNOO FO LÑÀÑL EF- ZO[ ôFSJLBÀUS A ha ·ÀHFO 2 h–x %JLEÌSUHFOYZ Y-Z=JTF 3 K yh M h-x = y &y= a ^h-xh YZ=Y Y- =-Y2+Y2 –+ x ha h -+Y=jY= YZ=-=-9 BL NC a \"MBO=ZY= a ^ hx - x2 h = A^ x h h/2 h +– A'^ x h = c ^ h - 2x h x = h a , y = h 22 xy Ü çgenin Alan› Alan = = 42 ÖRNEK 2 f ( x ) = x2 - 9x + 10 ÖRNEK 5 FôSJTJOJO IBOHJ OPLUBTOEBLJ LPPSEJOBUMBS UPQMBN Y - Z = EPôSVTV 0Y WF 0Z FLTFOMFSJOJO PMVö- NJOJNVNEVS UVSEVôVÑÀHFOJOJÀJOFÀJ[JMFOCJSLÌöFTJEPôSVÑ[F- SJOEF PMBO EJLEÌSUHFOMFSEFO FO CÑZÑL BMBOM PMBOO Y+G Y =Y2-Y+=5 Y 4 BMBOLBÀCS2EJS 5h Y =Y- –+ G =-+=- 6.8 PMEVôVOEBO - y \" \"#0 = = 24 O 2 –6 8 x 24 B A %JLEÌSUHFO= = 12 2 –9 m 89 36%PôSVEVS12

·/÷7&34÷5&:&)\";*3-*, .0%·- -÷.÷57&5·3&7 www.aydinyayinlari.com.tr ÖRNEK 6 ÖRNEK 8 :BS¿BQV[VOMVóVDNPMBOCJSEBJSFOJOJ¿FSJTJOEFOLË- NFUSFUFMJMFÀFWSJMFCJMFOEJLEÌSUHFOCJÀJNJOEFLJ õFMFSJEBJSFOJO¿FWSFZBZOOÐ[FSJOEFPMBDBLõFLJMEFCJS UBSMBOOBMBOFOÀPLLBÀN2EJS EJLEËSUHFOLFTJMFSFL¿LBSMZPS B+C=PMNBLÑ[FSF #VOB HÌSF HFSJZF LBMBO CÌMHFOJO BMBO FO B[ LBÀ \"MBO=BC=B -B =\" B DN2PMVS \"h B =-B-B=jB= \"MBO== a B2+C2=PMNBLÑ[FSF b BC OJO FO CÑZÑL EFôFSJOJ BS- 12 ZPSV[ 24 a.b = a 144 - a = 1440.a = f^ a h Gh B =B-B 0 62 ÖRNEK 9 .BLTJNVNEJLEÌSUHFOBMBO – +– JÀJO a = b = 6 2 #JS¿JGU¿JOJOFMJOEFNFUSFMJL¿JUZBQNBZBZFUFDFLNBM- [FNFWBSES#VNBM[FNFMFSMFCJSCJSMFSJZMFLFTJõNFZFO $FWBQCVEVSVNEBÖ-PMVS JLJUBOFLBSFõFLMJOEFLBQBMCËMHFPMVõUVSBDBLWFCVCËM- HFMFSFJLJBZSDJOTUBWVLLPZBDBLUS #VOB HÌSF CV CÌMHFMFSJO UPQMBN BMBO FO B[ LBÀ N2PMVS ,BSFMFSJOCJSBZSUOOV[VOMVôVBWFCNFUSFPMTVO ÖRNEK 7 4a + 4b = 32 _ b b Garaj Çardak :BOEBLJ LSPLJEF FW a+b = 8 2 + ^ 8 - a h2 = A^ a h bb j2+2= 3m HBSBK ¿BSEBLWFCBI- a2 + b2 = ` ¿F BZSUMBS N WF a b NPMBOCJSBSTBZB ZFSMFõUJSJMNJõUJS #BI- 2a + 2a - 16 = 0 b ¿FOJO LFOBSMBSOEBO bb õFLJMEFLJ V[VOMVLMBS- a=4 a EB NFTBGF CSBLMB- Havuz 18 m SBL ZÐ[NF IBWV[V Ev 1 m 1m CBI¿FZFFLMFONJõUJS 2m ÖRNEK 10 16 m 5BCBOZBSÀBQCS ZÑLTFLMJôJCSPMBOLPOJOJOJÀJ- )BWV[VO BMBO N2 PMEVôVOB HÌSF CBIÀFOJO BMB- OF ÀJ[JMFO FO CÑZÑL IBDJNMJ TJMJOEJSJO UBCBO ZBSÀB- OFOLÑÀÑLEFôFSJOJBMEôOEB HBSBKOBMBOLBÀN2 QLBÀCSEJS PMVS 6-h r = jI=-S YZ= 6–h 62 Y+ Z+ =# 6 2 h 40 200 r x x πr h = π r2 ^ 6 - 3r h=) S 2 33 #= Y+ d + 2 n=+Y+ + H'^ r h = π : 12r - 2 D = 0 200 3 9r #h Y =- 2 =jY=WFZ= 4 r= x 3 (BSBK= - - ==N2 ÖmN2 4 3

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 .0%·- ·/÷7&34÷5&:&)\";*3-*, ÖRNEK 11 ÖRNEK 13 G WF H JLJODJ EFSFDFEFO GPOLTJZPOMBSO HSBGJLMFSJ BSBTOB D2 C õFLJMEF WFSJMFO ZBNV- 2 ôVOBMBOOOFOÀPLPM- õFLJMEFLJHJCJEJLEËSUHFOZFSMFõUJSJMJZPS Ax 2 NBTJÀJOYLBÀPMNBM- y y = g(x) ES 4 DC B –2 O 2 x D2C 1 –1 B 2 4 – a2 2 A 2 –1 y = f(x) a a A B :FSMFöUJSJMFOEJLEÌSUHFOJOBMBOFOCÑZÑLPMBOJÀJO# (124+422+424a3) · 4 - 2 = A^ a h OPLUBTOOBQTJTJLBÀUS a 2 G Y =-Y2WFH Y =Y2- ^ a + 2 h,a 4 - 2 k = A^ a h a |AB|=BJÀJO 1. 4 - a2 + ^ a + 2 ha - 2 a k = A' (a) # B B2- $ B -B PMVS 2· 4 - 2 a #VEVSVNEB|BC|=-B2EJS -B2-B2-B= - B2+B- =jB= Y= \" B =B -B2 =B-B3 A'^ a h = 10 - 2 = 0 &a= 5 12a 6 ÖRNEK 14 :BSÀBQDNPMBOCJSLÑSFOJOJÀFSJTJOFZFSMFöUJSJMF- CJMFONBLTJNVNIBDJNMJEJLLPOJOJOZÑLTFLMJôJLBÀ DNEJS ÖRNEK 12 3 x2 - ( a - 2 ) x + 3 - a = 0 x3 r EFOLMFNJOJO LÌLMFSJOJO LBSFMFSJOJO UPQMBNOO NJOJ- NVNPMNBTJÀJOBHFSÀFMTBZTLBÀPMNBMES x +x =a-2 S2+Y2=I=+Y 12 π2 π x x =3-a ·r ·h = a 9 - 2 k^ x + 3 h =) Y 12 33 x 2 + 2 = 2 - 4a + 4 - 6 + 2a H'^ x h = π :^ - 2x h^ x + 3 h + a 9 - 2 kD x x a x 12 3 T^ a h = a2 - 2a + 4 = π a - 3x2 - 6x + 9 k 5h B =B-=jB= 3 = π ^ - 3 h^ x + 3 h^ x - 1 h 3 Y=I=4 5 91 44 1 6

·/÷7&34÷5&:&)\";*3-*, .0%·- -÷.÷57&5·3&7 www.aydinyayinlari.com.tr ÖRNEK 15 ÖRNEK 18 y = 2x3 - 6x2 - 5x + 1 FôSJTJOFÀJ[JMFOUFôFUMFSEFO 0OPLUBTOEBONFUSFEPóVEBCVMVOBO\"MQFSNEL FôJNJFOF[PMBOOUFNBTOPLUBTOOPSEJOBULBÀUS I[MBCBUZB NFUSFLV[FZEFCVMVOBO\"INFUJTF NELI[MBHÐOFZFHJUNFLUFEJS Z=G Y JÀJO Gh Y =Y2-Y- Kuzey Ghh Y =Y- V1=3 m/dk Y-=jY= 200m G =--+=-8 220m ÖRNEK 16 #BU %PóV 5BOFTJ5-EFOÐSFUJMFCJMFOCJSÐSÐOÐOTBUõGJZBU5- O V2=5 m/dk PMBSBLCFMJSMFOEJóJOEFBEFUTBUMBCJMJZPS Güney #VÑSÑOÑOGJZBUOEBZBQMBOIFS5-MJLJOEJSJNEF #VOBHÌSF BSBMBSOEBLJV[BLMLFOB[PMEVôVBOEB0 BEFU GB[MB TBUö ZBQEôOB HÌSF CV ÑSÑOÑO TBUöO- OPLUBTOBPMBOV[BLMLMBSOOUPQMBNLBÀUS EBOFOGB[MBLBÀ5-L»SFMEFFEJMJS E U = -U 2+ U- 2 ,»S Y = +Y -Y- Eh U = -U - + U- = +Y -Y Eh U = U- =jU= ,h Y = -Y + +Y - \"MQFSNFUSF ,h Y =-Y--Y \"INFUNFUSF ,h Y =-Y +=NFUSF -Y=jY= , = = ÖRNEK 19 ÖRNEK 17 ,BMESN Anne ,VNMVL BMBOEB PZOB- ZBO&MJG BOOFTJOJO¿B- LJõJMJLCJSTUBEZVNBTBIJQCJSGVUCPMUBLNOOTF- 80 m óSNBT Ð[FSJOF FO L- ZJSDJMFSJOEFONB¿CBõOBTBCJU5-ÐDSFUJTUFONJõUJS 170 m TB TÐSFEF BOOFTJOJO \"ODBLTFZJSDJMFSNB¿BJMHJHËTUFSNFNJõWFTFZJSDJ ZBOOBLPõBSBLHJUNFL NB¿BHJUNJõUJS#VOVOÐ[FSJOFZËOFUJNCJMFUGJZBUMBSOEB Elif Kumluk alan JTUJZPS &MJGhJO LVNEBLJ ZBQUóIFS5-MJLJOEJSJNMFSEFTFZJSDJTBZTOO ,BMESN I[ NEL LBMESN- LJõJBSUUóOCFMJSMFNJõUJS5BLNONB¿HFMJSMFSJOJOFOÐTU EBLJI[NELES TFWJZFZF HFMEJóJ BOEB CJS LF[ EBIB JOEJSJN LBSBS BMBO ZËOFUJNJOZBQUóIFS5-JOEJSJNEFTFZJSDJTBZT ,BMESNBV[BLMôN BOOFTJOFPMBOV[BLMô LJõJBSUNõUS NPMBO&MJGhJOLBMESNBJMLÀLUôBOEBBOOFTJOFPMBO #VOBHÌSF CVTF[POCPZVODBCJSNBÀUBFOGB[MBHF- V[BLMôNFUSFPMVS MJSFMEFFEJMEJôJOEFTUBEZVNEBFOB[LBÀLPMUVLCPö LBMNöUS .B¿MBSBCJMFUTJ[TFZJSDJBMONBNõUS f^ x h = 6400 + 2 + 150 - x x A 80 150–x +Y -Y =( Y x (h Y = -Y - +Y -Y=jY= 6 10 G2 Y = +Y = -Y G2h Y = -Y - +Y f' ( x) = 2x 1 G2h Y =-Y--Y - =0 =-Y 10 Y= UBNTBZPMNBEôJÀJOFOB[CPöLPMUVLJÀJO 2 6400 + 2 .6 Y=PMVSLPMUVLEPMV LPMUVLCPö 640+x2 x E Y= 6400 + 2 x 2 = 36.a 6400 + 2 k 100x x Y2=jY=j-Y= –8 92 NN

.BLTJNVN.JOJNVN1SPCMFNMFSJ TEST - 25 1. f ( x ) = x3 + 4x2 + 5x + 3 5. :BSÀBQ 8 2 DNPMBOCJSÀFNCFSÑ[FSJOEFLÌ GPOLTJZPOVOB ÀJ[JMFO UFôFUMFSEFO FôJNJ FO LÑ- öFMFSJCVMVOBOFOCÑZÑLBMBOBTBIJQ\"#$%EJL- ÀÑLPMBOOOFôJNJOFEJS EÌSUHFOJOJOBMBOLBÀDN2EJS A) 2 B) - 1 C) - 4 D) - 2 E) - 5 3 333 \" # $ % & 2. ¦FWSFTJ CS PMBO EJLEÌSUHFOMFSEFO LÌöFHFOJ 6. 4x - 3y = NJOJNVNPMBOOOBMBOLBÀCJSJNLBSFEJS EPôSVTVOVO Y WF Z FLTFOMFSJ JMF PMVöUVSEVôV ÑÀHFOJO JÀJOF ÀJ[JMFO CJS LÌöFTJ EPôSV Ñ[FSJO- \" # $ % & EF PMBO EJLEÌSUHFOMFSEFO FO CÑZÑL BMBOM PMB- OOBMBOLBÀCJSJNLBSFEJS \" # $ % & 3. DN V[VOMVóVOEB CJS UFM JLJ QBS¿BZB BZSMZPS 7. y = x FóSJTJÐ[FSJOEF CJSLFOBSY=EPóSVTV 1BS¿BMBSEBOCJSJOEFOLBSF CJSJOEFOFõLFOBSÐ¿HFO Ð[FSJOEFCVMVOBO\"#$%EJLEËSUHFOJWFSJMNJõUJS PMVõUVSVMVZPS #VJLJBMBOUPQMBNOONJOJNVNPMNBTJÀJOLB SFOJOCJSLFOBSLBÀDNPMNBMES 20 3 40 3 60 3 y y= x A) B) C) D C 9+4 3 9+4 3 9+4 3 80 3 90 3 D) E) 9+4 3 9+4 3 OA Bx x = 16 #VOB HÌSF \"#$% EJLEÌSUHFOJOJO BMBO FO ÀPL LBÀCJSJNLBSFEJS 4. ¦FWSFTJ CJSJN PMBO JLJ[LFOBS ÑÀHFOMFSEFO A) 64 B) 80 C) 128 33 33 33 BMBOFOCÑZÑLPMBOOBMBOLBÀCJSJNLBSFEJS D) 256 E) 100 400 200 100 33 3 A) B) C) 33 33 33 D) 50 E) 40 33 33 B A C A 93 C % C

TEST - 26 y = f(x) .BLTJNVN.JOJNVN1SPCMFNMFSJ 1. y 4. #JSLFOBSOOV[VOMVóVDNPMBOLBSFõFLMJOEF- AB x LJ CJS LBSUPOVO IFS LËõFTJOEFO Fõ LBSFMFS LFTJMJQ OC LBUMBO BSBLÐTUÐB¿LEJLEËSUHFOMFSQSJ[NBTõFLMJO- EFCJSLVUVZBQMBDBLUS #VLVUVOVOIBDNJFOGB[MBLBÀDN3UÑS \" # $ % & y = ( x - 3 )2 QBSBCPMÑOÑOÑ[FSJOEFLJCJS#OPL- UBTJMFPMVöUVSVMBOöFLJMEFLJ0\"#$EJLEÌSUHF- OJOJOBMBOFOÀPLLBÀCJSJNLBSFEJS \" # $ % & 2. D 6 C 5. 5BCBOZBSÀBQCS ZÑLTFLMJôJCSPMBOEJLLP 66 OJOJOJÀJOFZFSMFöUJSJMFOFOCÑZÑLIBDJNMJLÑSF- OJOZBSÀ BQLBÀCJSJNEJS A) 1 B) 2 C) 3 D) 4 E) 5 Ax B | | \"#$%ZBNVL AB =YDN | AD | = | DC | = | BC | =DN #VOBHÌSF ZBNVôVOBMBOOONBLTJNVNPMNB- TJÀJOYLBÀDNPMNBMES \" # $ % & 6. :BSÀBQ DN PMBO LÑSFOJO JÀJOF ZFSMFöUJSJMFCJ MFONBLTJNVNIBDJNMJEJLLPOJOJOIBDNJLBÀ rDN3UÑS 3. \" # $ 220 3 y D) 256 4 3 E) 100 AD –2 B x OC 2 y = f(x) õFLJMEFLJQBSBCPMJÀJOFZFSMFöUJSJMFO\"#$%EJL- 7. 5BCBOZBSÀBQCJSJN ZÑLTFLMJôJCJSJNPMBO EÌSUHFOJOJOBMBOFOÀPLLBÀCJSJNEJS LPOJOJO JÀJOF ZFSMFöUJSJMFO FO CÑZÑL IBDJNMJ TJ- 16 3 83 32 3 MJOEJSJOIBDNJLBÀrCJSJNLÑQUÑS A) B) C) \" # $ % & 9 3 9 16 3 40 3 E) D) 3 9 B B C 94 & C % B

.BLTJNVN.JOJNVN1SPCMFNMFSJ TEST - 27 1. /FIJSLFOBSOEBEJLEËSUHFOõFLMJOEFLJCJSCBI¿FOJO 4. #JSLJMJNÐSFUJDJTJ UBOFTJ- YMJSBPMBOLJMJN- FUSBGOFIJSUBSBGOEBLJLFOBSIBSJ¿NFUSFUFMJMF MFSEFOIBGUBEBYUBOFTBUZPSYUBOFLJMJNJOUPQMBN ¿FWSJMFDFLUJS NBMJZFUJY+MJSBES #V LJMJNDJOJO FO ZÑLTFL L»S FMEF FEFCJMNFTJ BAHÇE JÀJOIBGUBEBLBÀBEFULJMJNTBUMNBMES A) 600 B) 500 C) 450 D) 400 E) 300 5. V #VOB HÌSF UFM JMF ÀFWSJMFO CBIÀFOJO BMBO FO 2 = 1 km / sa ÀPLLBÀNFUSFLBSFPMBCJMJS \" # $ % & A 2. ôFLJMEFBMBODN2PMBOEJLEËSUHFOCJ¿JNJOEFLJ 6km LBSUPOVOJ¿FSJTJOFTBóWFTPMEBODN BMUWFÐTUT- 60° OSMBSEBODNV[BLMLPMBDBLõFLJMEFSFTJNZFSMFõ- B UJSJMJZPS V1 = 3km/sa 2 DC :VLBSEBLJõFLJMEF\"WF#OPLUBMBSOEBO71WF72 11 IBSFLFUMJMFSJ TSBTZMB LNTB WF LNTB I[MBSMB AB ZPMB¿LZPSMBS 2 \" WF # OPLUBMBS BSBT V[BLML LN PMEVôVOB #VOBHÌSF SFTNJOBMBOFOÀPLLBÀDN2PMVS \" # $ % & HÌSF CVJLJIBSFLFUMJOJOLBÀTBBUTPOSBBSBMBSO- EBLJNFT BGFFOLTBPMVS A) 2 3 C) 4 5 E) 1 7 B) 7 D) 7 7 6. %FOJ[EFZÐ[FOCJSLJõJOJOLZZBV[BLMóNFU SFEJS OTEL A 2000 m 500 m CB 3. #JS LBNZPO GBCSJLBT ZMEB UBOFTJ MJSBEBO | AB | =NFUSFEJS#VLJõJOJOZÐ[NFI[ LBNZPO TBUZPS )FS CJS LBNZPOVO MJSB NEL ZÐSÐNFI[NELES EBIB VDV[B TBUMNBT IBMJOEF ZMEB LBNZPO EBIBGB[MBTBUMBCJMJZPS #V LJöJ FO LTB TÑSFEF PUFMJOF WBSNBT JÀJO \" EBO LBÀ NFUSF V[BLMLUB $ OPLUBTOB ÀLNBM #VOBHÌSF ZMMLFOCÑZÑLLB[BODTBôMBZBDBL ES LBNZPOGJZBULBÀMJSBPMNBMES 100 150 200 250 300 A) 55000 B) 50000 C) 45000 A) B) C) D) E) D) 40000 E) 35000 6 6 666 A C % 95 B B %

TEST - 28 .BLTJNVN.JOJNVN1SPCMFNMFSJ 1. y = x 5. \"ó[B¿LTJMJOEJSõFLMJOEFLJUFOFLFLBWBOP[-TV FôSJTJOJO\" OPLUBTOBFOZBLOOPLUBTOO BMBCJMNFLUFEJS BQTJTJLBÀUS ÷NBMBUUBLVMMBOMBOUFOFLFOJONJOJNVNNJLUBS- A) 1 B) 2 5 E) 7 EBPMBCJMNFTJJÀJO TJMJOEJSJLLVUVOVOUBCBOZBS- C) D) 3 ÀBQLBÀDNPMNBMES 22 A) 2000 B) 3 2000 C) 3 1000 π π π 1000 E) 100r D) π 2. y = 4 - x2 QBSBCPMÑOÑOCJSJODJCÌMHFEFLJHSBGJ ôJOJOÑ[FSJOEFLJCJS\" B C OPLUBTOEBOÀJ[JMFO UFôFUJOLPPSEJOBUFLTFOMFSJJMFPMVöUVSEVôVÑÀ HFOJOBMBOONJOJNVNPMNBTJÀJOBLBÀPMNBM ES 3 3 23 6. :BS¿BQYDNPMBOCJSEBJSFEFOCJSEJMJNLFTJMJQLW A) B) C) SMBSBLCJSEJLLPOJOJOZBOZÐ[ÐLBQBUMBDBLUS 9 3 3 E) 2 3 D) 3 ,POJOJO FO CÑZÑL IBDJNMJ PMNBT JÀJO LFTJMFO EJMJNJO NFSLF[ BÀTOO ÌMÀÑTÑ LBÀ EFSFDF PM NBMES A) 180 # 0 C) 120 6 2 E) 360 D) 240 3. A ( 3, 4 ) OPLUBTOEBOHFÀFOWF*CÌMHFEFLPPS EJOBU FLTFOMFSJZMF NJOJNVN BMBOM CJS ÑÀHFO PMVöUVSBOEPôSVOVOFôJNJLBÀUS A) -2 B) - 4 C) -1 D) - 2 E) - 1 7. 0 1 x 1 3 4 PMNBLÐ[FSF 3 3 3 y = 1 x2 2 y= x 4. y = k FôSJMFSJOJOPMVöUVSEVôVLBQBMCÌMHFEF FôSJMFSF ÀJ[JMFOWFYFLTFOJOFEJLPMBOLJSJöMFSEFOV[VO x MVôVFOCÑZÑLPMBOOOV[VOMVôVLBÀCJSJNEJS GPOLTJZPOVOVO CBöMBOHÀ OPLUBTOB FO ZBL O A) 3 1 - 3 1 B) 3 1 OPLUBTOO CBöMBOHÀ OPLUBTOB V[BLMô 4 2 2 128 2 PMEVôVOBHÌSF LHFSÀFMTBZTLBÀUS C) 3 1 - 3 1 D) 3 1 - 3 1 A) 1 B) 4 C) 9 D) 16 E) 25 8 72 16 512 E) 3 1 16 & C B % 96 B C A

www.aydinyayinlari.com.tr -÷.÷57&5·3&7 .0%·- ·/÷7&34÷5&:&)\";*3-*, 10-÷/0.(3\"'÷,-&3÷ TANIM ÖRNEK 1 P ( x ) = anxn + an-1xn-1 ++ a0 f ( x ) = x2 - 3x + 2 QPMJOPNVOVOHSBGJôJOJÀJ[JOJ[ CJ¿JNJOEFLJ GPOLTJZPOMBSO QPMJOPN HSBGJLMFSJ ¿J[JMJSLFO G Y =jY2-Y+= y y = f(x) 1. P ( x ) = EFOLMFNJOJO WBSTB LËLMFSJ CVMV- Y=WFY= OVS %FSFDFTJ UFL PMBO LËLMFS FLTFOJ LFTFS- 23 x LFO ¿JGUPMBOMBSFLTFOFUFóFUPMVS Gh Y =Y- O 122 y J P(x) = Y-B MPLBM 3 – 1 PMBSBLHSBGJLWFYFLTFOJ- 2 4 –+ a O a x OJLFTFSHF¿FS 99 1 - +2 =- 42 4 y JJ P ( x ) = Y - b)2n ÖRNEK 2 (n ` /+ J¿JO MPLBM PMB- b SBL HSBGJL Y FLTFOJOF UF- f ( x ) = ( 2 - x ) ( x + 1 ) QPMJOPNVOVOHSBGJôJOJÀJ[JOJ[ Ob x óFUPMVS y JJJ P(x) = Y-D 2n + 1 G Y =JTFY1= Y2=-1 cO c (n `/+J¿JO MPLBMPMBSBL Gh Y =-Y-+-Y=-Y-1 x HSBGJLYFLTFOJOJIFNLF- 1 y TFS IFN EF Y FLTFOJOF 2 3 UFóFUPMVS +– 4 P ( x ) = k (x - a) (x - b)2 (x -D 3J¿JO 2 a < b <DWFL` R-JTF –1 2 x d2- 1 nd 1 +1n= 3 O1 2 22 4 y = f(x) ab x ekseni c 2. #BõMBOH¿WFCJUJõCËMHFMFSJJODFMFOJS BO O FOTPM FOTBô ÖRNEK 3 QP[JUJG UFL CËMHF CËMHF f ( x ) = ( x - 1) (x + 2) ( x + 1 ) QP[JUJG ¿JGU CËMHF CËMHF QPMJOPNVOVOHSBGJôJOJÀJ[JOJ[ OFHBUJG UFL CËMHF CËMHF Y= Y=-WFY=-UFLEFSFDFMJLÌLMFS y OFHBUJG ¿JGU CËMHF CËMHF y = f(x) 3. 5ÐSFW ZBSENZMB QPMJOPNVO BSUQ B[BMEó –2 –1 O 1 x ZFSMFS ZFSFM FLTUSFNVN WF NVUMBL FLTUSF- –2 NVNOPLUBMBSCVMVOVS 97

·/÷7&34÷5&:&)\";*3-*, .0%·- -÷.÷57&5·3&7 www.aydinyayinlari.com.tr ÖRNEK 4 ÖRNEK 6 f ( x ) = ( x - 1) ( 2 - x )2 ( 3 - x ) f ( x ) = x3 -YWFH Y = 4 - 2x2GPOLTJZPOMBSWFSJMJZPS QPMJOPNVOVOHSBGJôJOJÀJ[JOJ[ #VOBHÌSF G Y =H Y FöJUMJôJOJTBôMBZBOLBÀUBOFY HFSÀFMTBZTWBSES Y= Y=UFLLBU Y=ÀJGULBULÌLMFSEJSG =-12 G Y =Y3-Y –1 1 G Y = Y2-Y+ Y- 2 Gh Y =Y- + –+ =-[ Y- 4- Y- 2] 3 WF - 3 Gh Y =- Y- 2- Y- =- Y- Y- 2- UFLLBULÌL H Y =-Y2 x = 2 ve x = - 2 UFLLBULÌL ZF- 2– 1 2+ 1 SFMNBLTJNVN 2 22 + –+– y OPLUB FöJUMJôJ TBô- MBS BWFC y = f(x) y 1 3 x –3 bx O 2–1 2 2+1 aO 3 2 2 –12 y = f(x) ÖRNEK 5 ÖRNEK 7 f ( x ) = ( x - 1)3 (x + 1)2 f ( x ) = 2x3 - 6x2 -Y+ 5 QPMJOPNVOVOHSBGJôJOJÀJ[JOJ[ GPOLTJZPOVOVOHSBGJôJOJÀJ[JOJ[ Y= Y=EFSFDFEFOUFLLBU Y=-ÀJGULBULÌL Gh Y =Y2-Y-= –1 3 Y- Y+ + – + G =-1 Y- 2 Y+ 2+ Y+ Y- 3 f(–1) = 15 f(3) = 49 Y- 2 Y+ (134x4+4432+ 24x4-423) y 5x + 1 y 15 y = f(x) –1/5 O y = f(x) 5 3 x –1 x –1 O 1 –1 –1 –1/5 1 –49 + – ++ 98 UBOF

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AYT Matematik Ders İşleyiş Modülü Limit ve Türev

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AYT Matematik Ders İşleyiş Modülü Limit ve Türev

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