TUGAS KALKULUS

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Fungsi Invers dan Fungsi Komposisi1. Fungsi f(x) = 2x 3, maka f -1 (x) = Jawab : f(x) = y y = 2x 3 y + 3 = 2x 23 + y= xJadi, f -1 (x) = 23 + x2. Fungsi f(x) = 5 23 4+xx, maka f -1 (x) = Jawab : f(x) = y y = 5 23 4+xx2xy 5y = 4x + 32xy 4x = 3 + 5yx(2y 4) = 3 + 5y x = 4 25 3+y yJadi, f -1 (x) = 4 25 3+x x3. Fungsi f(x) = x2 + 1, maka f -1 (x) = Jawab : f(x) = y y = x2 + 1 y 1 = x21 1 y = xmaka f -1 (x) = 1 x4. Fungsi f(x) = 51 x, maka f -1 (x)= Jawab : f(x) = 51 x f(x) = y y = 51 x xy 5y = 1 xy = 1 + 5y x = y y 5 1+Jadi, f -1 (x) = x x 5 1+5. Fungsi f(x) = x 2 43, maka f -1 (5) = Jawab : f(x) = x 2 43 f(x) = y y = x 2 434y 2xy = 3 2xy = 3 4y x = yy24 3 f -1 (x) = xxatauxx23 424 3 2Jadi, f -1 (5) = ) 5 ( 23 ) 5 ( 4 = 103 20 = 10176. Fungsi f(x) = 32 2+xx, maka f -1 (3) = Jawab :f(x) = 32 2+xxf(x) = yy = 32 2+xxxy + 3y = 2x 2xy 2x = 2 3yx(y 2) = 2 3yx = 23 2 y yf -1 (x) = 23 2 x xJadi, f -1 (3) = 2 3) 3 ( 3 2 = 19 2 = 117. Sebuah fungsi f(x) = ax + b, jika f(-2) = 11 dan f(3) = -4. maka tentukan nilai a dan b?...Jawab :3f(x) = ax + bf(-2) = a(-2) + b = 11 -2a + b = 11 (1)F(3) = a(3) + b = -4 3a + b = -4 (2)Jadi : -2a + b = 11 (1) 3a + b = -4 (2) -5a = 15 a = -3a = -3, subtitusiak pers. 1-2a + b = 11 -2(-3) + b = 11 6 + b = 11 b = 5Jadi, nilai a = -3 dan b = 58. Jika f(x) = 4x2 + 2x 3, maka f(x 2) sama dengan Jawab :f(x) = 4x2 + 2x 3 = 4(x 2)2 + 2(x 2) 3 = 4(x2 4x + 4) + 2x 4 3 = 4x2 16x + 16 + 2x 4 3 = 4x2 14 x + 9Jadi, f(x) = 4x2 14 x + 99. Jika diketahui f(x) = 3x2 + 1 dan g(x) = 2x + 1, maka tentukan (f o g)(x) Jawab :(f o g)(x) = f(g(x))= f (2x + 1)= 3(2x + 1 )2 + 1= 3(4x2 + 4x + 1) + 1= 12x2 + 12x + 44Jadi, (f o g)(x) = 12x2 + 12x + 410. Jika diketahui f(x) = 3 62 5+xxdan g(x) = 4x2 + 2, maka (f o g)(x) = Jawab :(f o g)(x) = f(g(x))= 3 ) 2 4 ( 62 ) 2 4 ( 522+ + +xx= 3 12 242 10 2022+ + +xx= 15 248 2022++xxJadi, (f o g)(x) = 15 248 2022++xx11. Jika f(x) = 4 25+ xdan g(x) = x2 1, maka (f o g)(2) = Jawab :(f o g)(x) = f(g(x))= 4 ) 1 ( 252+ x= 4 2 252+ x= 2 252+ xJadi, (f o g)(2) = 2 ) 2 ( 252+= 2 85+= 21105512. Jika diketahui f(x) = 4x 1 dan g(x) = x2 + 1, maka (g o f)(x) = Jawab :(g o f)(x) = g(f(x))= g(4x 1)= (4x 1)2 + 1= 16x2 8x + 1 + 1= 16x2 8x + 2Jadi, (g o f)(x) = 16x2 8x + 213. Jika diketahui f(x) = 2x2, g(x) = 3x + 4, h(x) = x2 2, maka tentukan (f o g o h)(x) = Jawab :(f o g oh)(x) = f o g(h(x))= f o g(x2 2)= f(3(x2 2) + 4)= f(3x2 6 + 4)= f(3x2 2)= 2(3x2 2)= 2(9x4 12x2 + 4)= 18x4 24x2 + 8Jadi, (f o g o h)(x) = 18x4 24x2 + 814. Jika diketahui f(x) = 2x2, g(x) = 3x + 4, h(x) = x2 2, maka tentukan (g o f o h)(x) = Jawab :(g o f o h)(x) = g o f(h(x))= g o f(x2 2)= g(2(x2 2)2)= g(2(x4 4x2 + 4))= g(2x4 8x2 + 8)= 3(2x4 8x2 + 8) + 46= 6x4 24x2 + 24 + 4= 6x4 24x2 + 28Jadi, (g o f o h)(x) = 6x4 24x2 + 2815. Jika diketahui f(x) = 2x2, g(x) = 3x + 4, h(x) = x2 2, maka tentukan (g o h o f)(-2) = Jawab :(g o h o f)(x) = g o h(f(x))= g o h(2x2)= g((2x2)2 2)= g(4x4 2)= 3(4x4 2) + 4= 12x4 6 + 4= 12x4 2Jadi, (g o h o f)(-2) = 12(-2)4 2 = 12(16) 2 = 192 2 = 190 16. Jika diketahui fungsi f(x) = 3x2 + 1 dan g(x) = x + 2, maka tentukan nilai (g o f)(3) = Jawab :(g o f)(x) = g(f(x))= g(3x2 + 1)= (3x2 + 1) + 2= 3x2 +3Jadi, (g o f)(3) = 3(3)2 + 3= 3 + 9 + 3= 15717. Jika diketahui fungsi f(x) = 3x2 + 1 dan g(x) = x + 2, maka tentukan nilai (f o g)(5) = Jawab :(f o g)(x) = f(g(x))= f(x + 2)= 3(x + 2)2 + 1= 3(x2 + 4x + 4) + 1= 3x2 + 12x + 12 + 1= 3x2 + 12x + 13Jadi, (f o g)(5) = 3(5)2 + 12(5) + 13= 75 + 60 + 13= 14818. Jika diketahui fungsi f(x) = 3x2 + 1 dan g(x) = x + 2, maka tentukan nilai (g0 g)(-3) = Jawab :(g o g)(x) = g(g(x))= g(x + 2)= (x + 2) + 2= x + 4Jadi, (g o g)(-3) = (-3) + 4= 119. Jika diketahui fungsi f(x) = 3x2 + 1 dan g(x) = x + 2, maka tentukan nilai (f o f)(-5) = Jawab :(f o f)(x) = f(f(x))= f(3x2 + 1)= 3(3x2)2 + 1= 3(6x4) + 1= 18x4 + 18Jadi, (f o f)(-5) = 18(-5)4 + 1= 18(625) + 1= 1125120. Diketahui fungsi (f o g)(x) = -2x + 3, f(x) = 4x 1, maka tentukan g(x) = Jawab :(f o g)(x) = -2x + 3f(g(x)) = -2x + 34(g(x)) 1 = -2x + 34(g(x)) = -2x + 3 + 14(g(x)) = -2x + 4g(x) = 44 2 + xJadi, g(x) = 121+ x21. Diketahui fungsi (g o f)(x) = x2 6x + 3, g(x) = 2x 3, maka tentukan f(x) = Jawab :(g o f)(x) = x2 6x + 3g(f(x)) = x2 6x + 32(f(x)) 3 = x2 6x + 32(f(x)) = x2 6x + 3 + 32(f(x)) = x2 6x + 6f(x) = 26 62+ x xJadi, f(x) = 3 3212+ x x22. Diketahui (f o g)(x) = 4 2x dan g(x) = 6x + 1, maka tentukan f(x) = Jawab :9(f o g)(x) = 4 2xf(g(x)) = 4 2x g(x) = 6x + 1g(x) 1 = 6x61 ) ( x g= xf(g(x)) = 4 2 ,_

61 ) (x g= 4 626) ( 2+x g= 4 31) (31+ x g= ) (31313x g Jadi, f(x) = x3131323. Diketahui (g o f)(x) = 1 12 42+ x x dan f(x) = 2x 3, tentukan g(x) = Jawab :(g o f)(x) = 1 12 42+ x xg(f(x)) = 1 12 42+ x x f(x) = 2x 3f(x) + 3 = 2xxx f+23 ) (g(f(x)) = 4 123 ) (1223 ) (2+

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+

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+ x f x f= 1236) (21249 ) ( 6 ) (42+

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+ +x fx f x f= f(x)2 + 6f(x) + 9 6f(x) 18 + 110= f(x)2 8Jadi, g(x) = x2 824. Diketahui (f o g)(x) = x x +2dan g(x) = 2x + 1, maka tentukan f(x) = Jawab :(f o g)(x) = x x +2f(g(x)) = x x +2 g(x) = 2x + 1g(x) 1 = 2xxx g21 ) (f(g(x)) =

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+

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21 ) (21 ) (2x g x g= ,_

+

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+ 21 ) (41 ) ( 2 ) (2x g x g x g= ,_

+

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+ 42 ) ( 241 ) ( 2 ) (2x g x g x g= 41 ) (2 x gJadi, f(x) = 41414122x ataux25. Diketahui (f o g)(x) = 24 x dan g(x) = 12+ x , maka tentukan f(x) = Jawab :(f o g)(x) = 24 xf(g(x)) = 24 x g(x) = 12+ xg(x) 1 = 2xx = 1 ) ( x g11f(g(x)) = ( ) 2 1 ) (4 x g= 2 1 ) ( 2 ) (2 + x g x gJadi, f(x) = 1 22 x x 26. Jika f(x) = 2x 4 dan g(x) = 3x + 1, maka tentukan invers dari (g o f) -1 (x) = Jawab :Caranya, dikomposisikan dahulu kemudian di inverskan.(g o f)(x) = g(f(x))= g(2x 4)= 3(2x 4) + 1= 6x 12 + 1= 6x 11(g o f)(x) = yy = 6x 11y + 11 = 6x611 + y= xJadi, (g o f) -1 (x) = 611 + x27. Jika f(x) = 2x 4 dan g(x) = 3x + 1, maka tentukan invers dari (f o g) -1 (x) = Jawab :Caranya, dikomposisikan dahulu kemudian di inverskan.(f o g)(x) = f(g(x))= f(3x + 1)= 2(3x + 1) 4= 6x + 2 4= 6x 212(f o g)(x) = yy = 6x 2y + 2 = 6x62 + y= xJadi, (f o g) -1 (x) = 62 + x28. Jika f(x) = 3x 4 dan g(x) = 3x + 2, maka tentukan invers dari (f -1 o g -1) (x) = Jawab :Caranya, di inverskan masing-masing fungsi kemudian dikomposisiskan. f(x) = 3x 4f(x) = yy = 3x 4y + 4 = 3x34 + y= xMaka f -1(x) = 34 + x g(x) = 3x + 2g(x) = yy = 3x + 2y 2 = 3x32 y= xMaka g -1(x) = 32 xMaka, (f -1 o g -1) (x) = f -1(g -1(x))= f ,_

32 x13= 3432+ x= 331232+ x= 3310 + x= 910 + xJadi, (f -1 o g -1) (x) = 910 + x29. Jika f(x) = 3x 4 dan g(x) = 3x + 2, maka tentukan invers dari (g -1 o f -1) (x) = Jawab :Caranya, di inverskan masing-masing fungsi kemudian dikomposisiskan. f(x) = 3x 4f(x) = yy = 3x 4y + 4 = 3x34 + y= xMaka f -1(x) = 34 + x g(x) = 3x + 2g(x) = yy = 3x + 2y 2 = 3x32 y= xMaka g -1(x) = 32 xJadi (g -1 o f -1) (x) = g -1(f -1(x))14= g ,_

+34 x= 3234+ x= 33634+ x= 332 x= 92 xJadi, (g -1 o f -1) (x) = 92 xIntegral30. dx x adalah Jawab : x dx = 21x dx= 1212111 ++ x + c= 23231x + c= 2332x + c= 21.321x x + c= x x .321 + cJadi, x dx = x x .321 + c1531. dx x4 3adalah Jawab :dx x4 3= dx x43= c x +++1434311= c x +++444344431= c x +47471= c x +4774= c x x +4 374Jadi, dx x4 3 = c x x +4 37432. dxxx251adalah Jawab : dx x dx x dxxdxxxdxxx2 32 25251 1 = c x x ++ ++ + 1 2 1 31 211 31 = c x x +

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1 41141 = c x x + + 1 441= cxx + +1414Jadi, dxxx251 = cxx + +14141633. 21x dx adalah Jawab : 21x dx = 2xdx= 1 21 21 + + x + c= 111 x + c= 1 x + c atau cx + 1Jadi, 21x dx = 1 x + c atau cx + 134. + dx x x ) 2 (3 adalah Jawab : + dx x x ) 2 (3= + xdx dx x32= c x x +++++ + 1 1 1 31 111 32= c x x + +2 42142Jadi, + dx x x ) 2 (3 = c x x + +2 4214235. + dx x x x ) 2 4 (2 3 5 adalah Jawab : + dx x x x ) 2 4 (2 3 5= + dx x dx x dx x2 3 52 4= 1 2 1 3 1 51 221 341 51 + + +++++ x x x + c= c x x x + + 3 4 632446117= c x x x + + 4 4 63261Jadi, + dx x x x ) 2 4 (2 3 5 = c x x x + + 4 4 6326136. + dx x x2 2) 3 ( adalah Jawab : + dx x x2 2) 3 (= [ ] + +2 2 2 2) 3 ( ) 3 )( ( 2 ) ( x x x x dx= + + ) 9 6 (2 3 4x x x dx= c x x x +++++++ + + 1 2 1 3 1 41 291 361 41= c x x x + + +3 4 5392651= c x x x + + +3 4 532351Jadi, + dx x x2 2) 3 ( = c x x x + + +3 4 53235137. 213dx x adalah Jawab :213dx x = 214411]1x= ,_

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4 4) 1 (41) 2 (41= ,_

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) 1 (41) 16 (41= 41416= 415 atau 43318Jadi, 213dx x = 415 atau 43338. +212) 3 ( dx x x adalah Jawab :+212) 3 ( dx x x = 212 323311]1+ x x= ,_

+

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+2 3 2 3) 1 (23) 1 (31) 2 (23) 2 (31= ,_

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+ ) 1 (23) 1 (31) 4 (23) 8 (31= ,_

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+233121238= 233121238 + += 232123138 + += 2939+= 214 3 += 217Jadi, +212) 3 ( dx x x = 21739. dx x x ) 3 4 cos( sin 5 ( adalah Jawab : dx x x ) 3 4 cos( sin 5 ( = dx x xdx ) 3 4 cos( sin 519 = c x x + )) 3 4 sin( ( ) sin ( 531 = c x x + + ) 3 4 sin( cos 531Maka dx x x ) 3 4 cos( sin 5 ( = c x x + + ) 3 4 sin( cos 53140. dx x x ) 2 cos (cos21 adalah Jawab : dx x x ) 2 cos (cos21= xdx