SOAL MATEMATIKA FUNGSI KOMPOSISI.docx

1.

Jika g(x) = x + 1 dan (f o g)(x) = x2 + 3x + 1 maka f(x) sama dengan ... A. x2 + 5x + 5 B. x2 + x - 1 C. x2 + 4x + 3 D. x2 + 6x + 1 E. x2 + 3x - 1

D. x + 5 E. x - 3 Pembahasan

f(x) = x + 2 *(x) = x2 - 2 (f o g o *)(x) = 2x 2 + 4 2 ⇒ f(g(*(x))) = 2x  + 4 2 2 ⇒ f(g(x  - 2)) = 2x  + 4 2 2 ⇒ g(x  - 2) + 2 = 2x  + 4 2 2 ⇒ g(x  - 2) = 2x  + 2 misa!kan x2 - 2 = a maka x = (a + 2) 2 ⇒ g(a) = 2'√(a + 2)  + 2 ⇒ g(a) = 2.(a + 2) + 2 ⇒ g(a) = 2a + 4 + 2 ⇒ g(a) = 2a + 6  Jadi# g(x) = 2x + 6

Pembahasan

g(x) = x + 1 (f o g)(x) = x 2 + 3x + 1 2 ⇒ (f o g)(x) = x  + 3x + 1 2 ⇒ f (g(x)) = x  + 3x + 1 2 ⇒ f(x + 1) = x  + 3x + 1 isa! x + 1 = "# maka x = " - 1. 2 ⇒ f(") = (" - 1)  + 3(" - 1) + 1 2 ⇒ f(") = "  - 2" + 1 + 3" - 3 + 1 2 ⇒ f(") = "  + " - 1  Jadi f(x) = x2 + x - 1 ---$ o"si B. 2.

4.

Jika g( g(x + 1) = 2x - 1 da dan f( f(g(x + 1) 1)) = 2x + 4# maka f(%) sama dengan ... A. 6 B. 5 C. 3 D. -4 E. -6

Pembahasan

f(x) = 1& (2x - 1) (f o g)(x) = x& (3x - 2) ⇒ f(g(x)) = x& (3x - 2) ⇒ 1& (2(g(x)) - 1) = x& (3x - 2) ⇒ (3x - 2) = x(2(g(x)) - 1) ⇒ (3x - 2) = x(2(g(x)) - 1) ⇒ (3x - 2)&x = (2(g(x)) - 1) ⇒ 2(g(x)) = (3x - 2)&x + 1 ---$ samakan "en/e0 ⇒ 2(g(x)) = (3x - 2 + x) & x ⇒ g(x) = (3x - 2 + x) & 2x ⇒ g(x) = (4x - 2) & 2x ⇒ g(x) = 2 - 1&x ---$ o"si C.

Pembahasan

g(x + 1) = 2x - 1 f(g(x + 1)) = 2x + 4 ⇒ f(2x - 1) = 2x + 4 misa! 2x - 1 = "# maka x = (" + 1)&2 ⇒ f(") = 2'(" + 1)&2 + 4 ⇒ f(") = " + 1 + 4 maka f(x) = x + 5 ⇒ f(%) = % + 5 = 5 ---$ o"si B. 3.

Dikea*i f , x → x + 2 dan * , x → x2 - 2. Jika (f o g o *)(x) = 2x 2 + 4# maka g(x) ada!a* ... A. 2x + 3 B. 2x + 6 C. 2x + 

Jika f( f(xx) = 1& (2x - 1) da dan n (f o g)(x) = x& (3x - 2)# maka g(x) sama dengan ... A.2 + 1&x B. 1 + 2&x C. 2 - 1&x D. 1 - 1&x E. 2 - 2&x

5.

Jika f(x) = (x + 1) 1) da dan (f (f o g)(x) = 2 (x - 1)# maka fngsi g(x) ada!a* ...

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A. 2x - 1 B. 2x - 3 C. 4x - 5 D. 4x - 3 E. 5x - 4

C. 11 D. 14 E. 1 Pembahasan

(g o f)(x) = g(f(x)) 2 ⇒ (g o f)(x) = 2(3x - 1)  + 3 2 ⇒ (g o f)(x) = 2(x  - 6x + 1) + 3 2 ⇒ (g o f)(x) = 1x  - 12x + 2 + 3 2 ⇒ (g o f)(x) = 1x  - 12x + 5 2 ⇒ (g o f)(1) = 1(1)  - 12(1) + 2 + 3 ⇒ (g o f)(1) = 1 - 12 + 2 + 3 ⇒ (g o f)(1) = 11 ---$o"si C.

Pembahasan

f(x) = (x + 1) (f o g)(x) = 2 (x - 1) ⇒ f (g(x)) = 2 √(x - 1) ⇒ √(g(x) + 1) = 2 √(x - 1) ---$ dikadakan ⇒ (g(x) + 1) = 4(x - 1) ⇒ g(x) = 4x - 4 - 1 ⇒ g(x) = 4x - 5 ---$ o"si C. 2

6. Jika Jika f(x) f(x) = x - 4# maka maka ni!a ni!aii f(x) f(x) + (f(x)) (f(x))  3f(x) nk x = 3 ada!a* ... A. 3 B. 1% C. 12 D. 14 E. 16

. Dikea Dikea*i *i f(x) f(x) = (x + 4)& (6x (6x - 5) # x  5&6 dan fngsi in7es dai f(x) ada!a* f 1 (x). i!ai dai f -1(2) sama dengan ... A. 14&3 B. 1&14 C. 6&21 D. -1&14 E. -14&3

Pembahasan

Pembahasan

f(x) = x - 4 (f(x))2 = (x - 4)(x - 4) = x 2 - x + 16 3f(x) = 3(x - 4) = 3x - 12

f(x) = (x + 4)& (6x - 5) ⇒ / = (x + 4)& (6x - 5) ⇒ /(6x - 5) = (x + 4) ⇒ 6x/ - 5/ = x + 4 ⇒ 6x/ - x = 5/ + 4 ⇒ (6/ - )x = 5/ + 4 ⇒ x = (5/ + 4)& (6/ - ) aka di"eo!e* , -1 ⇒ f  (x) = (5x + 4)& (6x - ) dengan x  &6 -1 ⇒ f  (2) = (5.2 + 4)& (6.2 - ) -1 ⇒ f  (2) = 14& 3 ---$ o"si A.

aka , f(x) + (f(x))2 - 3f(x) = x - 4 + x 2 - x + 16 (3x - 12) 2 2 - x + ⇒ f(x) + (f(x))  - 3f(x) = x - 4 + x 16 - 3x + 12 2 2 - 1%x + 24 ⇒ f(x) + (f(x))  - 3f(x) = x nk x = 3 di"eo!e* , 2 - 1%(3) + 24 =  - 3% + 24 = 3 ---$ ⇒ 3 o"si A

. Jik Jika a g(x g(x + 1) = 2x - 1 dan f(g f(g(x (x + 1)) 1)) = 2

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E. -6 Pembahasan

f(g(x + 1)) = 2x + 4 ⇒ f(2x - 1)) = 2x + 4 misa! 2x - 1 = %# maka x = ½ ⇒ f(%) = 2(8) + 4 = 5 ---$ B.

1%.Jika f(x + 1) = x - 3 dan g(x) = x 2 - 2x maka ni!ai (f -1 o g)(3) ada!a* ... A. -3 B. -1

C. 1 D. 3 E.  Pembahasan

f(x + 1) = x - 3 ⇒ f(x) = x - 4 -1 ⇒ f  (x) = x + 4 -1 -1 ⇒ (f   o g)(x) = f  (g(x)) -1 2 ⇒ (f   o g)(x) = x  - 2x + 4 -1 2 ⇒ (f   o g)(3) = 3  - 2(3) + 4 -1 ⇒ (f   o g)(3) =  - 6 + 4 -1 ⇒ (f   o g)(x) = ---$ o"si E.