Pertanyaan

Fungsi invers f : R → R , g : R → R ditentukan dengan rumus f ( x ) = x + 3 , g ( x ) = − 4 x , dan h ( x ) = 5 x + 1. Tentukan. a. ( f − 1 ∘ g − 1 ∘ h − 1 ) ( x )

Fungsi invers $f : R \to R$ , $g : R \to R$ ditentukan dengan rumus $f (x) = x + 3,$ $g (x) = - 4 x,$ dan $h (x) = 5 x + 1.$

Tentukan.

a. $(f^{- 1} \circ g^{- 1} \circ h^{- 1}) (x)$

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

Klaim

R. RGFLLIMA

Master Teacher

Jawaban terverifikasi

Jawaban

$begin mathsize 14px style left parenthesis straight f to the power of negative 1 end exponent ring operator straight g to the power of negative 1 end exponent ring operator straight h to the power of negative 1 end exponent right parenthesis left parenthesis straight x right parenthesis equals fraction numerator x plus 59 over denominator negative 20 end fraction. end style$

Pembahasan

Dengan sifat invers fungsi komposisi: Dengan: Misalkan: Maka: Jadi,

Dengan sifat invers fungsi komposisi:

Dengan:

Misalkan:

begin mathsize 14px style straight y equals left parenthesis straight h ring operator straight g ring operator straight f right parenthesis left parenthesis straight x right parenthesis end style

Maka:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row straight y equals cell negative 20 straight x minus 59 end cell row cell straight y plus 59 end cell equals cell negative 20 straight x end cell row cell fraction numerator straight y plus 59 over denominator negative 20 end fraction end cell equals straight x row cell fraction numerator straight y plus 59 over denominator negative 20 end fraction end cell equals cell left parenthesis straight h ring operator straight g ring operator straight f right parenthesis to the power of negative 1 end exponent left parenthesis straight y right parenthesis space end cell row cell fraction numerator straight x plus 59 over denominator negative 20 end fraction end cell equals cell left parenthesis straight h ring operator straight g ring operator straight f right parenthesis to the power of negative 1 end exponent left parenthesis straight x right parenthesis end cell row cell fraction numerator straight x plus 59 over denominator negative 20 end fraction end cell equals cell left parenthesis straight f to the power of negative 1 end exponent ring operator straight g to the power of negative 1 end exponent ring operator straight h to the power of negative 1 end exponent right parenthesis left parenthesis straight x right parenthesis end cell end table end style$

Jadi, $begin mathsize 14px style left parenthesis straight f to the power of negative 1 end exponent ring operator straight g to the power of negative 1 end exponent ring operator straight h to the power of negative 1 end exponent right parenthesis left parenthesis straight x right parenthesis equals fraction numerator x plus 59 over denominator negative 20 end fraction. end style$

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

0.0 (0 rating)

Pertanyaan serupa

f ( x ) = x + 3 2 x − 5 , x  = − 3 dan g ( x ) = x 2 , maka ( f ∘ g ) − 1 ( x ) = ....

5.0

Jawaban terverifikasi

Jika fungsi f : R → R dan g : R → R didefinisikan dengan f ( x ) = 4 x + 8 dan g ( x ) = 2 x − 2 , tentukan: c. ( f ∘ g ) ( x ) d. ( f ∘ g ) − 1 ( 2 )

0.0

Jawaban terverifikasi

Jika f dan g adalah fungsi dalam R dengan f ( x ) = x + 5 dan g ( x ) = 3 x + 4 , tentukan: a. ( f ∘ g ) − 1 ( x )

5.0

Jawaban terverifikasi

Jika f ( x ) = 4 x + 6 , g ( x ) = x − 1 x ; x  = 1 , dan h ( x ) = ( f ∘ g ) ( x ) , h − 1 ( x ) adalah ...

5.0

Jawaban terverifikasi

Jika f : R → R dan g : R → R ditentukan oleh f ( x ) = x + 5 dan g ( x ) = 3 x − 7 ,tentukan: b. ( g ∘ f ) − 1 ( x )

0.0

Jawaban terverifikasi