Iklan

Iklan

Pertanyaan

Diketahi f − 1 , g − 1 dan h − 1 berturut-turut menyatakan invers fungsi f , g dan h . Jika ( f − 1 ∘ g − 1 ∘ h − 1 ) ( x ) = 2 x − 4 dan , nilai f ( 8 ) adalah ....

Diketahi ,  dan  berturut-turut menyatakan invers fungsi ,  dan . Jika  dan Error converting from MathML to accessible text., nilai  adalah ....

  1. begin mathsize 14px style negative 3 over 11 end style 

  2. size 14px minus size 14px 4 over size 14px 5 

  3. size 14px minus size 14px 9 over size 14px 11 

  4. size 14px minus size 14px 12 over size 14px 11 

  5. size 14px minus size 14px 5 over size 14px 11 

Iklan

R. RGFLLIMA

Master Teacher

Jawaban terverifikasi

Iklan

Pembahasan

Dengan konsep invers komposisi fungsi: Misalkan maka: Jika , Maka: Dengan menerapkan sifat : Misalkan , Maka Jadi, NIlai dari .

Dengan konsep invers komposisi fungsi:

Misalkan begin mathsize 14px style straight y equals left parenthesis straight h ring operator straight g right parenthesis left parenthesis straight x right parenthesis rightwards arrow straight x equals left parenthesis straight h ring operator straight g right parenthesis to the power of negative 1 end exponent left parenthesis straight y right parenthesis end style maka:

Error converting from MathML to accessible text. 

Jika 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 2 straight x minus 4 end style, Maka:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row 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 equals cell 2 straight x minus 4 end cell row cell straight f to the power of negative 1 end exponent left parenthesis left parenthesis 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 right parenthesis end cell equals cell 2 straight x minus 4 end cell row cell straight f to the power of negative 1 end exponent left parenthesis fraction numerator negative straight x minus 3 over denominator 2 straight x minus 1 end fraction right parenthesis end cell equals cell 2 straight x minus 4 end cell row cell Misalkan space straight p end cell equals cell fraction numerator negative straight x minus 3 over denominator 2 straight x minus 1 end fraction end cell row cell straight p left parenthesis 2 straight x minus 1 right parenthesis end cell equals cell negative straight x minus 3 end cell row cell 2 xp minus straight p end cell equals cell negative straight x minus 3 end cell row cell 2 xp plus straight x end cell equals cell straight p minus 3 end cell row cell straight x left parenthesis 2 straight p plus 1 right parenthesis end cell equals cell straight p minus 3 end cell row straight x equals cell fraction numerator straight p minus 3 over denominator 2 straight p plus 1 end fraction end cell row cell rightwards double arrow straight f to the power of negative 1 end exponent left parenthesis fraction numerator negative straight x minus 3 over denominator 2 straight x minus 1 end fraction right parenthesis end cell equals cell 2 x minus 4 end cell row cell straight f to the power of negative 1 end exponent left parenthesis straight p right parenthesis end cell equals cell 2 left parenthesis fraction numerator straight p minus 3 over denominator 2 straight p plus 1 end fraction right parenthesis minus 4 end cell row cell straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis end cell equals cell 2 left parenthesis fraction numerator straight x minus 3 over denominator 2 straight x plus 1 end fraction right parenthesis minus 4 end cell end table end style 

Dengan menerapkan sifat begin mathsize 14px style left parenthesis left parenthesis straight f right parenthesis to the power of negative 1 end exponent right parenthesis to the power of negative 1 end exponent left parenthesis straight x right parenthesis equals straight f left parenthesis straight x right parenthesis end style:

Misalkan begin mathsize 14px style straight m equals straight f to the power of negative 1 end exponent left parenthesis straight x right parenthesis rightwards arrow straight x equals straight f left parenthesis straight m right parenthesis end style, Maka

Error converting from MathML to accessible text. 

Jadi, NIlai dari begin mathsize 14px style straight f left parenthesis 8 right parenthesis equals negative 9 over 11 end style.

 

 

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

2

Iklan

Iklan

Pertanyaan serupa

Jika f ( x ) = x − 1 dan g ( x ) = 2 x + 4 , maka yang tepat adalah ....

2

4.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia