Iklan

Iklan

Pertanyaan

Diketahui dan b. ( g − 1 ∘ f − 1 ) ( x ) ?

Diketahui begin mathsize 14px style f open parentheses x close parentheses equals x minus 3 end style dan begin mathsize 14px style g open parentheses x close parentheses equals 2 x plus 4 end style 

b.  ?  

   space undefined

Iklan

S. Nur

Master Teacher

Jawaban terverifikasi

Jawaban

diperoleh bahwa

diperoleh bahwa begin mathsize 14px style left parenthesis g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis equals fraction numerator x minus 1 over denominator 2 end fraction end style                undefined 

Iklan

Pembahasan

Dengan menggunakan rumus invers fungsi Diperoleh Nilai adalah dimasukkan ke Jadi, diperoleh bahwa

Dengan menggunakan rumus invers fungsi

begin mathsize 14px style z open parentheses x close parentheses equals a x plus b blank semicolon a not equal to 0 z to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x minus b over denominator a end fraction blank semicolon a not equal to 0 end style 

Diperoleh 

begin mathsize 14px style f open parentheses x close parentheses equals x minus 3 f to the power of negative 1 end exponent open parentheses x close parentheses equals x plus 3 g open parentheses x close parentheses equals 2 x plus 4 g to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x minus 4 over denominator 2 end fraction end style  

Nilai begin mathsize 14px style left parenthesis g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis end style adalah begin mathsize 14px style f to the power of negative 1 end exponent end style dimasukkan ke begin mathsize 14px style g to the power of negative 1 end exponent space colon space left parenthesis g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis equals g to the power of negative 1 end exponent left parenthesis f to the power of negative 1 end exponent left parenthesis x right parenthesis right parenthesis end style 

begin mathsize 14px style left parenthesis g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis equals fraction numerator left parenthesis x plus 3 right parenthesis minus 4 over denominator 2 end fraction left parenthesis g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis equals fraction numerator x minus 1 over denominator 2 end fraction end style  

Jadi, diperoleh bahwa begin mathsize 14px style left parenthesis g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis equals fraction numerator x minus 1 over denominator 2 end fraction end style                undefined 

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

38

Iklan

Iklan

Pertanyaan serupa

Diketahui dan c. Apakah ( f − 1 ∘ g − 1 ) ( x ) = ( g − 1 ∘ f − 1 ) ( x ) ?

1

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