Iklan

Pertanyaan

Diketahui fungsi f ( x ) = x − 2 dan ( g ∘ f ) ( x ) = 2 x − 3 . Tentukan: a . g ( x ) b . ( g ∘ f ) − 1 ( x )

Diketahui fungsi .

Tentukan:

 

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

02

:

20

:

36

:

22

Klaim

Iklan

E. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sebelas Maret

Jawaban terverifikasi

Pembahasan

Diketahui fungsi . Menentukan Misalkan , sehingga a. Jadi, . Menentukan b. Jadi, .

Diketahui fungsi begin mathsize 14px style f left parenthesis x right parenthesis equals x minus 2 space dan space left parenthesis g ring operator f right parenthesis left parenthesis x right parenthesis equals 2 x minus 3 end style.

Menentukanbegin mathsize 14px style g left parenthesis x right parenthesis end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis g ring operator f right parenthesis left parenthesis x right parenthesis end cell equals cell 2 x minus 3 end cell row cell g left parenthesis f left parenthesis x right parenthesis right parenthesis end cell equals cell 2 x minus 3 end cell row cell g left parenthesis x minus 2 right parenthesis end cell equals cell 2 x minus 3 end cell row blank blank blank end table end style 

Misalkan begin mathsize 14px style a equals x minus 2 space maka space x equals a plus 2 end style, sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell g left parenthesis a right parenthesis end cell equals cell 2 left parenthesis a plus 2 right parenthesis minus 3 end cell row cell g left parenthesis a right parenthesis end cell equals cell 2 a plus 4 minus 3 end cell row cell g left parenthesis a right parenthesis end cell equals cell 2 a plus 1 end cell row cell g left parenthesis x right parenthesis end cell equals cell 2 x plus 1 end cell end table end style 

a. Jadi, begin mathsize 14px style g left parenthesis x right parenthesis equals 2 x plus 1 end style.

Menentukan begin mathsize 14px style left parenthesis g ring operator f right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis g ring operator f right parenthesis left parenthesis x right parenthesis end cell equals cell 2 x minus 3 end cell row y equals cell 2 x minus 3 end cell row cell y plus 3 end cell equals cell 2 x end cell row cell fraction numerator y plus 3 over denominator 2 end fraction end cell equals x row cell fraction numerator y plus 3 over denominator 2 end fraction end cell equals cell left parenthesis g ring operator f right parenthesis to the power of negative 1 end exponent left parenthesis y right parenthesis space end cell row cell fraction numerator x plus 3 over denominator 2 end fraction end cell equals cell left parenthesis g ring operator f right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis end cell row blank blank blank row blank blank blank end table end style 

b. Jadi, begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis g ring operator f right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell fraction numerator x plus 3 over denominator 2 end fraction end cell end table end style

 

 

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

1

Mohammad Jas Miko Maullana

Ini yang aku cari!

Iklan

Pertanyaan serupa

Diketahui jika fungsi f ( x ) = x + 3 dan ( f ∘ g ) − 1 ( x ) = 2 x − 7 , maka fungsi g ( x ) adalah...

3

4.3

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