Iklan

Iklan

Pertanyaan

F u n g s i f : R → d an g : R → R d i t e n t u kan o l e h f ( x ) = 3 x − 2 d an g ( x ) = x − 1 x ​ u n t u k x  = 1 , maka ( f o g ) ( x ) a d a l ah ...

  1. fraction numerator 5 x minus 2 over denominator x minus 1 end fraction

  2. fraction numerator 5 x plus 2 over denominator x minus 1 end fraction

  3. fraction numerator x plus 1 over denominator x minus 1 end fraction

  4. fraction numerator x minus 2 over denominator x minus 1 end fraction

  5. fraction numerator x plus 2 over denominator x minus 1 end fraction

Iklan

D. Kamilia

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Iklan

Pembahasan

left parenthesis f o g right parenthesis left parenthesis x right parenthesis space equals space f left parenthesis g left parenthesis x right parenthesis right parenthesis  left parenthesis f o g right parenthesis left parenthesis x right parenthesis equals f open parentheses fraction numerator x over denominator x minus 1 end fraction close parentheses  space space space space space space space space space space space space space equals 3 open parentheses fraction numerator x over denominator x minus 1 end fraction close parentheses minus 2  space space space space space space space space space space space space space equals fraction numerator 3 x over denominator x minus 1 end fraction minus 2  space space space space space space space space space space space space space equals fraction numerator 3 x minus 2 left parenthesis x minus 1 right parenthesis over denominator x minus 1 end fraction  space space space space space space space space space space space space space equals fraction numerator x plus 2 over denominator x minus 1 end fraction

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

2

jesika

Pembahasan lengkap banget Ini yang aku cari! Bantu banget Makasih ❤️ Mudah dimengerti

Iklan

Iklan

Pertanyaan serupa

Diketahui g ( x ) = 2 x + 3 dan f ( x ) = x 2 − 4 x + 6 , maka ( f ∘ g ) ( x ) = ...

3

4.8

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