Iklan

Iklan

Pertanyaan

Jika f ( x ) = 2 x + 3 dan g ( x ) = 3 x − 1 1 ​ , maka ( f ∘ g ) − 1 ( x ) =

Jika dan , maka

Iklan

F. Ayudhita

Master Teacher

Jawaban terverifikasi

Jawaban

jawabannya A

jawabannya A

Iklan

Pembahasan

Pembahasan
lock

Jika dan , Kita tentukan dulu fungsi komposisinya: maka adalah Jadi jawabannya A

Jika dan ,

Kita tentukan dulu fungsi komposisinya:

size 14px left parenthesis size 14px f size 14px ring operator size 14px g size 14px right parenthesis begin mathsize 14px style left parenthesis x right parenthesis end style size 14px equals size 14px f size 14px left parenthesis size 14px g size 14px left parenthesis size 14px x size 14px right parenthesis size 14px right parenthesis size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px equals size 14px f begin mathsize 14px style left parenthesis fraction numerator 1 over denominator 3 x minus 1 end fraction right parenthesis end style size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px equals size 14px 2 begin mathsize 14px style left parenthesis fraction numerator 1 over denominator 3 x minus 1 end fraction right parenthesis end style size 14px plus size 14px 3 size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px equals fraction numerator size 14px 2 over denominator size 14px 3 size 14px x size 14px minus size 14px 1 end fraction size 14px plus fraction numerator size 14px 3 size 14px left parenthesis size 14px 3 size 14px x size 14px minus size 14px 1 size 14px right parenthesis over denominator size 14px 3 size 14px x size 14px minus size 14px 1 end fraction size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px equals fraction numerator size 14px 2 size 14px plus size 14px 9 size 14px x size 14px minus size 14px 3 over denominator size 14px 3 size 14px x size 14px minus size 14px 1 end fraction size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px space size 14px equals fraction numerator size 14px 9 size 14px x size 14px minus size 14px 1 over denominator size 14px 3 size 14px x size 14px minus size 14px 1 end fraction

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

begin mathsize 14px style y equals fraction numerator 9 x minus 1 over denominator 3 x minus 1 end fraction y left parenthesis 3 x minus 1 right parenthesis equals 9 x minus 1 3 x y minus y equals 9 x minus 1 3 x y minus 9 x equals y minus 1 x left parenthesis 3 y minus 9 right parenthesis equals y minus 1 x equals fraction numerator y minus 1 over denominator 3 y minus 9 end fraction end style

size 14px left parenthesis size 14px f size 14px ring operator size 14px g size 14px right parenthesis to the power of size 14px minus size 14px 1 end exponent size 14px left parenthesis size 14px x size 14px right parenthesis size 14px equals fraction numerator size 14px x size 14px minus size 14px 1 over denominator size 14px 3 size 14px x size 14px minus size 14px 9 end fraction

Jadi jawabannya A

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

120

krisna betricia sinaga

Makasih ❤️

Bima Aditya Suta Nandho

Makasih ❤️

Dealova Lumempow

Jawaban tidak sesuai

Iklan

Iklan

Iklan

Iklan

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

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2023 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia