Iklan

Iklan

Pertanyaan

Diketahui dan Jika dan maka adalah....

Diketahui f open parentheses x close parentheses equals fraction numerator x over denominator 2 x minus 3 end fraction dan g open parentheses x close parentheses equals fraction numerator 2 minus x over denominator 3 x end fraction. Jika D subscript g equals open curly brackets x vertical line x not equal to 0 comma space x element of straight real numbers close curly brackets dan D subscript f equals open curly brackets x vertical line x not equal to 3 over 2 comma space x element of straight real numbers close curly brackets comma maka open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses adalah ....

  1. left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 4 x plus 2 over denominator 11 x plus 1 end fraction comma space x not equal to negative 1 over 11 

  2. left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 4 x minus 2 over denominator 11 x minus 1 end fraction comma space x not equal to 1 over 11 

  3. left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 4 x plus 1 over denominator 11 x plus 2 end fraction comma space x not equal to negative 2 over 11

  4. left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 4 x minus 1 over denominator 11 x minus 2 end fraction comma space x not equal to 2 over 11

  5. left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 4 over denominator 11 x end fraction plus 2 comma space x not equal to 0

Iklan

N. Syafriah

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

jawaban yang tepat adalah B.

Iklan

Pembahasan

Diketahui bahwa dan serta dan , maka dapat diperiksa bahwa terdefinisi dan memiliki invers. Ingat bahwa invers dari fungsi f ( x ) = c x + d a x + b ​ adalah f ( x ) − 1 = c x − a − d x + b ​ . Akan dicariinvers dari dan g ( x ) terlebih dahulu sebagai berikut. Selanjutnya, Ingat kembali bahwa ( f ∘ g ) − 1 ( x ) = ( g − 1 ∘ f − 1 ) ( x ) sehingga ( f ∘ g ) − 1 ( x ) dapat ditentukan sebagai berikut. Dengan demikian, . Jadi, jawaban yang tepat adalah B.

Diketahui bahwa f open parentheses x close parentheses equals fraction numerator x over denominator 2 x minus 3 end fraction dan g open parentheses x close parentheses equals fraction numerator 2 minus x over denominator 3 x end fraction serta D subscript g equals open curly brackets x vertical line x not equal to 0 comma space x element of straight real numbers close curly brackets dan D subscript f equals open curly brackets x vertical line x not equal to 3 over 2 comma x element of straight real numbers close curly brackets, maka dapat diperiksa bahwa open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses terdefinisi dan memiliki invers.

Ingat bahwa invers dari fungsi  adalah .

Akan dicari invers dari f open parentheses x close parentheses dan terlebih dahulu sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell fraction numerator x over denominator 2 x minus 3 end fraction rightwards arrow f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator 3 x over denominator 2 x minus 1 end fraction end cell row blank blank blank row cell g left parenthesis x right parenthesis end cell equals cell fraction numerator 2 minus x over denominator 3 x end fraction equals fraction numerator negative x plus 2 over denominator 3 x end fraction rightwards arrow g to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 2 over denominator 3 x plus 1 end fraction end cell end table


Selanjutnya, Ingat kembali bahwa  sehingga  dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell open parentheses g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses open parentheses x close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses f to the power of negative 1 end exponent open parentheses x close parentheses close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses fraction numerator 3 x over denominator 2 x minus 1 end fraction close parentheses end cell row blank equals cell fraction numerator 2 over denominator 3 open parentheses begin display style fraction numerator 3 x over denominator 2 x minus 1 end fraction end style close parentheses plus 1 end fraction end cell row blank equals cell fraction numerator 2 over denominator begin display style fraction numerator 9 x over denominator 2 x minus 1 end fraction end style plus 1 end fraction end cell row blank equals cell fraction numerator 2 over denominator begin display style fraction numerator 9 x over denominator 2 x minus 1 end fraction end style plus begin display style fraction numerator 2 x minus 1 over denominator 2 x minus 1 end fraction end style end fraction end cell row blank equals cell fraction numerator 2 over denominator begin display style fraction numerator 11 x minus 1 over denominator 2 x minus 1 end fraction end style end fraction end cell row blank equals cell fraction numerator 2 open parentheses 2 x minus 1 close parentheses over denominator 11 x minus 1 end fraction end cell row blank equals cell fraction numerator 4 x minus 2 over denominator 11 x minus 1 end fraction end cell end table  


Dengan demikian, left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 4 x minus 2 over denominator 11 x minus 1 end fraction comma space x not equal to 1 over 11.

Jadi, jawaban yang tepat adalah B.

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

1

Iklan

Iklan

Pertanyaan serupa

Diketahui f(x) = 2x - 5 dan dengan . Maka, adalah ....

7

5.0

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

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

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