Roboguru
SD

Diketahui f(x)=2−x dan g(x)=1+x. Jika  dengan , maka fungsi invers dari h(x) adalah ....

Pertanyaan

Diketahui begin mathsize 14px style f left parenthesis x right parenthesis equals 2 minus x end style dan begin mathsize 14px style g left parenthesis x right parenthesis equals 1 plus x end style. Jika begin mathsize 14px style h open parentheses x close parentheses equals fraction numerator f open parentheses x close parentheses over denominator g open parentheses x close parentheses end fraction end style dengan begin mathsize 14px style D subscript h equals open curly brackets x vertical line x not equal to negative 1 comma x element of straight real numbers close curly brackets end style, maka fungsi invers dari begin mathsize 14px style h left parenthesis x right parenthesis end style adalah ....

  1. begin mathsize 14px style h to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x plus 1 over denominator x plus 2 end fraction end style 

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

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

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

  5. begin mathsize 14px style h to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator 2 plus x over denominator x minus 1 end fraction end style 

S. Luke

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah D.

Pembahasan

Diketahui begin mathsize 14px style f left parenthesis x right parenthesis equals 2 minus x end style dan begin mathsize 14px style g left parenthesis x right parenthesis equals 1 plus x end style, maka didapat perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell h open parentheses x close parentheses end cell equals cell fraction numerator 2 minus x over denominator 1 plus x end fraction end cell row y equals cell fraction numerator 2 minus x over denominator 1 plus x end fraction end cell row cell y open parentheses 1 plus x close parentheses end cell equals cell 2 minus x end cell row cell y plus x y end cell equals cell 2 minus x end cell row cell x y plus x end cell equals cell 2 minus y end cell row cell x open parentheses y plus 1 close parentheses end cell equals cell 2 minus y end cell row x equals cell fraction numerator 2 minus y over denominator y plus 1 end fraction end cell end table end style

Kemudian, menggunakan konsep invers begin mathsize 14px style h to the power of negative 1 end exponent open parentheses y close parentheses equals x end style, didapat perhitungan sebagai berikut.

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

Jadi, jawaban yang tepat adalah D.

46

5.0 (3 rating)

Pertanyaan serupa

Fungsi invers dari f(x) = a - bx adalah ....

90

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

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia