Roboguru

f(x)=x2−1​f(x)=x2+1​  maka (f∘g)−1(x)+(g∘f)−1(x) adalah ....

Pertanyaan

begin mathsize 14px style f left parenthesis x right parenthesis equals square root of x squared minus 1 end root f left parenthesis x right parenthesis equals square root of x squared plus 1 end root end style 

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

  1. begin mathsize 14px style 2 x end style 

  2. size 14px 2 size 14px x size 14px plus size 14px 1 

  3. begin mathsize 14px style x squared end style 

  4. begin mathsize 14px style x squared plus 1 end style 

  5. begin mathsize 14px style x to the power of 4 minus 1 end style 

Pembahasan Soal:

Pada soal yang diberikan, kedua fungsinya ditulis begin mathsize 14px style f left parenthesis x right parenthesis end style seharusnya salah satunya adalah fungsi size 14px g size 14px left parenthesis size 14px x size 14px right parenthesis. Kita misalkan fungsi yang kedua adalah undefined, maka diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis x right parenthesis end cell equals cell square root of x squared minus 1 end root end cell row y equals cell square root of x squared minus 1 end root end cell row cell y squared end cell equals cell x squared minus 1 end cell row cell x squared end cell equals cell y squared plus 1 end cell row x equals cell square root of y squared plus 1 end root end cell row cell f to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell square root of x squared plus 1 end root end cell row blank blank blank row cell g left parenthesis x right parenthesis end cell equals cell square root of x squared plus 1 end root end cell row y equals cell square root of x squared plus 1 end root end cell row cell y squared end cell equals cell x squared plus 1 end cell row cell x squared end cell equals cell y squared minus 1 end cell row x equals cell square root of y squared minus 1 end root end cell row cell g to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell square root of x squared minus 1 end root end cell end table end style  

sehingga diperoleh:

begin mathsize 14px style 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 left parenthesis x right parenthesis 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 left parenthesis x right parenthesis end cell row blank equals cell square root of open parentheses square root of x squared plus 1 end root close parentheses squared minus 1 end root end cell row blank equals cell square root of x squared plus 1 minus 1 end root end cell row blank equals x row cell open parentheses g ring operator f close parentheses to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell open parentheses f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses left parenthesis x right parenthesis end cell row blank equals cell square root of open parentheses square root of x squared minus 1 end root close parentheses squared plus 1 end root end cell row blank equals cell square root of x squared minus 1 plus 1 end root end cell row blank equals x end table end style 

sehingga diperoleh:

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

Jadi, jawaban yang benar adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

P. Afrisno

Mahasiswa/Alumni Universitas Sebelas Maret

Terakhir diupdate 05 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Diketahui fungsi f(x)=3+2x, g(x)=x+43x−1​,x=−4. Invers dari (f∘g)(x)= ...

Pembahasan Soal:

Tentukan terlebih dahulu komposisi fungsi open parentheses f ring operator g close parentheses open parentheses x close parentheses, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g close parentheses open parentheses x close parentheses end cell equals cell f open parentheses g open parentheses x close parentheses close parentheses end cell row blank equals cell f open parentheses fraction numerator 3 x minus 1 over denominator x plus 4 end fraction close parentheses end cell row blank equals cell 3 plus 2 open parentheses fraction numerator 3 x minus 1 over denominator x plus 4 end fraction close parentheses end cell row blank equals cell 3 plus fraction numerator 6 x minus 2 over denominator x plus 4 end fraction end cell row blank equals cell fraction numerator 3 open parentheses x plus 4 close parentheses over denominator x plus 4 end fraction plus fraction numerator 6 x minus 2 over denominator x plus 4 end fraction end cell row blank equals cell fraction numerator 3 x plus 12 over denominator x plus 4 end fraction plus fraction numerator 6 x minus 2 over denominator x plus 4 end fraction end cell row blank equals cell fraction numerator 9 x plus 10 over denominator x plus 4 end fraction end cell end table

Kemudian tentukan invers dari open parentheses f ring operator g close parentheses open parentheses x close parentheses.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g close parentheses open parentheses x close parentheses end cell equals cell fraction numerator 9 x plus 10 over denominator x plus 4 end fraction end cell row y equals cell fraction numerator 9 x plus 10 over denominator x plus 4 end fraction end cell row cell y open parentheses x plus 4 close parentheses end cell equals cell 9 x plus 10 end cell row cell x y plus 4 y end cell equals cell 9 x plus 10 end cell row cell x y minus 9 x end cell equals cell negative 4 y plus 10 end cell row cell x open parentheses y minus 9 close parentheses end cell equals cell negative 4 y plus 10 end cell row x equals cell fraction numerator negative 4 y plus 10 over denominator y minus 9 end fraction end cell row cell open parentheses f ring operator g close parentheses to the power of negative 1 end exponent end cell equals cell fraction numerator negative 4 x plus 10 over denominator x minus 9 end fraction end cell end table

Maka, invers dari open parentheses f ring operator g close parentheses open parentheses x close parentheses equals fraction numerator negative 4 x plus 10 over denominator x minus 9 end fraction.

0

Roboguru

Diketahui: f(x)=3x+4 dan g(x)=2x+14x−5​;x=21​. Tentukan invers dari f∘g(x)!

Pembahasan Soal:

Ingat bahwa:

Jika undefined maka begin mathsize 14px style f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator negative d x plus b over denominator c x minus a end fraction end style 

Komposisi f(x)dang(x) adalah

fg(x)=====f(g(x))3(2x+14x5)+42x+112x15+42x+112x15+8x+42x+120x11 

Sehingga diperoleh inversnya adalah

fg1(x)==2x20x112x+20x+11

Dengan demikian, invers dari fg(x) adalah fg1(x)=2x+20x+11.

0

Roboguru

Jika f(x)=4x+6,g(x)=x−1x​;x=1, dan  h(x)=(f∘g)(x),h−1(x) adalah ...

Pembahasan Soal:

Dari soal diketahui

f open parentheses x close parentheses equals 4 x plus 6 g open parentheses x close parentheses equals fraction numerator x over denominator x minus 1 end fraction comma x not equal to 1      

 Fungsi h open parentheses x close parentheses equals left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis dapat ditentukan dengan cara berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell h open parentheses x close parentheses end cell equals cell left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis end cell row blank equals cell f open parentheses g open parentheses x close parentheses close parentheses end cell row blank equals cell f open parentheses fraction numerator x over denominator x minus 1 end fraction close parentheses end cell row blank equals cell 4 open parentheses fraction numerator x over denominator x minus 1 end fraction close parentheses plus 6 end cell row blank equals cell fraction numerator 4 x over denominator x minus 1 end fraction plus 6 end cell row blank equals cell fraction numerator 4 x plus 6 open parentheses x minus 1 close parentheses over denominator x minus 1 end fraction end cell row blank equals cell fraction numerator 4 x plus 6 x minus 6 over denominator x minus 1 end fraction end cell row blank equals cell fraction numerator 10 x minus 6 over denominator x minus 1 end fraction end cell end table 

Invers fungsi, merupakan fungsi kebalikan, sehingga h to the power of negative 1 end exponent open parentheses x close parentheses 

table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator 10 x minus 6 over denominator x minus 1 end fraction end cell row cell y x minus y end cell equals cell 10 x minus 6 end cell row cell y x minus 10 x end cell equals cell y minus 6 end cell row cell x open parentheses y minus 10 close parentheses end cell equals cell y minus 6 end cell row x equals cell fraction numerator y minus 6 over denominator y minus 10 end fraction end cell row cell h to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator x minus 6 over denominator x minus 10 end fraction comma x not equal to 10 end cell end table  

 Oleh karena itu, jawaban yang tepat adalah A.

0

Roboguru

Diketahui fungsi f(x)=x−3x+2​,x=3 dan g(x)=x+4. Invers fungsi (f∘g)(x) adalah ....

Pembahasan Soal:

Dengan menggunakan konsep fungsi komposisi, kemudian invers fungsi, maka didapatkan:

(fg)(x)yyx+yyxxx(y1)x(fg)1(x)=========f(g(x))x+43x+4+2x+1x+6x+1x+6x+6y+6y+6y1y+6x1x+6

Jadi, invers fungsi left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator negative x plus 6 over denominator x minus 1 end fraction end cell end table.

0

Roboguru

Fungsi fdang ditentukan oleh f(x)=3x−1 dan g(x)=x−1x+2​,x=1. Fungsi invers (g∘f) adalah (g∘f)−1. Nilai dari (g∘f)−1(−2)=....

Pembahasan Soal:

Diketahui:

table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis x right parenthesis end cell equals cell 3 x minus 1 end cell row cell g open parentheses x close parentheses end cell equals cell fraction numerator x plus 2 over denominator x minus 1 end fraction comma space x not equal to 1 end cell end table 

Sehingga diperoleh open parentheses g ring operator f close parentheses open parentheses x close parentheses sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses g ring operator f close parentheses open parentheses x close parentheses end cell equals cell g open parentheses f open parentheses x close parentheses close parentheses end cell row blank equals cell fraction numerator f open parentheses x close parentheses plus 2 over denominator f open parentheses x close parentheses minus 1 end fraction end cell row blank equals cell fraction numerator 3 x minus 1 plus 2 over denominator 3 x minus 1 minus 1 end fraction end cell row blank equals cell fraction numerator 3 x plus 1 over denominator 3 x minus 2 end fraction end cell end table 

Selanjutnya dapat ditentukan invers fungsi open parentheses g ring operator f close parentheses open parentheses x close parentheses sebagai berikut.

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

Sehingga diperoleh nilai dari Error converting from MathML to accessible text. sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses g ring operator f close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator 2 x plus 1 over denominator 3 x minus 3 end fraction end cell row cell open parentheses g ring operator f close parentheses to the power of negative 1 end exponent open parentheses negative 2 close parentheses end cell equals cell fraction numerator 2 open parentheses negative 2 close parentheses plus 1 over denominator 3 open parentheses negative 2 close parentheses minus 3 end fraction end cell row blank equals cell fraction numerator negative 4 plus 1 over denominator negative 6 minus 3 end fraction end cell row blank equals cell fraction numerator negative 3 over denominator negative 9 end fraction end cell row blank equals cell 1 third end cell end table 

Jadi, jawaban yang benar adalah E.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved