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Pertanyaan

Diketahui fungsi begin mathsize 14px style f colon R rightwards arrow R end style dan begin mathsize 14px style g colon R rightwards arrow R end style. Jika fungsi begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator 2 x plus 7 over denominator 3 x minus 5 end fraction comma x not equal to 5 over 3 end style dan begin mathsize 14px style g to the power of negative 1 end exponent open parentheses x close parentheses equals 4 x minus 1 end style, nilai fungsi begin mathsize 14px style open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses 2 close parentheses end style adalah...

  1. begin mathsize 14px style negative 8 end style

  2. begin mathsize 14px style 8 end style

  3. begin mathsize 14px style 10 end style

  4. begin mathsize 14px style 12 end style

  5. begin mathsize 14px style 16 end style

Pembahasan Soal:

Dengan menggunakan konsep invers fungsi,

Misalkan begin mathsize 14px style f open parentheses x close parentheses equals y end style maka

 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals y row cell fraction numerator 2 x plus 7 over denominator 3 x minus 5 end fraction end cell equals y row cell 2 x plus 7 end cell equals cell y open parentheses 3 x minus 5 close parentheses end cell row cell 2 x plus 7 end cell equals cell 3 x y minus 5 y end cell row cell 2 x minus 3 x y end cell equals cell negative 5 y minus 7 end cell row cell x open parentheses 2 minus 3 y close parentheses end cell equals cell negative 5 y minus 7 end cell row x equals cell fraction numerator negative 5 y minus 7 over denominator 2 minus 3 y end fraction comma y not equal to 2 over 3 end cell row x equals cell fraction numerator 5 y plus 7 over denominator 3 y minus 2 end fraction comma y not equal to 2 over 3 end cell end table end style

Sehingga diperoleh begin mathsize 14px style f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator 5 x plus 7 over denominator 3 x minus 2 end fraction comma x not equal to 2 over 3 end style.

Selanjutnya dengan menggunakan sifat invers komposisi fungsi maka

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 open parentheses x close parentheses end cell equals cell g to the power of negative 1 end exponent open parentheses x close parentheses ring operator f to the power of negative 1 end exponent open parentheses x close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses fraction numerator 5 x plus 7 over denominator 3 x minus 2 end fraction close parentheses end cell row blank equals cell 4 open parentheses fraction numerator 5 x plus 7 over denominator 3 x minus 2 end fraction close parentheses minus 1 end cell row cell open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses 2 close parentheses end cell equals cell 4 open parentheses fraction numerator 5 open parentheses 2 close parentheses plus 7 over denominator 3 open parentheses 2 close parentheses minus 2 end fraction close parentheses minus 1 end cell row blank equals cell 4 open parentheses 17 over 4 close parentheses minus 1 end cell row blank equals cell 17 minus 1 end cell row blank equals 16 end table end style

Dengan demikian nilai dari begin mathsize 14px style open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses 2 close parentheses end style adalah begin mathsize 14px style 16 end style.

Oleh karena itu, jawaban yang benar adalah E.

 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

E. Nur

Mahasiswa/Alumni Institut Teknologi Sepuluh Nopember

Terakhir diupdate 28 Maret 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui fungsi  dan . Jika fungsi   dan , Nilai fungsi  adalah ...

Pembahasan Soal:

Pertama akan dicari invers dari fungsi  begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator 2 x plus 7 over denominator 3 x minus 5 end fraction comma space x not equal to 5 over 3 end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator 2 x plus 7 over denominator 3 x minus 5 end fraction end cell row cell 3 y x minus 5 y end cell equals cell 2 x plus 7 end cell row cell 3 y x minus 2 x end cell equals cell 5 y plus 7 end cell row cell x open parentheses 3 y minus 2 close parentheses end cell equals cell 5 y plus 7 end cell row x equals cell fraction numerator 5 y plus 7 over denominator 3 y minus 2 end fraction end cell row cell f to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator 5 x plus 7 over denominator 3 x minus 2 end fraction end cell end table end style 

Dengan menggunakan sifat invers akan ditentukan  begin mathsize 14px style left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent open parentheses x close parentheses end style 

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 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 cell space space space space space space space space space space space space space space space space end cell equals cell g to the power of negative 1 end exponent open parentheses fraction numerator 5 x plus 7 over denominator 3 x minus 2 end fraction close parentheses end cell row cell space space space space space space space space space space space space space space space space end cell equals cell 4 open parentheses fraction numerator 5 x plus 7 over denominator 3 x minus 2 end fraction close parentheses minus 1 end cell row cell space space space space space space space space space space space space space space space space end cell equals cell fraction numerator 20 x plus 28 minus 3 x plus 2 over denominator 3 x minus 2 end fraction end cell row cell space space space space space space space space space space space space space space space space end cell equals cell fraction numerator 17 x plus 30 over denominator 3 x minus 2 end fraction end cell end table end style        

Nilai fugsi begin mathsize 14px style left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent open parentheses 2 close parentheses end style adalah  

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 open parentheses 2 close parentheses end cell equals cell fraction numerator 17 open parentheses 2 close parentheses plus 30 over denominator 3 open parentheses 2 close parentheses minus 2 end fraction end cell row cell space space space space space space space space space space space space space space space space space end cell equals cell fraction numerator 34 plus 30 over denominator 6 minus 2 end fraction end cell row cell space space space space space space space space space space space space space space space space space end cell equals cell 64 over 4 end cell row cell space space space space space space space space space space space space space space space space space end cell equals 16 end table end style  

Jadi, jawaban yang tepat  adalah E

1

Roboguru

Jika dan dengan dan . NIlai dari adalah....

Pembahasan Soal:

Pertama kita menetukan fungsi komposisi:

begin mathsize 14px style 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 cell open parentheses f ring operator g close parentheses open parentheses x close parentheses end cell equals cell f open parentheses x squared minus 1 close parentheses end cell row cell open parentheses f ring operator g close parentheses open parentheses x close parentheses end cell equals cell square root of 2 open parentheses x squared minus 1 close parentheses end root end cell end table end style

kemudian kita menetukan invers dari fungsi komposisi tersebut:

begin mathsize 14px style 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 square root of 2 open parentheses x squared minus 1 close parentheses end root end cell row y equals cell square root of 2 open parentheses x squared minus 1 close parentheses end root end cell row cell y squared end cell equals cell 2 open parentheses x squared minus 1 close parentheses end cell row cell y squared end cell equals cell 2 x squared minus 2 end cell row cell 2 x squared end cell equals cell y squared plus 2 end cell row x equals cell square root of fraction numerator y squared plus 2 over denominator 2 end fraction end root end cell 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 square root of fraction numerator x squared plus 2 over denominator 2 end fraction end root end cell end table end style

Sehingga nilai dari Error converting from MathML to accessible text. adalah:

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 end cell equals cell square root of fraction numerator x squared plus 2 over denominator 2 end fraction end root end cell row cell left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent open parentheses square root of 3 over 2 end root close parentheses end cell equals cell square root of fraction numerator square root of begin display style 3 over 2 end style end root squared plus 2 over denominator 2 end fraction end root end cell row cell left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent open parentheses square root of 3 over 2 end root close parentheses end cell equals cell square root of fraction numerator begin display style 3 over 2 end style plus 2 over denominator 2 end fraction end root end cell row cell left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent open parentheses square root of 3 over 2 end root close parentheses end cell equals cell square root of 7 over 4 end root end cell row blank blank blank end table end style

Jadi, nilai undefined adalah begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 7 over 4 end root end cell end table end style

0

Roboguru

Diketahui  dan , maka tentukan hasil dari !

Pembahasan Soal:

Diketahui begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator 2 x plus 2 over denominator 3 x minus 1 end fraction end style dan begin mathsize 14px style g open parentheses x close parentheses equals fraction numerator x plus 2 over denominator x end fraction end style, maka hasil dari begin mathsize 14px style open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end style dapat ditentukan sebagai berikut:

  • Dengan menerapkan konsep komposisi dua fungsi, diperoleh perhitungan sebagai berikut.

begin mathsize 14px style 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 fraction numerator 2 g open parentheses x close parentheses plus 2 over denominator 3 g open parentheses x close parentheses minus 1 end fraction end cell row blank equals cell fraction numerator 2 open parentheses begin display style fraction numerator x plus 2 over denominator x end fraction end style close parentheses plus 2 over denominator 3 open parentheses begin display style fraction numerator x plus 2 over denominator x end fraction end style close parentheses minus 1 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 2 x plus 4 over denominator x end fraction end style plus begin display style fraction numerator 2 x over denominator x end fraction end style over denominator begin display style fraction numerator 3 x plus 6 over denominator x end fraction end style minus begin display style x over x end style end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 6 x plus 4 over denominator x end fraction end style over denominator begin display style fraction numerator 2 x plus 6 over denominator x end fraction end style end fraction end cell row blank equals cell fraction numerator 6 x plus 4 over denominator up diagonal strike x end fraction cross times fraction numerator up diagonal strike x over denominator 2 x plus 6 end fraction end cell row blank equals cell fraction numerator 6 x plus 4 over denominator 2 x plus 6 end fraction end cell row blank equals cell fraction numerator up diagonal strike 2 open parentheses 3 x plus 2 close parentheses over denominator up diagonal strike 2 open parentheses x plus 3 close parentheses end fraction end cell row blank equals cell fraction numerator 3 x plus 2 over denominator x plus 3 end fraction end cell end table end style 

  • Selanjutnya, dengan menerapkan konsep invers fungsi, diperoleh perhitungan sebagai berikut.

begin mathsize 14px style 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 3 x plus 2 over denominator x plus 3 end fraction end cell row y equals cell fraction numerator 3 x plus 2 over denominator x plus 3 end fraction end cell row cell y open parentheses x plus 3 close parentheses end cell equals cell 3 x plus 2 end cell row cell x y plus 3 y end cell equals cell 3 x plus 2 end cell row cell x y minus 3 x end cell equals cell negative 3 y plus 2 end cell row cell x open parentheses y minus 3 close parentheses end cell equals cell negative 3 y plus 2 end cell row x equals cell fraction numerator negative 3 y plus 2 over denominator y minus 3 end fraction end cell 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 fraction numerator negative 3 x plus 2 over denominator x minus 3 end fraction end cell end table end style 

Jadi, hasil dari begin mathsize 14px style open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end style adalah begin mathsize 14px style open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative 3 x plus 2 over denominator x minus 3 end fraction end style.

0

Roboguru

Diketahui , maka fungsi dari adalah ...

Pembahasan Soal:

Diketahui :

straight f left parenthesis straight x right parenthesis equals space 3 straight x space minus space 2 space dan space straight g left parenthesis straight x right parenthesis equals space straight x space plus space 6,

Ditanya :

maka fungsi dari left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis adalah ...

Penyelesaian

Akan dicari nilai 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 straight f ring operator straight g close parentheses open parentheses straight x close parentheses end cell equals cell straight f open parentheses straight g open parentheses straight x close parentheses close parentheses end cell row blank equals cell 3 open parentheses straight g open parentheses straight x close parentheses close parentheses minus 2 end cell row blank equals cell 3 open parentheses straight x plus 6 close parentheses minus 2 end cell row blank equals cell 3 straight x plus 18 minus 2 end cell row blank equals cell 3 straight x plus 16 end cell end table

Akan dicari nilai dari left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses straight f ring operator straight g close parentheses open parentheses straight x close parentheses end cell equals cell 3 straight x plus 16 space end cell row cell straight y space end cell equals cell 3 straight x plus 16 end cell row cell straight y minus 16 end cell equals cell 3 straight x end cell row cell fraction numerator straight y minus 16 over denominator 3 end fraction end cell equals straight x end table

Jadi nai left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x minus 16 over denominator 3 end fraction.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Jika  dan  berturut-turut adalah invers dari fungsi ,  dan . Maka nilai ....

Pembahasan Soal:

Dengan menerapkan sifat invers fungsi komposisi, diperoleh:

open parentheses f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals open parentheses g ring operator f close parentheses to the power of negative 1 end exponent open parentheses x close parentheses 

Dengan menerapkan konsep komposisi dua fungsi, diperoleh perhitungan 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 2 f open parentheses x close parentheses minus 3 end cell row blank equals cell 2 open parentheses fraction numerator x minus 2 over denominator 3 x minus 1 end fraction close parentheses minus 3 end cell row blank equals cell fraction numerator 2 x minus 4 over denominator 3 x minus 1 end fraction minus fraction numerator 9 x minus 3 over denominator 3 x minus 1 end fraction end cell row blank equals cell fraction numerator 2 x minus 4 minus 9 x plus 3 over denominator 3 x minus 1 end fraction end cell row blank equals cell fraction numerator negative 7 x minus 1 over denominator 3 x minus 1 end fraction end cell end table 

Dengan menerapkan konsep invers fungsi, diperoleh perhitungan 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 negative 7 x minus 1 over denominator 3 x minus 1 end fraction end cell row y equals cell fraction numerator negative 7 x minus 1 over denominator 3 x minus 1 end fraction end cell row cell y open parentheses 3 x minus 1 close parentheses end cell equals cell negative 7 x minus 1 end cell row cell 3 x y minus y end cell equals cell negative 7 x minus 1 end cell row cell 3 x y plus 7 x end cell equals cell y minus 1 end cell row cell x open parentheses 3 y plus 7 close parentheses end cell equals cell y minus 1 end cell row x equals cell fraction numerator y minus 1 over denominator 3 y plus 7 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 x minus 1 over denominator 3 x plus 7 end fraction end cell end table 

Selanjutnya, dengan cara substitusi diperoleh:

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 x minus 1 over denominator 3 x plus 7 end fraction end cell row cell open parentheses g ring operator f close parentheses to the power of negative 1 end exponent open parentheses 1 close parentheses end cell equals cell fraction numerator 1 minus 1 over denominator 3 times 1 plus 7 end fraction end cell row blank equals cell 0 over 10 end cell row blank equals 0 end table 

Jadi, nilai dari open parentheses f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses open parentheses 1 close parentheses adalah 0.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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