Roboguru

Diketahui , dan . Tentukan: d. ,

Pertanyaan

Diketahui f open parentheses x close parentheses equals fraction numerator 2 x minus 1 over denominator x plus 1 end fraction comma space g open parentheses x close parentheses equals x squared minus 1, dan h open parentheses x close parentheses equals 3 x plus 5. Tentukan:

d. open parentheses h to the power of negative 1 end exponent ring operator 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

Pembahasan Soal:

Diketahui f open parentheses x close parentheses equals fraction numerator 2 x minus 1 over denominator x plus 1 end fraction comma space g open parentheses x close parentheses equals x squared minus 1, dan h open parentheses x close parentheses equals 3 x plus 5. Maka:

- Menentukan f to the power of negative 1 end exponent open parentheses x close parentheses 

space space space space space space f open parentheses x close parentheses equals fraction numerator 2 x minus 1 over denominator x plus 1 end fraction space space space space space space space space space space y equals fraction numerator 2 x minus 1 over denominator x plus 1 end fraction y open parentheses x plus 1 close parentheses equals 2 x minus 1 space space space x y plus y equals 2 x minus 1 space x y minus 2 x equals negative 1 minus y x open parentheses y minus 2 close parentheses equals negative 1 minus y space space space space space space space space space space x equals fraction numerator negative 1 minus y over denominator y minus 2 end fraction space space f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative 1 minus x over denominator x minus 2 end fraction 

- Menentukan g to the power of negative 1 end exponent open parentheses x close parentheses 

space space space space g open parentheses x close parentheses equals x squared minus 1 space space space space space space space space space y equals x squared minus 1 space space space space y plus 1 equals x squared space space space space space space space space x squared equals y plus 1 space space space space space space space space space x equals plus-or-minus square root of y plus 1 end root g to the power of negative 1 end exponent open parentheses x close parentheses equals plus-or-minus square root of x plus 1 end root 

- Menentukan h to the power of negative 1 end exponent open parentheses x close parentheses 

space space space space space h open parentheses x close parentheses equals 3 x plus 5 space space space space space space space space space space y equals 3 x plus 5 space space space space y minus 5 equals 3 x space space space space space space space 3 x equals y minus 5 space space space space space space space space space x equals fraction numerator y minus 5 over denominator 3 end fraction h to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x minus 5 over denominator 3 end fraction 

Sehingga:

 open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses open parentheses f to the power of negative 1 end exponent open parentheses x close parentheses close parentheses open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses open parentheses fraction numerator negative 1 minus x over denominator x minus 2 end fraction close parentheses open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals open parentheses h to the power of negative 1 end exponent close parentheses open parentheses plus-or-minus square root of fraction numerator negative 1 minus x over denominator x minus 2 end fraction plus 1 end root close parentheses open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals open parentheses h to the power of negative 1 end exponent close parentheses open parentheses plus-or-minus square root of fraction numerator negative 1 minus x plus x minus 2 over denominator x minus 2 end fraction end root close parentheses open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals open parentheses h to the power of negative 1 end exponent close parentheses open parentheses plus-or-minus square root of fraction numerator negative 3 over denominator x minus 2 end fraction end root close parentheses open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals fraction numerator plus-or-minus square root of fraction numerator negative 3 over denominator x minus 2 end fraction end root minus 5 over denominator 3 end fraction 

Jadi, open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals fraction numerator plus-or-minus square root of fraction numerator negative 3 over denominator x minus 2 end fraction end root minus 5 over denominator 3 end fraction.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

W. Lestari

Mahasiswa/Alumni Universitas Sriwijaya

Terakhir diupdate 02 Mei 2021

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Pertanyaan yang serupa

Jika =...

Pembahasan Soal:

Penyelesaian:

f left parenthesis x right parenthesis equals fraction numerator 2 x minus 1 over denominator 3 x plus 4 end fraction space d a n space g left parenthesis x right parenthesis equals fraction numerator x plus 2 over denominator 2 end fraction  left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis equals f left parenthesis g left parenthesis x right parenthesis right parenthesis  equals f open parentheses fraction numerator x plus 5 over denominator 2 end fraction close parentheses  equals 2 fraction numerator open parentheses begin display style fraction numerator x plus 5 over denominator 2 end fraction end style close parentheses minus 1 over denominator x open parentheses begin display style fraction numerator x plus 5 over denominator 2 end fraction end style close parentheses plus 4 end fraction  equals fraction numerator x plus 5 minus 1 over denominator begin display style fraction numerator 3 x plus 15 plus 8 over denominator 2 end fraction end style end fraction equals fraction numerator x plus 4 over denominator begin display style fraction numerator 3 x plus 23 over denominator 2 end fraction end style end fraction equals fraction numerator 2 x plus 8 over denominator 3 x plus 23 end fraction  left parenthesis f ring operator g right parenthesis 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 equals fraction numerator negative 23 plus 8 over denominator 3 x minus 2 end fraction  left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis negative 1 right parenthesis equals fraction numerator negative 23 left parenthesis negative 1 right parenthesis plus 8 over denominator 3 left parenthesis negative 1 right parenthesis minus 2 end fraction  equals fraction numerator 23 plus 8 over denominator negative 3 minus 2 end fraction equals negative 31 over 5

0

Roboguru

Fungsi  dan  didefinisakan oleh  dan , Tentukan: b.

Pembahasan Soal:

Menentukan begin mathsize 14px style open parentheses g ring operator f close parentheses 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 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 g open parentheses 2 x plus 11 close parentheses end cell row blank equals cell 1 minus open parentheses 2 x plus 11 close parentheses end cell row blank equals cell negative 2 x minus 10 end cell end table end style 

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

begin mathsize 14px style 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 negative 2 x minus 10 end cell row y equals cell negative 2 x minus 10 end cell row cell negative 2 x minus 10 end cell equals y row cell negative 2 x end cell equals cell y plus 10 end cell row x equals cell fraction numerator y plus 10 over denominator negative 2 end fraction end cell row x equals cell negative 1 half y minus 5 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 negative 1 half x minus 5 end cell row blank blank blank end table end style

 

0

Roboguru

2. Fungsi  didefinisikan oleh  dan . Tentukan: e.

Pembahasan Soal:

Perhatikan penghitungan berikut!

 f open parentheses x close parentheses equals 2 x plus 11 misal colon space y equals f open parentheses x close parentheses space maka space x equals f to the power of negative 1 end exponent open parentheses x close parentheses f open parentheses x close parentheses equals 2 x plus 11 y equals 2 x plus 11 y minus 11 equals 2 x fraction numerator y minus 11 over denominator 2 end fraction equals x f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x minus 11 over denominator 2 end fraction  g open parentheses x close parentheses equals 1 minus x misal colon space z equals g open parentheses x close parentheses space maka space x equals g to the power of negative 1 end exponent open parentheses x close parentheses g open parentheses x close parentheses equals 1 minus x z equals 1 minus x x equals 1 minus z g to the power of negative 1 end exponent open parentheses x close parentheses equals 1 minus x  

sehingga, open parentheses f ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses yaitu:

 open parentheses f ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals f open parentheses g to the power of negative 1 end exponent open parentheses x close parentheses close parentheses open parentheses f ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals f open parentheses 1 minus x close parentheses open parentheses f ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals 2 open parentheses 1 minus x close parentheses plus 11 open parentheses f ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals 2 minus 2 x plus 11 open parentheses f ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals 13 minus 2 x               

Jadi, open parentheses f ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals 13 minus 2 x .space 

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

Diketahui dan , maka

Pembahasan Soal:

Pertama kita tentukan funsgi komposisi begin mathsize 14px style open parentheses f ring operator g close parentheses 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 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 3 x plus 4 close parentheses end cell row blank equals cell fraction numerator open parentheses 3 x plus 4 close parentheses plus 1 over denominator left parenthesis 3 x plus 4 right parenthesis minus 5 end fraction end cell row blank equals cell fraction numerator 3 x plus 5 over denominator 3 x minus 1 end fraction end cell end table end style 

Sehingga :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis end cell equals cell fraction numerator 3 x plus 5 over denominator 3 x minus 1 end fraction end cell row y equals cell fraction numerator 3 x plus 5 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 3 x plus 5 end cell row cell 3 x y minus y end cell equals cell 3 x plus 5 end cell row blank blank cell 3 x y minus 3 x equals 5 plus y end cell row cell x left parenthesis 3 y minus 3 right parenthesis end cell equals cell 5 plus y end cell row y equals cell fraction numerator y plus 5 over denominator 3 y minus 3 end fraction end cell 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 fraction numerator x plus 5 over denominator 3 x minus 3 end fraction end cell end table end style

Jadi, jawaban yang tepat adalah D

0

Roboguru

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