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Jika f(x)=x+36​,x=−3, dan g(x)=x2, untuk x≥0, daerah asal dari (f∘g)−1(x) adalah ...

Pertanyaan

Jika begin mathsize 14px style f left parenthesis x right parenthesis equals fraction numerator 6 over denominator x plus 3 end fraction comma space x not equal to negative 3 end style, dan begin mathsize 14px style g left parenthesis x right parenthesis equals x squared end style, untuk begin mathsize 14px style x greater or equal than 0 end style, daerah asal dari begin mathsize 14px style left parenthesis f space ring operator space g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis end style adalah ...

  1. begin mathsize 14px style x less than 2 end style 

  2. begin mathsize 14px style x less or equal than 2 end style 

  3. begin mathsize 14px style 0 less or equal than x less than 2 end style 

  4. begin mathsize 14px style 0 less or equal than x less or equal than 2 end style 

  5. begin mathsize 14px style 0 less than x less than 2 end style 

Pembahasan Soal:

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 f left parenthesis g open parentheses x close parentheses right parenthesis end cell row blank equals cell f left parenthesis x squared right parenthesis end cell row cell left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis end cell equals cell fraction numerator 6 over denominator x squared plus 3 end fraction end cell end table end style  

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 6 over denominator x squared plus 3 end fraction end cell row y equals cell fraction numerator 6 over denominator x squared plus 3 end fraction end cell row cell x squared y plus 3 y end cell equals 6 row cell x squared y end cell equals cell negative 3 y plus 6 end cell row cell x squared end cell equals cell fraction numerator negative 3 y plus 6 over denominator y end fraction end cell row x equals cell square root of fraction numerator negative 3 y plus 6 over denominator y end fraction end root 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 square root of fraction numerator negative 3 x plus 6 over denominator x end fraction end root end cell end table end style  

daerah asal begin mathsize 14px style fraction numerator negative 3 x plus 6 over denominator x end fraction greater than 0 end style dan begin mathsize 14px style x not equal to 0 end style 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis negative 3 x plus 6 right parenthesis x end cell greater than 0 row cell negative 1 left parenthesis 3 x minus 6 right parenthesis x end cell greater than 0 row cell left parenthesis 3 x minus 6 right parenthesis x end cell less than 0 row cell 3 x minus 6 end cell less than cell 0 space atau space x greater than 0 end cell row x less than cell 2 space atau space x greater than 0 end cell end table end style  

dan undefined 

daerah asal begin mathsize 14px style 0 less than x less than 2 end style 

 Jadi, jawaban yang tepat adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 05 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Jika f:R→R  didefinisikan dengan   , maka nilai x  yang memenuhi  adalah ....

Pembahasan Soal:

Dengan menggunakan sifat

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 comma end cell row cell open parentheses f to the power of negative 1 end exponent close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell f open parentheses x close parentheses comma end cell row cell open parentheses f to the power of negative 1 end exponent ring operator f close parentheses open parentheses x close parentheses end cell equals cell x comma end cell end table end style 

maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f to the power of negative 1 end exponent ring operator f close parentheses to the power of negative 1 end exponent open parentheses 2 x close parentheses end cell equals cell negative 4 end cell row cell open parentheses f to the power of negative 1 end exponent ring operator open parentheses f to the power of negative 1 end exponent close parentheses to the power of negative 1 end exponent close parentheses open parentheses 2 x close parentheses end cell equals cell negative 4 end cell row cell open parentheses f to the power of negative 1 end exponent ring operator f close parentheses open parentheses 2 x close parentheses end cell equals cell negative 4 end cell row cell 2 x end cell equals cell negative 4 end cell row x equals cell negative 4 over 2 end cell row x equals cell negative 2 end cell end table end style 

Jadi, jawabannya D.

0

Roboguru

Jika , maka rumus fungsi (f∘g)(x) adalah ....

Pembahasan Soal:

Ingat bahwa begin mathsize 14px style 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 equals open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end style sehingga diperoleh hasil berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row 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 equals cell x over 2 minus 3 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 x over 2 minus 3 end cell end table end style 

Berdasarkan sifat begin mathsize 14px style open parentheses f to the power of negative 1 end exponent close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals f open parentheses x close parentheses end style, kita perlu melakukan invers pada begin mathsize 14px style open parentheses f ring operator g close parentheses to the power of negative 1 end exponent end style untuk mendapatkan table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell size 14px left parenthesis size 14px f size 14px ring operator size 14px g size 14px right parenthesis end cell end table.

 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 x over 2 minus 3 end cell row y equals cell x over 2 minus 3 end cell row cell y plus 3 end cell equals cell x over 2 end cell row cell 2 left parenthesis y plus 3 right parenthesis end cell equals x row cell 2 y plus 6 end cell equals x row x equals cell 2 y plus 6 end cell end table end style

Dari hasil di atas, didapat rumus fungsi size 14px left parenthesis size 14px f size 14px ring operator size 14px g size 14px right parenthesis size 14px left parenthesis size 14px x size 14px right parenthesis adalah sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses open parentheses f ring operator g close parentheses to the power of negative 1 end exponent close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell 2 x plus 6 end cell row cell open parentheses f ring operator g close parentheses open parentheses x close parentheses end cell equals cell 2 x plus 6 end cell end table end style

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Diketahui f(x) = 4x + 2 dan g(x)=x+1x−3​,x=−1. Invers dari (g o f)(x) adalah ....

Pembahasan Soal:

begin mathsize 14px style open parentheses straight g ring operator straight f close parentheses open parentheses straight x close parentheses equals straight g open parentheses straight f open parentheses straight x close parentheses close parentheses  equals straight g open parentheses 4 straight x plus 2 close parentheses  equals fraction numerator left parenthesis 4 straight x plus 2 right parenthesis minus 3 over denominator left parenthesis 4 straight x plus 2 right parenthesis plus 1 end fraction  equals fraction numerator 4 straight x minus 1 over denominator 4 straight x plus 3 end fraction    Invers space dari space straight h left parenthesis straight x right parenthesis equals fraction numerator ax plus straight b over denominator cx plus straight d end fraction space adalah space straight h to the power of negative 1 end exponent left parenthesis straight x right parenthesis equals fraction numerator negative dx plus straight b over denominator cx minus straight a end fraction  Sehingga space invers space dari space left parenthesis straight g ring operator straight f right parenthesis left parenthesis straight x right parenthesis equals fraction numerator 4 straight x minus 1 over denominator 4 straight x plus 3 end fraction space adalah  open parentheses straight g ring operator straight f close parentheses to the power of negative 1 end exponent open parentheses straight x close parentheses equals fraction numerator negative 3 straight x minus 1 over denominator 4 straight x minus 4 end fraction  open parentheses straight g ring operator straight f close parentheses to the power of negative 1 end exponent open parentheses straight x close parentheses equals fraction numerator 3 straight x plus 1 over denominator 4 minus 4 straight x end fraction end style

0

Roboguru

Jika f(x)=4x−1 dan g(x)=2x−3, nilai (f−1∘g)−1(2)adalah...

Pembahasan Soal:

Ingat kembali sifat invers fungsi komposisi berikut.

(f1)1(x)(fg)1(x)==f(x)(f1g1)(x)

Dengan menggunakan konsep invers fungsi. Misalkan begin mathsize 14px style g open parentheses x close parentheses equals y end style.

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

Selanjutnya dengan menggunakan sifat invers dari kompisisi fungsi diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f to the power of negative 1 end exponent 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 open parentheses f to the power of negative 1 end exponent close parentheses 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 x close parentheses ring operator f open parentheses x close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses f open parentheses x close parentheses close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses 4 x minus 1 close parentheses end cell row blank equals cell fraction numerator 4 x minus 1 plus 3 over denominator 2 end fraction end cell row blank equals cell fraction numerator 4 x plus 2 over denominator 2 end fraction end cell row cell open parentheses f to the power of negative 1 end exponent ring operator g close parentheses to the power of negative 1 end exponent open parentheses 2 close parentheses end cell equals cell fraction numerator 4 open parentheses 2 close parentheses plus 2 over denominator 2 end fraction end cell row blank equals cell 10 over 2 end cell row blank equals 5 end table end style

Dengan demikian nilai dari begin mathsize 14px style open parentheses f to the power of negative 1 end exponent 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 5 end style.

0

Roboguru

Jika   , maka g(x) =  ....

Pembahasan Soal:

Dengan menggunakan sifat

begin mathsize 14px style open parentheses f to the power of negative 1 end exponent ring operator f ring operator g close parentheses open parentheses x close parentheses equals g open parentheses x close parentheses end style 

maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f to the power of negative 1 end exponent ring operator f ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses end cell equals cell 6 minus 3 x end cell row cell g to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell 6 minus 3 x end cell end table end style 

Kita perlu melakukan invers pada begin mathsize 14px style g to the power of negative 1 end exponent end style untuk mencari g .

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell g to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell 6 minus 3 x end cell row y equals cell 6 minus 3 x 3 end cell row x equals cell 6 minus y end cell row x equals cell fraction numerator 6 minus y over denominator 3 end fraction end cell row cell g to the power of negative 1 end exponent open parentheses x close parentheses end cell equals y row cell g open parentheses y close parentheses end cell equals x row cell g open parentheses y close parentheses end cell equals cell fraction numerator 6 minus y over denominator 3 end fraction end cell row cell g open parentheses x close parentheses end cell equals cell fraction numerator 6 minus x over denominator 3 end fraction end cell row blank blank blank end table end style 

Jadi, jawabannya C.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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

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