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Diketahui  dan  . Fungsi  adalah...

Pertanyaan

Diketahui begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator 2 x minus 5 over denominator x plus 3 end fraction comma x not equal to negative 3 end style dan  begin mathsize 14px style g open parentheses x close parentheses equals 2 over x end style. Fungsi 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...

  1. begin mathsize 14px style fraction numerator 2 minus 4 x over denominator 3 x plus 5 end fraction comma x not equal to negative 5 over 3 end style

  2. begin mathsize 14px style fraction numerator 2 x plus 4 over denominator 3 x plus 5 end fraction comma x not equal to negative 5 over 3 end style

  3. begin mathsize 14px style fraction numerator 4 minus 2 x over denominator 3 x minus 5 end fraction comma x not equal to 5 over 3 end style

  4. begin mathsize 14px style fraction numerator 4 minus 2 x over denominator 3 x plus 5 end fraction comma x not equal to negative 5 over 3 end style

  5. begin mathsize 14px style fraction numerator 3 x plus 5 over denominator 2 x minus 4 end fraction comma x not equal to 2 end style

Pembahasan Soal:

Dengan menggunakan konsep komposisi fungsi 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 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 2 over x close parentheses end cell row blank equals cell fraction numerator 2 open parentheses begin display style 2 over x end style close parentheses minus 5 over denominator begin display style 2 over x end style plus 3 end fraction end cell row blank equals cell fraction numerator begin display style 4 over x end style minus begin display style fraction numerator 5 x over denominator x end fraction end style over denominator begin display style 2 over x end style plus begin display style fraction numerator 3 x over denominator x end fraction end style end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 4 minus 5 x over denominator x end fraction end style over denominator begin display style fraction numerator 2 plus 3 x over denominator x end fraction end style end fraction end cell row blank equals cell fraction numerator 4 minus 5 x over denominator 2 plus 3 x end fraction comma x not equal to negative 2 over 3 end cell end table end style

Dengan menggunkan konsep invers fungsi

Misalkan begin mathsize 14px style open parentheses f ring operator g close parentheses 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 open parentheses f ring operator g close parentheses open parentheses x close parentheses end cell equals y row cell fraction numerator 4 minus 5 x over denominator 2 plus 3 x end fraction end cell equals y row cell 4 minus 5 x end cell equals cell y open parentheses 2 plus 3 x close parentheses end cell row cell 4 minus 5 x end cell equals cell 2 y plus 3 x y end cell row cell negative 5 x minus 3 x y end cell equals cell 2 y minus 4 end cell row cell x open parentheses negative 5 minus 3 y close parentheses end cell equals cell 2 y minus 4 end cell row x equals cell fraction numerator 2 y minus 4 over denominator negative 5 minus 3 y end fraction comma y not equal to negative 5 over 3 end cell row x equals cell fraction numerator 4 minus 2 y over denominator 3 y plus 5 end fraction comma y not equal to negative 5 over 3 end cell end table end style

Dengan demikian 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 4 minus 2 x over denominator 3 x plus 5 end fraction comma x not equal to negative 5 over 3 end style.

Oleh karena itu, jawaban yang benar adalah D.

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 . Fungsi invers dari  adalah ...

Pembahasan Soal:

 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 3 x plus 1 close parentheses end cell row cell space space space space space space space space space space space space space end cell equals cell fraction numerator 2 open parentheses 3 x plus 1 close parentheses over denominator open parentheses 3 x plus 1 close parentheses plus 3 end fraction end cell row cell space space space space space space space space space space space space space end cell equals cell fraction numerator 6 x plus 2 over denominator 3 x plus 4 end fraction end cell row cell m i s a l space y end cell equals cell fraction numerator 6 x plus 2 over denominator 3 x plus 4 end fraction end cell row cell y open parentheses 3 x plus 4 close parentheses end cell equals cell 6 x plus 2 end cell row cell 3 x y plus 4 y space end cell equals cell 6 x plus 2 end cell row cell 3 x y minus 6 x space end cell equals cell 2 minus 4 y end cell row cell x open parentheses 3 y minus 6 close parentheses end cell equals cell 2 minus 4 y end cell row x equals cell fraction numerator 2 minus 4 y over denominator 3 y minus 6 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 minus 4 x over denominator 3 x minus 6 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 minus 4 open parentheses negative 2 close parentheses over denominator 3 open parentheses negative 2 close parentheses minus 6 end fraction end cell row cell space space space space space space space space space space space space space space space space space space space end cell equals cell fraction numerator 2 plus 8 over denominator negative 6 minus 6 end fraction end cell row cell space space space space space space space space space space space space space space space space space space space end cell equals cell negative 10 over 12 end cell row cell space space space space space space space space space space space space space space space space space space space end cell equals cell negative 5 over 6 end cell row blank blank blank end table end style  

0

Roboguru

Domain fungsi  dan  adalah ...

Pembahasan Soal:

Dari soal diketahui

f open parentheses x close parentheses equals 8 x minus 1 g open parentheses x close parentheses equals x minus 2  

Fungsi komposisi 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 f open parentheses g open parentheses x close parentheses close parentheses end cell row blank equals cell f open parentheses x minus 2 close parentheses end cell row blank equals cell 8 open parentheses x minus 2 close parentheses minus 1 end cell row blank equals cell 8 x minus 16 minus 1 end cell row blank equals cell 8 x minus 17 end cell end table      

Invers dari fungsi open parentheses f ring operator g close parentheses open parentheses x close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row y equals cell 8 x minus 17 end cell row cell 8 x end cell equals cell y plus 17 end cell row x equals cell fraction numerator y plus 17 over denominator 8 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 x plus 17 over denominator 8 end fraction end cell end table  

Jadi, Invers dari  open parentheses f ring operator g close parentheses open parentheses x close parentheses adalah Error converting from MathML to accessible text. 

0

Roboguru

Jika dan . Nilai

Pembahasan Soal:

Lakukan operasi komposisi terlebih dahulu.

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis g ring operator f right parenthesis left parenthesis x right parenthesis end cell equals cell g left parenthesis f left parenthesis x right parenthesis right parenthesis end cell row blank equals cell g open parentheses 5 x plus 1 half close parentheses end cell row blank equals cell 3 over 4 open parentheses 5 x plus 1 half close parentheses minus 6 end cell row blank equals cell open parentheses fraction numerator 15 x over denominator 4 end fraction plus 3 over 8 close parentheses minus 6 end cell row blank equals cell 1 fourth left parenthesis 15 x minus 45 right parenthesis end cell end table

Tentukan invers fungsi komposisi tersebut.

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis g ring operator f right parenthesis left parenthesis x right parenthesis end cell equals cell 1 fourth left parenthesis 15 x minus 45 right parenthesis end cell row y equals cell 1 fourth left parenthesis 15 x minus 45 right parenthesis end cell row y equals cell fraction numerator 15 x over denominator 4 end fraction minus 45 over 4 end cell row cell 4 y end cell equals cell 15 x minus 45 end cell row cell 15 x end cell equals cell 4 y plus 45 end cell row x equals cell 4 over 15 y plus 45 over 3 end cell row cell left parenthesis g ring operator f right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell 1 over 15 left parenthesis 4 x plus 45 right parenthesis end cell end table

Sehingga diperoleh :

left parenthesis g ring operator f right parenthesis to the power of negative 1 end exponent left parenthesis negative 5 right parenthesis equals 1 over 15 left parenthesis 4 x plus 45 right parenthesis space space space space space space space space space space space space space space space space space space space space space equals 1 over 15 left parenthesis 4 open parentheses negative 5 close parentheses plus 45 right parenthesis space space space space space space space space space space space space space space space space space space space space space space equals 5 over 3

Nilai dari left parenthesis g ring operator f right parenthesis to the power of negative 1 end exponent left parenthesis negative 5 right parenthesis adalah 5 over 3.

Oleh karena itu, jawaban yang tepat adalah A.

0

Roboguru

Jika fungsi  dan g(x) = 3x + 1 maka = ....

Pembahasan Soal:

begin mathsize 14px style straight i right parenthesis space Cari space straight g ring operator straight f  space space space 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 3 straight f open parentheses straight x close parentheses plus 1 equals 3. fraction numerator 2 straight x plus 3 over denominator straight x minus 5 end fraction plus 1 equals fraction numerator 6 straight x plus 9 over denominator straight x minus 5 end fraction plus fraction numerator straight x minus 5 over denominator straight x minus 5 end fraction  space space space equals fraction numerator 6 straight x plus 9 plus straight x minus 5 over denominator straight x minus 5 end fraction equals fraction numerator 7 straight x plus 4 over denominator straight x minus 5 end fraction    ii right parenthesis space Cari space invers space dengan space formula space berikut  space space space space straight h open parentheses straight x close parentheses equals fraction numerator ax plus straight b over denominator cx plus straight d end fraction rightwards arrow straight h to the power of negative 1 end exponent open parentheses straight x close parentheses equals fraction numerator negative dx plus straight b over denominator cx minus straight a end fraction    space space space space open parentheses straight g ring operator straight f close parentheses open parentheses straight x close parentheses equals fraction numerator 7 straight x plus 4 over denominator straight x minus 5 end fraction  space space space space Dengan space straight a equals 7 comma space straight b equals 4 comma space straight c equals 1 comma space straight d equals negative 5 comma space maka  space space space space 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 5 straight x plus 4 over denominator straight x minus 7 end fraction end style

0

Roboguru

Diketahui . Tentukan .

Pembahasan Soal:

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

Sehingga, begin mathsize 14px style f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator 1 minus 5 x over denominator 3 x plus 2 end fraction end style.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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