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Misalkan f ( x ) = x − 2 2 x + 3 ​ dan g ( x ) = 5 x + 3 ​ ,tentukanlah : ( f ∘ g ) − 1 ( x )

Misalkan   dan  ,  tentukanlah :
 

   

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R. Novianto

Master Teacher

Mahasiswa/Alumni Universitas Tanjungpura Pontianak

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Jawaban

 begin mathsize 14px style left parenthesis straight f ring operator straight g right parenthesis to the power of negative 1 end exponent left parenthesis straight x right parenthesis equals fraction numerator straight x squared plus 24 straight x minus 3 over denominator 5 straight x squared minus 20 straight x plus 20 end fraction end style 

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Pembahasan

Diketahui: dan , Ditanya : = .... Penyelesaian : kita cari invers dari terlebih dahulu maka untuk menentukan , kita misalkan : kita juga bisa gunakan cara cepat mencari invers fungsi Jadi, dapat disimpulkan kemudian menentukan , kitamisalkan Jadi, dapat disimpulkan maka dengan menggunakan sifat invers komposisi fungsi Jadi,

Diketahui:
begin mathsize 14px style straight f open parentheses straight x close parentheses equals fraction numerator 2 straight x plus 3 over denominator straight x minus 2 end fraction end style dan  begin mathsize 14px style straight g open parentheses straight x close parentheses equals square root of 5 straight x plus 3 end root end style,  

Ditanya : begin mathsize 14px style open parentheses straight f ring operator straight g close parentheses to the power of negative 1 end exponent open parentheses straight x close parentheses end style = ....

Penyelesaian :

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

kita cari invers dari undefined terlebih dahulu

maka untuk menentukan  begin mathsize 14px style straight f to the power of negative 1 end exponent open parentheses straight x close parentheses end style, kita misalkan undefined :
 

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

kita juga bisa gunakan cara cepat mencari invers fungsi

begin mathsize 14px style straight f open parentheses straight x close parentheses equals fraction numerator ax plus straight b over denominator cx plus straight d end fraction rightwards arrow straight f 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 end style   
 

Jadi, dapat disimpulkan begin mathsize 14px style straight f open parentheses straight x close parentheses equals fraction numerator 2 straight x plus 3 over denominator straight x minus 2 end fraction rightwards arrow straight f to the power of negative 1 end exponent open parentheses straight x close parentheses equals fraction numerator 2 straight x plus 3 over denominator straight x minus 2 end fraction end style 

  

kemudian menentukan begin mathsize 14px style straight g to the power of negative 1 end exponent open parentheses straight x close parentheses end style, kita misalkan begin mathsize 14px style straight g to the power of negative 1 end exponent open parentheses straight x close parentheses equals straight y end style 
 

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row straight y equals cell square root of 5 straight x plus 3 end root end cell row cell straight y squared end cell equals cell 5 straight x plus 3 end cell row cell straight y squared minus 3 end cell equals cell 5 straight x end cell row cell fraction numerator straight y squared minus 3 over denominator 5 end fraction end cell equals straight x end table end style 
 

Jadi, dapat disimpulkan begin mathsize 14px style straight g to the power of negative 1 end exponent left parenthesis straight x right parenthesis equals fraction numerator straight x squared minus 3 over denominator 5 end fraction end style  

maka dengan menggunakan sifat invers komposisi fungsi

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell open parentheses straight f ring operator straight g close parentheses to the power of negative 1 end exponent open parentheses straight x close parentheses end cell equals cell open parentheses straight g to the power of negative 1 end exponent ring operator straight f to the power of negative 1 end exponent close parentheses open parentheses straight x close parentheses end cell row blank equals cell fraction numerator open parentheses straight f open parentheses straight x close parentheses close parentheses squared minus 3 over denominator 5 end fraction end cell row blank equals cell fraction numerator open parentheses begin display style fraction numerator 2 straight x plus 3 over denominator straight x minus 2 end fraction end style close parentheses squared minus 3 over denominator 5 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 4 straight x squared plus 12 straight x plus 9 over denominator straight x squared minus 4 straight x plus 4 end fraction end style minus 3 over denominator 5 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 4 straight x squared plus 12 straight x plus 9 minus 3 open parentheses straight x squared minus 4 straight x plus 4 close parentheses over denominator straight x squared minus 4 straight x plus 4 end fraction end style over denominator 5 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 4 straight x squared plus 12 straight x plus 9 minus 3 straight x squared plus 12 straight x minus 12 over denominator straight x squared minus 4 straight x plus 4 end fraction end style over denominator 5 end fraction end cell row blank equals cell fraction numerator straight x squared plus 24 straight x minus 3 over denominator 5 open parentheses straight x squared minus 4 straight x plus 4 close parentheses end fraction end cell row blank equals cell fraction numerator straight x squared plus 24 straight x minus 3 over denominator 5 straight x squared minus 20 straight x plus 20 end fraction end cell end table end style 
 

Jadi, begin mathsize 14px style left parenthesis straight f ring operator straight g right parenthesis to the power of negative 1 end exponent left parenthesis straight x right parenthesis equals fraction numerator straight x squared plus 24 straight x minus 3 over denominator 5 straight x squared minus 20 straight x plus 20 end fraction end style 

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