Roboguru

Diketahui , dan . Tentukan: e. ,

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:

e. open parentheses g ring operator h ring operator f close parentheses to the power of negative 1 end exponent open parentheses x close parentheses,

Pembahasan Soal:

Ingat kembali:

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

Berarti:

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

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

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

Diketahui , , dan  berturut-turut menyatakan invers fungsi , , dan . Jika  dan , nilai  adalah ...

Pembahasan Soal:

begin mathsize 14px style open parentheses f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator h to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals 2 x minus 4 end style dengan menggunakan konsep invers fungsi komposisi maka:

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

Misalkan begin mathsize 14px style open parentheses table attributes columnalign right center left columnspacing 0px end attributes row blank blank h end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank ring operator end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank g end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank ring operator end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank f end table close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals y end style maka begin mathsize 14px style x equals open parentheses table attributes columnalign right center left columnspacing 0px end attributes row blank blank h end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank ring operator end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank g end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank ring operator end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank f end table close parentheses open parentheses y close parentheses end style sehingga:

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

begin mathsize 14px style open parentheses h ring operator g close parentheses open parentheses x close parentheses equals fraction numerator x minus 3 over denominator 2 x plus 1 end fraction comma space x not equal to 1 half end style maka berdasarkan konsep komposisi fungsi:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses h ring operator g ring operator f close parentheses open parentheses x close parentheses end cell equals cell fraction numerator x plus 4 over denominator 2 end fraction end cell row cell open parentheses h ring operator g close parentheses open parentheses f open parentheses x close parentheses close parentheses end cell equals cell fraction numerator x plus 4 over denominator 2 end fraction end cell row cell fraction numerator f open parentheses x close parentheses minus 3 over denominator 2 f open parentheses x close parentheses plus 1 end fraction end cell equals cell fraction numerator x plus 4 over denominator 2 end fraction end cell row cell 2 open parentheses f open parentheses x close parentheses minus 3 close parentheses end cell equals cell open parentheses x plus 4 close parentheses open parentheses 2 f open parentheses x close parentheses plus 1 close parentheses end cell row cell 2 f open parentheses x close parentheses minus 6 end cell equals cell 2 x f open parentheses x close parentheses plus 8 f open parentheses x close parentheses plus x plus 4 end cell row cell 2 f open parentheses x close parentheses minus 8 f open parentheses x close parentheses minus 2 x f open parentheses x close parentheses end cell equals cell x plus 4 plus 6 end cell row cell negative 6 f open parentheses x close parentheses minus 2 x f open parentheses x close parentheses end cell equals cell x plus 10 end cell row cell f open parentheses x close parentheses open parentheses negative 6 minus 2 x close parentheses end cell equals cell x plus 10 end cell row cell f open parentheses x close parentheses end cell equals cell fraction numerator x plus 10 over denominator negative 6 minus 2 x end fraction end cell row cell f open parentheses 8 close parentheses end cell equals cell fraction numerator 8 plus 10 over denominator negative 6 minus 2 open parentheses 8 close parentheses end fraction end cell row blank equals cell fraction numerator 18 over denominator negative 22 end fraction end cell row blank equals cell negative 9 over 11 end cell end table end style 

Oleh karena itu, jawaban yang benar adalah C.

0

Roboguru

Diketahui fungsi , dan  Jika fungsi  dan  Tentukan fungsi

Pembahasan Soal:

Diketahui

(fgh)1(x)=3xdany=3xmakay2=3xx=3y2

Lebih lanjut, diperoleh

(fgh)(x)f(g(h(x)))f(h(x)h(x)+2)43(h(x)h(x)+2)24h(x)3h(x)+6h(x)3h(x)+6h(x)3h(x)+69h(x)+18(x212+9)h(x)+18(x212+9)h(x)h(x)===========3x23x23x23x23x243x2312x212h(x)x2h(x)018(x212+9)18

Dengan demikian, diperoleh nilai h(x)=(x212+9)18.

1

Roboguru

Jika diketahui fungsi =...

Pembahasan Soal:

Penyelesaian:

  • h left parenthesis x right parenthesis equals 2 x squared minus 1  y equals 2 x squared minus 1  y plus 1 equals 2 x squared  x equals square root of fraction numerator y plus 1 over denominator 2 end fraction end root  h to the power of negative 1 end exponent equals square root of fraction numerator x plus 1 over denominator 2 end fraction end root
  • f left parenthesis x right parenthesis equals 2 x minus 7 space d a n space g left parenthesis x right parenthesis equals x squared plus 5  left parenthesis f ring operator g ring operator h to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis equals left parenthesis f ring operator left parenthesis g ring operator h to the power of negative 1 end exponent right parenthesis right parenthesis left parenthesis x right parenthesis  m i s a l equals left parenthesis g ring operator h to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis equals p left parenthesis x right parenthesis  left parenthesis g ring operator h to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis equals g open parentheses fraction numerator square root of x plus 1 end root over denominator 2 end fraction close parentheses  equals open parentheses square root of fraction numerator x plus 1 over denominator 2 end fraction end root close parentheses squared plus 5  equals fraction numerator x plus 1 over denominator 2 end fraction plus 5  equals fraction numerator x plus 1 over denominator 2 end fraction plus 10 over 2 equals fraction numerator x plus 11 over denominator 2 end fraction
  • left parenthesis f ring operator left parenthesis g ring operator h to the power of negative 1 end exponent right parenthesis right parenthesis left parenthesis x right parenthesis equals open parentheses f ring operator p close parentheses left parenthesis x right parenthesis  equals f open parentheses fraction numerator x plus 11 over denominator 2 end fraction close parentheses  equals 2 open parentheses fraction numerator x plus 11 over denominator 2 end fraction close parentheses minus 7  equals x plus 11 minus 7  equals x plus 4

0

Roboguru

Diketahui ,  dan , tentukan nilai x sehingga .

Pembahasan Soal:

Perhatikan perhitungan berikut 

(hgf)(x)====h(g(f(x)))h(g(2x1))h(3(2x1))h(42x)

Lebih lanjut, diperoleh    

(hgf)(x)y(42x)y4y2xy2xy2xxx========42x442x44444yy44y2y44y2y+2

sehingga

(hgf)1(x)11x====2x+22x+22x21

Dengan demikian, nilai x sehingga bold left parenthesis bold italic h bold ring operator bold italic g bold ring operator bold italic f bold right parenthesis to the power of bold minus bold 1 end exponent bold left parenthesis bold italic x bold right parenthesis bold equals bold 1 adalah bold 1 over bold 2 bold.  

0

Roboguru

Diketahui , , dan  berturut-turut menyatakan invers fungsi , , dan . Jika  dan , , nilai  adalah ....

Pembahasan Soal:

Diketahui: begin mathsize 14px style left parenthesis f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator h to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis equals 2 x minus 4 end style dan begin mathsize 14px style left parenthesis h ring operator g right parenthesis left parenthesis x right parenthesis equals fraction numerator x minus 3 over denominator 2 x plus 1 end fraction end style

perhatikan bahwa begin mathsize 14px style open parentheses h ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals open parentheses g to the power of negative 1 end exponent ring operator h to the power of negative 1 end exponent close parentheses open parentheses x close parentheses end style, misalkan begin mathsize 14px style open parentheses g to the power of negative 1 end exponent ring operator h to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals y space end style maka:

begin mathsize 14px style y equals fraction numerator x minus 3 over denominator 2 x plus 1 end fraction end style

Karena begin mathsize 14px style left parenthesis f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator h to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis equals 2 x minus 4 end style maka 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f to the power of negative 1 end exponent open parentheses open parentheses g to the power of negative 1 end exponent ring operator h to the power of negative 1 end exponent close parentheses open parentheses x close parentheses close parentheses end cell equals cell 2 x minus 4 end cell row cell f to the power of negative 1 end exponent open parentheses y close parentheses end cell equals cell 2 open parentheses fraction numerator y minus 3 over denominator 2 y plus 1 end fraction close parentheses minus 4 end cell row blank equals cell fraction numerator 2 y minus 6 over denominator 2 y plus 1 end fraction minus fraction numerator 4 open parentheses 2 y plus 1 close parentheses over denominator 2 y plus 1 end fraction end cell row blank equals cell fraction numerator negative 4 y minus 10 over denominator 2 y plus 1 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 negative 4 x minus 10 over denominator 2 x plus 1 end fraction end cell end table end style

misalkan : begin mathsize 14px style y equals f to the power of negative 1 end exponent open parentheses x close parentheses end style dan begin mathsize 14px style open parentheses f to the power of negative 1 end exponent open parentheses x close parentheses close parentheses to the power of negative 1 end exponent equals f open parentheses x close parentheses end style  maka 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator negative 4 x minus 10 over denominator 2 x plus 1 end fraction end cell row y equals cell fraction numerator negative 4 x minus 10 over denominator 2 x plus 1 end fraction end cell row cell 2 x y plus y end cell equals cell negative 4 x minus 10 end cell row cell y plus 10 end cell equals cell x open parentheses negative 4 minus 2 y close parentheses end cell row x equals cell fraction numerator y plus 10 over denominator negative 4 minus 2 y end fraction end cell row cell f open parentheses x close parentheses end cell equals cell fraction numerator x plus 10 over denominator negative 4 minus 2 x end fraction end cell row cell f open parentheses 8 close parentheses end cell equals cell fraction numerator 8 plus 10 over denominator negative 4 minus 2 open parentheses 8 close parentheses end fraction end cell row blank equals cell negative 4 over 5 end cell end table end style   

Jadi, jawaban yang tepat adalah B. 

 

0

Roboguru

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