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Diketahui , dan . Tentukan: b. ,

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:

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

Pembahasan Soal:

Ingat kembali:

f open parentheses x close parentheses equals a x squared plus b x plus c rightwards arrow f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative b plus-or-minus square root of b squared minus 4 a open parentheses c minus x close parentheses end root over denominator 2 a end fraction 

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:

open parentheses g ring operator h close parentheses open parentheses x close parentheses equals g open parentheses h open parentheses x close parentheses close parentheses open parentheses g ring operator h close parentheses open parentheses x close parentheses equals g open parentheses 3 x plus 5 close parentheses open parentheses g ring operator h close parentheses open parentheses x close parentheses equals open parentheses 3 x plus 5 close parentheses squared minus 1 open parentheses g ring operator h close parentheses open parentheses x close parentheses equals 9 x squared plus 30 x plus 25 minus 1 open parentheses g ring operator h close parentheses open parentheses x close parentheses equals 9 x squared plus 30 x plus 24 

Sehingga:

 open parentheses g ring operator h close parentheses open parentheses x close parentheses equals 9 x squared plus 30 x plus 24 open parentheses g ring operator h close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative 30 plus-or-minus square root of 30 squared minus 4 times 9 times open parentheses 24 minus x close parentheses end root over denominator 2 times 9 end fraction open parentheses g ring operator h close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative 30 plus-or-minus square root of 900 minus 36 open parentheses 24 minus x close parentheses end root over denominator 18 end fraction open parentheses g ring operator h close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative 30 plus-or-minus square root of 900 minus 864 plus 36 x end root over denominator 18 end fraction open parentheses g ring operator h close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative 30 plus-or-minus square root of 36 plus 36 x end root over denominator 18 end fraction 

Jadi, open parentheses g ring operator h close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative 30 plus-or-minus square root of 36 plus 36 x end root over denominator 18 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

Tentukan  jika  dan  ?

Pembahasan Soal:

Terlebih dahulu tentukan komposisi undefined, sebagai berikut:

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 5 x plus 7 close parentheses end cell row blank equals cell 3 open parentheses 5 x plus 7 close parentheses minus 1 end cell row blank equals cell 15 x plus 21 minus 1 end cell row blank equals cell 15 x plus 20 end cell end table end style 

Setelah diperoleh komposisi begin mathsize 14px style open parentheses g ring operator f close parentheses open parentheses x close parentheses equals 15 x plus 20 end style, maka tentukan invers dari fungsi komposisi tersebut. Dengan demikian:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 15 x plus 20 end cell equals y row cell 15 x end cell equals cell y minus 20 end cell row x equals cell fraction numerator y minus 20 over denominator 15 end fraction end cell end table end style

Berdasarkan perhitungan di atas, diperoleh begin mathsize 14px style open parentheses g ring operator f close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x minus 20 over denominator 15 end fraction end style. Kemudian substitusikan nilai begin mathsize 14px style x equals 2 end style  pada fungsi invers tersebut. Dengan demikian:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes 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 x minus 20 over denominator 15 end fraction end cell row cell open parentheses g ring operator f close parentheses to the power of negative 1 end exponent open parentheses 2 close parentheses end cell equals cell fraction numerator 2 minus 20 over denominator 15 end fraction end cell row blank equals cell fraction numerator negative 18 over denominator 15 end fraction end cell row blank equals cell negative 6 over 5 end cell end table end style  

Dengan demikian, nilai begin mathsize 14px style open parentheses g ring operator f close parentheses to the power of negative 1 end exponent open parentheses 2 close parentheses equals negative 6 over 5 end style.

0

Roboguru

Jika diketahui fungsi =....

Pembahasan Soal:

Penyelesaian:

  • f left parenthesis x right parenthesis equals 3 x minus 2  y equals 3 x minus 2  3 x equals y plus 2 left right double arrow x equals fraction numerator y plus 2 over denominator 3 end fraction  f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x plus 2 over denominator 3 end fraction
  • g left parenthesis x right parenthesis equals square root of 2 x plus 1 end root  y equals square root of 2 x plus 1 end root  y squared equals 2 x plus 1  2 x equals y squared minus 1  x equals fraction numerator y squared minus 1 over denominator 2 end fraction  g to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x squared minus 1 over denominator 2 end fraction
  • h left parenthesis x right parenthesis equals 5 x squared plus 2  y equals 5 x squared plus 2  y minus 2 equals 5 x squared  x squared equals fraction numerator y minus 2 over denominator 5 end fraction  x equals square root of fraction numerator y minus 2 over denominator 5 end fraction end root  h to the power of negative 1 end exponent left parenthesis x right parenthesis equals square root of fraction numerator x minus 2 over denominator 5 end fraction end root
  • M i s a l space left parenthesis 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 p left parenthesis x right parenthesis  left parenthesis 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 left parenthesis g to the power of negative 1 end exponent left parenthesis h to the power of negative 1 end exponent left parenthesis x right parenthesis right parenthesis right parenthesis  equals g to the power of negative 1 end exponent open parentheses square root of fraction numerator x minus 2 over denominator 5 end fraction end root close parentheses squared minus 1  equals fraction numerator begin display style fraction numerator x minus 2 over denominator 5 end fraction end style minus begin display style 5 over 5 end style over denominator 2 end fraction equals fraction numerator x minus 7 over denominator 5 end fraction.1 half  equals fraction numerator x minus 7 over denominator 10 end fraction    p left parenthesis x right parenthesis equals fraction numerator x minus 7 over denominator 10 end fraction  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 left parenthesis f to the power of negative 1 end exponent ring operator p right parenthesis left parenthesis x right parenthesis  left parenthesis f to the power of negative 1 end exponent left parenthesis p right parenthesis right parenthesis left parenthesis x right parenthesis equals f to the power of negative 1 end exponent open parentheses fraction numerator x minus 7 over denominator 10 end fraction close parentheses  equals fraction numerator begin display style fraction numerator x minus 7 over denominator 10 end fraction end style plus 2 over denominator 3 end fraction  equals fraction numerator x plus 13 over denominator 10 end fraction.1 third equals fraction numerator x plus 13 over denominator 30 end fraction

0

Roboguru

Diketahui fungsi  dan . Tentukan .

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 left parenthesis x right parenthesis right parenthesis end cell row blank equals cell f open parentheses fraction numerator 1 over denominator 3 x end fraction close parentheses end cell row blank equals cell fraction numerator 6 times begin display style fraction numerator 1 over denominator 3 x end fraction end style plus 11 over denominator 4 times begin display style fraction numerator 1 over denominator 3 x end fraction end style minus 9 end fraction end cell row blank equals cell fraction numerator 6 plus 33 x over denominator 4 minus 27 x end fraction. end cell end table end style

Dengan memisalkan begin mathsize 14px style left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis equals y end style, maka :

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator 6 plus 33 x over denominator 4 minus 27 x end fraction end cell row cell y times left parenthesis 4 minus 27 x right parenthesis end cell equals cell 6 plus 33 x end cell row cell 4 y minus 27 x y end cell equals cell 6 plus 33 x end cell row cell 33 x plus 27 x y end cell equals cell 6 minus 4 y end cell row x equals cell fraction numerator 6 plus 4 y over denominator 27 y plus 33 end fraction. end cell end table end style

Sehingga diperoleh begin mathsize 14px style left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 6 plus 4 x over denominator 27 x plus 33 end fraction. end style

0

Roboguru

Diketahui  dengan  dan . Tentukan !

Pembahasan Soal:

Menentukan fungsi komposisi.

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 x plus 2 close parentheses end cell row blank equals cell 4 minus 2 open parentheses x plus 2 close parentheses end cell row blank equals cell 4 minus 2 x minus 4 end cell row blank equals cell negative 2 x end cell end table end style 

Menentukan invers fungsi komposisi.

begin mathsize 14px style 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 y row cell negative 2 x end cell equals y row x equals cell negative y over 2 end cell row cell left parenthesis g ring operator f right parenthesis to the power of negative 1 end exponent left parenthesis y right parenthesis end cell equals cell negative y over 2 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 negative x over 2 end cell end table end style  

Jadi, begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell left parenthesis g ring operator f right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell x over 2 end cell end table end style 

0

Roboguru

Diketahui fungsi  dan . Hitunglah nilai dari !

Pembahasan Soal:

Diketahui fungsi begin mathsize 14px style f left parenthesis x right parenthesis equals 3 x minus 5 end style dan size 14px g size 14px left parenthesis size 14px x size 14px right parenthesis size 14px equals size 14px 2 size 14px x size 14px plus size 14px 1, sehingga dapat ditentukan fungsi komposisi berikut.

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 3 open parentheses 2 x plus 1 close parentheses minus 5 end cell row blank equals cell 6 x plus 3 minus 5 end cell row blank equals cell 6 x minus 2 end cell end table end style 

Selanjutnya, dapat ditentukan invers dari fungsi komposisi yang diperoleh sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses f ring operator g close parentheses left parenthesis x right parenthesis equals 6 x minus 2 end cell row y equals cell 6 x minus 2 end cell row cell 6 x end cell equals cell y plus 2 end cell row x equals cell fraction numerator y plus 2 over denominator 6 end fraction 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 fraction numerator x plus 2 over denominator 6 end fraction end cell end table end style 

Sehingga dapat ditentukan nilai dari Error converting from MathML to accessible text. sebagai berikut.

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 fraction numerator x plus 2 over denominator 6 end fraction end cell row cell open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses 2 close parentheses end cell equals cell fraction numerator 2 plus 2 over denominator 6 end fraction end cell row blank equals cell 4 over 6 end cell row blank equals cell 2 over 3 end cell end table end style 

Dengan demikian diperoleh nilai dari Error converting from MathML to accessible text..

0

Roboguru

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