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Diketahui  . Invers dari  adalah  maka nilai  adalah ...

Pertanyaan

Diketahui  h open parentheses x close parentheses equals 2 x plus 11. Invers dari h open parentheses x close parentheses adalah h to the power of negative 1 end exponent open parentheses x close parentheses comma maka nilai h to the power of negative 1 end exponent open parentheses 4 close parentheses adalah ...

Pembahasan Soal:

Dari soal diketahui 

h open parentheses x close parentheses equals 2 x plus 11

Invers dari fungsi h open parentheses x close parentheses 

table attributes columnalign right center left columnspacing 0px end attributes row cell h open parentheses x close parentheses end cell equals cell 2 x plus 11 end cell row y equals cell 2 x plus 1 end cell row cell y minus 1 end cell equals cell 2 x end cell row x equals cell fraction numerator y minus 1 over denominator 2 end fraction end cell row cell h to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator x minus 1 over denominator 2 end fraction end cell end table  

Jadi, invers dari h open parentheses x close parentheses adalah Error converting from MathML to accessible text.  

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Roy

Mahasiswa/Alumni Universitas Negeri Surabaya

Terakhir diupdate 02 Mei 2021

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Pertanyaan yang serupa

Diketahui fungsi  dan . Jika , maka nilai  adalah ...

Pembahasan Soal:

Tentukan invers fungsinya terlebih dahulu.

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell x cubed end cell row y equals cell x cubed end cell row x equals cell cube root of y end cell row cell f to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell cube root of x end cell row blank blank blank row cell g open parentheses x close parentheses end cell equals cell 3 x minus 4 end cell row y equals cell 3 x minus 4 end cell row x equals cell fraction numerator y plus 4 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 cell fraction numerator x plus 4 over denominator 3 end fraction end cell end table

Kemudian dari fungsi-fungsi invers tersebut gunakan rumus fungsi komposisi untuk mengetahui nilai a.

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 8 close parentheses end cell equals a row cell g to the power of negative 1 end exponent open parentheses f to the power of negative 1 end exponent open parentheses 8 close parentheses close parentheses end cell equals a row cell g to the power of negative 1 end exponent open parentheses cube root of 8 close parentheses end cell equals a row cell fraction numerator cube root of 8 plus 4 over denominator 3 end fraction end cell equals a row cell fraction numerator 2 plus 4 over denominator 3 end fraction end cell equals a row cell 6 over 3 end cell equals a row 2 equals a end table

Kemudian diperoleh:

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 to the power of negative 1 end exponent close parentheses open parentheses 10 a close parentheses end cell equals cell f to the power of negative 1 end exponent open parentheses g to the power of negative 1 end exponent open parentheses 10 open parentheses 2 close parentheses close parentheses close parentheses end cell row blank equals cell f to the power of negative 1 end exponent open parentheses g to the power of negative 1 end exponent open parentheses 20 close parentheses close parentheses end cell row blank equals cell f to the power of negative 1 end exponent open parentheses fraction numerator 20 plus 4 over denominator 3 end fraction close parentheses end cell row blank equals cell cube root of fraction numerator 20 plus 4 over denominator 3 end fraction end root end cell row blank equals cell cube root of 24 over 3 end root end cell row blank equals cell cube root of 8 end cell row blank equals 2 end table

Maka, nilai open parentheses f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses open parentheses 10 a close parentheses adalah 2.

0

Roboguru

Jika  dan , tentukan: a.

Pembahasan Soal:

Diketahui:

 begin mathsize 14px style f open parentheses x close parentheses equals 2 x minus 4 end style 

begin mathsize 14px style g open parentheses x close parentheses equals 3 x plus 9 end style

Ditanyakan:

begin mathsize 14px style f to the power of negative 1 end exponent left parenthesis x right parenthesis equals... end style 

jawab:

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

Jadi, begin mathsize 14px style f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x plus 4 over denominator 2 end fraction end style

0

Roboguru

Diketahui  dan   b.  ?

Pembahasan Soal:

Dengan menggunakan rumus invers fungsi

begin mathsize 14px style z open parentheses x close parentheses equals a x plus b blank semicolon a not equal to 0 z to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x minus b over denominator a end fraction blank semicolon a not equal to 0 end style 

Diperoleh 

begin mathsize 14px style f open parentheses x close parentheses equals x minus 3 f to the power of negative 1 end exponent open parentheses x close parentheses equals x plus 3 g open parentheses x close parentheses equals 2 x plus 4 g to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x minus 4 over denominator 2 end fraction end style  

Nilai begin mathsize 14px style left parenthesis g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis end style adalah begin mathsize 14px style f to the power of negative 1 end exponent end style dimasukkan ke begin mathsize 14px style g to the power of negative 1 end exponent space colon space left parenthesis g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis equals g to the power of negative 1 end exponent left parenthesis f to the power of negative 1 end exponent left parenthesis x right parenthesis right parenthesis end style 

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

Jadi, diperoleh bahwa begin mathsize 14px style left parenthesis g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent right parenthesis left parenthesis x right parenthesis equals fraction numerator x minus 1 over denominator 2 end fraction end style                undefined 

0

Roboguru

Jika  adalah invers dari , maka nilai  adalah ....

Pembahasan Soal:

Jika begin mathsize 14px style g open parentheses x close parentheses equals fraction numerator 2 x plus 6 over denominator 3 minus x end fraction comma space x not identical to 3 end style dan begin mathsize 14px style y 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 y equals cell fraction numerator 2 x plus 6 over denominator 3 minus x end fraction rightwards double arrow 3 y minus x y equals 2 x plus 6 end cell row blank rightwards double arrow cell 3 y minus 6 equals 2 x plus x y end cell row blank rightwards double arrow cell 3 y minus 6 equals open parentheses 2 plus y close parentheses x end cell row blank rightwards double arrow cell x equals fraction numerator open parentheses 3 y minus 6 close parentheses over denominator open parentheses 2 plus y close parentheses end fraction end cell row cell text Jadi, end text space g to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator open parentheses 3 x minus 6 close parentheses over denominator open parentheses 2 plus x close parentheses end fraction end cell row cell space g to the power of negative 1 end exponent open parentheses negative 1 close parentheses end cell equals cell fraction numerator open parentheses 3 open parentheses negative 1 close parentheses minus 6 close parentheses over denominator open parentheses 2 plus open parentheses negative 1 close parentheses close parentheses end fraction end cell row blank equals cell fraction numerator negative 9 over denominator 1 end fraction end cell row blank equals cell negative 9 end cell end table end style 

0

Roboguru

Diketahui fungsi sebagai berikut:   Tentukan: f.

Pembahasan Soal:

Ada 3 langkah untuk menentukan fungsi invers, yaitu:

  1. Ubahlah bentuk begin mathsize 14px style y equals f open parentheses x close parentheses end style menjadi bentuk begin mathsize 14px style x equals f open parentheses y close parentheses end style.
  2. Tuliskan begin mathsize 14px style x end style sebagai begin mathsize 14px style f to the power of negative 1 end exponent open parentheses y close parentheses end style sehingga begin mathsize 14px style f to the power of negative 1 end exponent open parentheses y close parentheses equals f open parentheses y close parentheses end style.
  3. Ubahlah variabel begin mathsize 14px style y end style dengan begin mathsize 14px style x end style sehingga diperoleh rumus fungsi invers begin mathsize 14px style f to the power of negative 1 end exponent open parentheses x close parentheses end style.

Diketahui:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell f open parentheses x close parentheses end cell equals cell 2 x plus 1 end cell row cell h open parentheses x close parentheses end cell equals cell x minus 4 end cell row cell g open parentheses x close parentheses end cell equals cell x squared minus 3 x plus 2 end cell row cell m open parentheses x close parentheses end cell equals cell fraction numerator x plus 4 over denominator 2 x minus 5 end fraction end cell end table end style

Akan ditentukan begin mathsize 14px style f to the power of negative 1 end exponent open parentheses x close parentheses end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell f open parentheses x close parentheses equals y end cell equals cell 2 x plus 1 end cell row cell negative 2 x end cell equals cell negative y plus 1 end cell row x equals cell fraction numerator negative y plus 1 over denominator negative 2 end fraction end cell row cell x equals f open parentheses y close parentheses end cell equals cell fraction numerator y minus 1 over denominator 2 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 x minus 1 over denominator 2 end fraction end cell end table end style 

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

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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