Roboguru

Jika diketahui  dan  . Rumus komposisi fungsi  adalah ...

Pertanyaan

Jika diketahui begin mathsize 14px style f left parenthesis x right parenthesis equals 4 x squared space plus space 3 comma space g left parenthesis x right parenthesis equals x space minus space 5 end style dan begin mathsize 14px style h left parenthesis x right parenthesis equals square root of 2 x end root end style . Rumus komposisi fungsi begin mathsize 14px style open parentheses f ring operator g ring operator h close parentheses open parentheses x close parentheses end style adalah ...

  1. ...

  2. ...space

Pembahasan Soal:

1. Tentukan begin mathsize 14px style open parentheses f ring operator g close parentheses open parentheses x close parentheses end style

begin mathsize 14px style open parentheses f ring operator g close parentheses open parentheses x close parentheses equals 4 open parentheses x minus 5 close parentheses squared plus 3 space space space space space space space space space space space space space space equals 4 open parentheses x squared minus 10 x plus 25 close parentheses plus 3 space space space space space space space space space space space space space space equals 4 x squared minus 40 x plus 100 plus 3 space space space space space space space space space space space space space space equals 4 x squared minus 40 x plus 103 end style

2. Tentukan begin mathsize 14px style open parentheses f ring operator g ring operator h close parentheses open parentheses x close parentheses end style

begin mathsize 14px style open parentheses f ring operator g ring operator h close parentheses open parentheses x close parentheses equals 4 open parentheses square root of 2 x end root close parentheses squared minus 40 open parentheses square root of 2 x end root close parentheses plus 103 space space space space space space space space space space space space space space space space space space space space equals 4 open parentheses 2 x close parentheses minus 40 square root of 2 x end root plus 103 space space space space space space space space space space space space space space space space space space space space equals 8 x minus 40 square root of 2 x end root plus 103 end style

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Azizatul

Mahasiswa/Alumni Universitas Negeri Yogyakarta

Terakhir diupdate 28 Maret 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui ,  dan . Tentukan nilai  dari fungsi berikut!

Pembahasan Soal:

Tentukan bentuk open parentheses f ring operator g ring operator h 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 ring operator h close parentheses end cell equals cell f open parentheses g open parentheses h open parentheses x close parentheses close parentheses close parentheses end cell row blank equals cell f open parentheses g open parentheses square root of 1 minus 2 x end root close parentheses close parentheses end cell row blank equals cell f open parentheses open parentheses square root of 1 minus 2 x end root close parentheses squared minus 2 close parentheses end cell row blank equals cell f open parentheses 1 minus 2 x minus 2 close parentheses end cell row blank equals cell f open parentheses negative 2 x minus 1 close parentheses end cell row blank equals cell square root of open parentheses negative 2 x minus 1 close parentheses plus 1 end root end cell row blank equals cell square root of negative 2 x end root end cell end table 

a. Kuadratkan kedua ruas. 

 table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g ring operator h close parentheses open parentheses x close parentheses end cell equals 2 row cell open parentheses square root of negative 2 x end root close parentheses squared end cell equals cell 2 squared end cell row cell negative 2 x end cell equals 4 row x equals cell fraction numerator 4 over denominator negative 2 end fraction end cell row x equals cell negative 2 end cell row blank blank blank end table 

Jadi, diperoleh nilai x equals negative 2.

b. Kuadratkan kedua ruas. 

 table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g ring operator h close parentheses open parentheses x close parentheses end cell equals 5 row cell open parentheses square root of negative 2 x end root close parentheses squared end cell equals cell 5 squared end cell row cell negative 2 x end cell equals 25 row x equals cell negative 25 over 2 end cell end table 

Jadi, diperoleh nilai x equals negative 25 over 2.

0

Roboguru

DIketahui ,  dan . Maka nilai dari  adalah ....

Pembahasan Soal:

Berdasarkan sifat komposisi fungsi, maka didapatkan :

table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis x right parenthesis end cell equals cell x plus 2 end cell row cell left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis end cell equals cell open parentheses x squared minus 2 x plus 3 close parentheses plus 2 end cell row blank equals cell x squared minus 2 x plus 5 end cell row cell open parentheses f ring operator g ring operator h close parentheses left parenthesis x right parenthesis end cell equals cell open parentheses negative 2 x plus 5 close parentheses squared minus 2 open parentheses negative 2 x plus 5 close parentheses plus 5 end cell row blank equals cell 4 x squared minus 20 x plus 25 plus 4 x minus 10 plus 5 end cell row blank equals cell 4 x squared minus 20 x plus 4 x plus 25 minus 10 plus 5 end cell row blank equals cell 4 x squared minus 16 x plus 20 end cell end table

Maka nilai dari left parenthesis f ring operator g ring operator h right parenthesis left parenthesis x right parenthesis adalah Error converting from MathML to accessible text.

Oleh karena itu, jawaban yang benar adalah D

0

Roboguru

Jika f(x) = 4 - 2x, g(x) = 3x + 6, dan h(x) = 2 - 4x dengan  , maka nilai x yang memenuhi  adalah ....

Pembahasan Soal:

Pertama, kita cari undefined .

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses straight f ring operator straight h close parentheses open parentheses straight x close parentheses end cell equals cell straight f open parentheses straight h open parentheses straight x close parentheses close parentheses end cell row blank equals cell straight f open parentheses 2 minus 4 straight x close parentheses end cell row blank equals cell 4 minus 2 open parentheses 2 minus 4 straight x close parentheses end cell row blank equals cell 4 minus 4 plus 8 straight x end cell row blank equals cell 8 straight x end cell end table end style 

Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses straight g ring operator straight f ring operator straight h close parentheses open parentheses straight x close parentheses end cell equals 2 row cell straight g open parentheses open parentheses straight f ring operator straight h close parentheses open parentheses straight x close parentheses close parentheses end cell equals 2 row cell straight g open parentheses 8 straight x close parentheses end cell equals 2 row cell 3 open parentheses 8 straight x close parentheses plus 6 end cell equals 2 row cell 24 straight x plus 6 end cell equals 2 row cell 24 straight x end cell equals cell 2 minus 6 end cell row cell 24 straight x end cell equals cell negative 4 end cell row straight x equals cell fraction numerator negative 4 over denominator 24 end fraction end cell row straight x equals cell negative 1 over 6 end cell end table end style 

Jadi, jawabannya D.

0

Roboguru

Diketahui , , dan  dengan . Tentukan nilai  jika .

Pembahasan Soal:

Diketahui:

 begin mathsize 14px style f left parenthesis x right parenthesis equals x squared minus x plus 5 end style

begin mathsize 14px style g left parenthesis x right parenthesis equals x over 2 end style

begin mathsize 14px style h left parenthesis x right parenthesis equals 3 x plus p end style

maka begin mathsize 14px style open parentheses f ring operator g ring operator h close parentheses open parentheses x close parentheses end style adalah 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g ring operator h close parentheses open parentheses x close parentheses end cell equals cell f open parentheses g open parentheses h open parentheses x close parentheses close parentheses close parentheses end cell row blank equals cell f open parentheses g open parentheses 3 x plus p close parentheses close parentheses end cell row blank equals cell f open parentheses fraction numerator 3 x plus p over denominator 2 end fraction close parentheses end cell row blank equals cell open parentheses fraction numerator 3 x plus p over denominator 2 end fraction close parentheses squared minus open parentheses fraction numerator 3 x plus p over denominator 2 end fraction close parentheses plus 5 end cell end table end style  

maka nilai begin mathsize 14px style p end style jika begin mathsize 14px style left parenthesis f ring operator g ring operator h right parenthesis left parenthesis negative 1 right parenthesis equals 7 end style adalah 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis f ring operator g ring operator h right parenthesis left parenthesis negative 1 right parenthesis end cell equals 7 row cell open parentheses fraction numerator negative 3 plus p over denominator 2 end fraction close parentheses squared minus open parentheses fraction numerator negative 3 plus p over denominator 2 end fraction close parentheses plus 5 end cell equals 7 row cell open parentheses fraction numerator p squared minus 6 p plus 9 over denominator 4 end fraction close parentheses minus fraction numerator 2 open parentheses negative 3 plus p close parentheses over denominator 4 end fraction end cell equals 2 row cell fraction numerator p squared minus 6 p plus 9 plus 6 minus 2 p over denominator 4 end fraction end cell equals 2 row cell p squared minus 8 p plus 15 end cell equals 8 row cell p blank squared minus 8 p plus 7 end cell equals 0 row cell open parentheses p minus 7 close parentheses open parentheses p minus 1 close parentheses end cell equals 0 end table end style

Jadi, nilai begin mathsize 14px style p equals 1 end style atau begin mathsize 14px style p equals 7 end style.

3

Roboguru

DIketahui ,  dan . Fungsi komposisi  =....

Pembahasan Soal:

Berdasarkan sifat komposisi fungsi, maka didapatkan :

table attributes columnalign right center left columnspacing 0px end attributes row cell h left parenthesis x right parenthesis end cell equals cell open parentheses fraction numerator 1 over denominator x plus 1 end fraction close parentheses end cell row cell open parentheses h ring operator g close parentheses left parenthesis x right parenthesis end cell equals cell open parentheses fraction numerator 1 over denominator left parenthesis x squared minus 1 right parenthesis plus 1 end fraction close parentheses end cell row blank equals cell open parentheses 1 over x squared close parentheses end cell row cell open parentheses h ring operator g ring operator f close parentheses left parenthesis x right parenthesis end cell equals cell open parentheses 1 over left parenthesis x plus 2 right parenthesis squared close parentheses end cell row blank equals cell fraction numerator 1 over denominator x squared plus 4 x plus 4 end fraction end cell end table

Maka :

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis f ring operator g ring operator h right parenthesis left parenthesis a right parenthesis end cell equals 11 row cell 4 a squared plus 30 a plus 61 end cell equals 11 row cell 4 a squared plus 30 a plus 61 minus 11 end cell equals 0 row cell 4 a squared plus 30 a plus 50 end cell equals 0 row cell 2 a squared plus 15 a plus 25 end cell equals 0 row cell left parenthesis 2 a plus 5 right parenthesis left parenthesis a plus 5 right parenthesis end cell equals 0 end table

Maka fungsi komposisi left parenthesis h ring operator g ring operator f right parenthesis left parenthesis x right parenthesis adalah fraction numerator 1 over denominator x squared plus 4 x plus 4 end fraction

Oleh karena itu, jawaban yang benar adalah D

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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