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Diketahui f ( x ) = x + 1 2 x − 1 ​ , g ( x ) = x 2 − 1 , dan h ( x ) = 3 x + 5 . Tentukan: a. ( f − 1 ∘ h − 1 ) ( x ) ,

Diketahui , dan . Tentukan:

a. 

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W. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sriwijaya

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Jawaban

.

 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 x close parentheses equals fraction numerator 3 minus x over denominator x minus 11 end fraction.

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Pembahasan

Ingat kembali: Berarti: Diketahui , dan . Maka: Sehingga: Jadi, .

Ingat kembali:

open parentheses f 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 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 close parentheses open parentheses x close parentheses equals open parentheses 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:

open parentheses h ring operator f close parentheses open parentheses x close parentheses equals h open parentheses f open parentheses x close parentheses close parentheses open parentheses h ring operator f close parentheses open parentheses x close parentheses equals h open parentheses fraction numerator 2 x minus 1 over denominator x plus 1 end fraction close parentheses open parentheses h ring operator f close parentheses open parentheses x close parentheses equals 3 open parentheses fraction numerator 2 x minus 1 over denominator x plus 1 end fraction close parentheses plus 5 open parentheses h ring operator f close parentheses open parentheses x close parentheses equals fraction numerator 6 x minus 2 over denominator x plus 1 end fraction plus 5 open parentheses h ring operator f close parentheses open parentheses x close parentheses equals fraction numerator 6 x minus 2 plus 5 open parentheses x plus 1 close parentheses over denominator x plus 1 end fraction open parentheses h ring operator f close parentheses open parentheses x close parentheses equals fraction numerator 6 x minus 2 plus 5 x plus 5 over denominator x plus 1 end fraction open parentheses h ring operator f close parentheses open parentheses x close parentheses equals fraction numerator 11 x plus 3 over denominator x plus 1 end fraction 

Sehingga:

space space space space open parentheses h ring operator f close parentheses open parentheses x close parentheses equals fraction numerator 11 x plus 3 over denominator x plus 1 end fraction space space space space space space space space space space space space space space space space y equals fraction numerator 11 x plus 3 over denominator x plus 1 end fraction space space space space y times open parentheses x plus 1 close parentheses equals 11 x plus 3 space space space space space space space space space x y plus y equals 11 x plus 3 space space space space space x y minus 11 x equals 3 minus y space space space space x open parentheses y minus 11 close parentheses equals 3 minus y space space space space space space space space space space space space space space space space x equals fraction numerator 3 minus y over denominator y minus 11 end fraction open parentheses h ring operator f close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator 3 minus x over denominator x minus 11 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 close parentheses open parentheses x close parentheses equals fraction numerator 3 minus x over denominator x minus 11 end fraction.

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Diketahui fungsi f : R → R dan g : R → R dirumuskan oleh f ( x ) = x − 2 2 x + 1 ​ , x  = 2 dan g ( x ) = x + 3 . Jika h ( x ) = ( f ∘ g ) ( x ) . Tentukanlah invers h ( x ) .

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