Pertanyaan

Fungsi f : R → R dan g : R → R ditentukan oleh rumus f ( x ) = 7 x − 5 dan g ( x ) = 4 x + 9 . Tentukan Apakah ( g ∘ f ) − 1 ( x ) = ( f − 1 ∘ g − 1 ) ( x ) (Ya / Tidak)

Fungsi $f : R \to R$ dan $g : R \to R$ ditentukan oleh rumus $f (x) = 7 x - 5$ dan $g (x) = 4 x + 9$ . Tentukan

Apakah $(g \circ f)^{- 1} (x) =$ $(f^{- 1} \circ g^{- 1}) (x)$ (Ya / Tidak)

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Pertama kita tentukan invers dari Selanjutnya kita tentukan invers dari Maka sebagai berikut : maka diperoleh, Kemudian kita tentukan maka diperoleh, dan hasilnya sama dengan Jadi, ya terbuktibahwa

Pertama kita tentukan invers dari

$begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell straight f open parentheses straight x close parentheses end cell equals cell 7 straight x minus 5 end cell row straight y equals cell 7 straight x minus 5 end cell row cell negative 7 straight x end cell equals cell negative straight y minus 5 end cell row straight x equals cell fraction numerator negative straight y minus 5 over denominator negative 7 end fraction end cell row sehingga colon blank row cell straight f to the power of negative 1 end exponent open parentheses straight x close parentheses end cell equals cell fraction numerator negative straight x minus 5 over denominator negative 7 end fraction end cell end table end style$

Selanjutnya kita tentukan invers dari

$begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell straight g open parentheses straight x close parentheses end cell equals cell 4 straight x plus 9 end cell row straight y equals cell 4 straight x plus 9 end cell row cell negative 4 straight x end cell equals cell negative straight y plus 9 end cell row straight x equals cell fraction numerator negative straight y plus 9 over denominator negative 4 end fraction end cell row sehingga colon blank row cell straight g to the power of negative 1 end exponent open parentheses straight x close parentheses end cell equals cell fraction numerator negative straight x plus 9 over denominator negative 4 end fraction end cell end table end style$

Maka sebagai berikut :

$begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell open parentheses straight f to the power of negative 1 end exponent ring operator straight g to the power of negative 1 end exponent close parentheses open parentheses straight x close parentheses end cell equals cell fraction numerator negative open parentheses straight g to the power of negative 1 end exponent open parentheses straight x close parentheses close parentheses minus 5 over denominator negative 7 end fraction end cell row blank equals cell fraction numerator negative open parentheses begin display style fraction numerator negative straight x plus 9 over denominator negative 4 end fraction end style close parentheses minus 5 over denominator negative 7 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator straight x minus 9 over denominator negative 4 end fraction minus open parentheses fraction numerator negative 20 over denominator negative 4 end fraction close parentheses end style over denominator negative 7 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator straight x minus 9 plus 20 over denominator negative 4 end fraction end style over denominator negative 7 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator straight x plus 11 over denominator negative 4 end fraction end style over denominator negative 7 end fraction end cell row blank equals cell fraction numerator straight x plus 11 over denominator open parentheses negative 4 close parentheses times open parentheses negative 7 close parentheses end fraction end cell row blank equals cell fraction numerator straight x plus 11 over denominator 28 end fraction end cell end table end style$

maka diperoleh, $begin mathsize 14px style open parentheses straight f to the power of negative 1 end exponent ring operator straight g to the power of negative 1 end exponent close parentheses open parentheses straight x close parentheses equals fraction numerator straight x plus 11 over denominator 28 end fraction end style$

Kemudian kita tentukan begin mathsize 14px style open parentheses straight g ring operator straight f to the power of negative 1 end exponent close parentheses open parentheses straight x close parentheses end style

$begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell open parentheses straight g ring operator straight f close parentheses open parentheses straight x close parentheses end cell equals cell 4 open parentheses straight f open parentheses straight x close parentheses close parentheses plus 9 end cell row blank equals cell 4 open parentheses 7 straight x minus 5 close parentheses plus 9 end cell row blank equals cell 28 straight x minus 20 plus 9 end cell row blank equals cell 28 straight x minus 11 end cell end table Dijadikan space invers table attributes columnalign right center left columnspacing 2px end attributes row straight y equals cell 28 straight x minus 11 end cell row cell negative 28 straight x end cell equals cell negative straight y minus 11 end cell row straight x equals cell fraction numerator negative straight y minus 11 over denominator negative 28 end fraction end cell row straight x equals cell fraction numerator straight y plus 11 over denominator 28 end fraction end cell row maka colon blank row cell open parentheses straight g ring operator straight f close parentheses to the power of negative 1 end exponent open parentheses straight x close parentheses end cell equals cell fraction numerator straight x plus 11 over denominator 28 end fraction end cell end table end style$

maka diperoleh, $begin mathsize 14px style open parentheses straight g ring operator straight f close parentheses to the power of negative 1 end exponent open parentheses straight x close parentheses equals fraction numerator straight x plus 11 over denominator 28 end fraction end style$ dan hasilnya sama dengan $begin mathsize 14px style open parentheses straight f to the power of negative 1 end exponent ring operator straight g to the power of negative 1 end exponent close parentheses open parentheses straight x close parentheses equals fraction numerator straight x plus 11 over denominator 28 end fraction end style$

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