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Diketahui dan a. ( f − 1 ∘ g − 1 ) ( x ) ?

Diketahui begin mathsize 14px style f open parentheses x close parentheses equals x minus 3 end style dan begin mathsize 14px style g open parentheses x close parentheses equals 2 x plus 4 end style 

a.  ?  

   space undefined

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P. Tessalonika

Master Teacher

Mahasiswa/Alumni Universitas Negeri Medan

Jawaban terverifikasi

Jawaban

diperoleh bahwa .

diperoleh bahwa begin mathsize 14px style 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 x close parentheses equals fraction numerator x plus 2 over denominator 2 end fraction end style.

Pembahasan

Dengan menggunakan rumus invers fungsi Diperoleh Nilai adalah dimasukkan ke Dengan demikian, diperoleh bahwa .

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

begin mathsize 14px style 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 x close parentheses equals fraction numerator x minus 4 over denominator 2 end fraction plus 3 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 x close parentheses equals fraction numerator x minus 4 over denominator 2 end fraction plus 6 over 2 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 x close parentheses equals fraction numerator x plus 2 over denominator 2 end fraction  end style  

Dengan demikian, diperoleh bahwa begin mathsize 14px style 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 x close parentheses equals fraction numerator x plus 2 over denominator 2 end fraction end style.

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