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Pertanyaan

Fungsi g: R -> R dan h: R -> R dinyatakan oleh g open parentheses x close parentheses equals fraction numerator 2 x minus 1 over denominator x plus 5 end fraction comma x not equal to negative 5 dan h(x) = 4x - 3. Bentuk
komposisi (g o h)(x) adalah ....
 

  1. fraction numerator 8 x plus 7 over denominator 4 x plus 8 end fraction; x not equal to-2
     

  2. fraction numerator 8 x minus 7 over denominator 4 x plus 8 end fraction;x not equal to-2

  3. fraction numerator 8 x minus 5 over denominator 4 x plus 2 end fraction;x not equal tonegative 1 half

  4. fraction numerator 8 x minus 7 over denominator 4 x plus 2 end fraction;x not equal tonegative 1 half

  5. fraction numerator 8 x plus 7 over denominator 4 x plus 2 end fraction;x not equal tonegative 1 half

I. Roy

Master Teacher

Mahasiswa/Alumni Universitas Negeri Surabaya

Jawaban terverifikasi

Jawaban

(goh)(x) = fraction numerator 8 x minus 7 over denominator 4 x plus 2 end fraction comma space x equals space minus 1 half

Pembahasan

(goh)(x) = g(h(x))

 

           = g(4x-3)

 

           = fraction numerator 2 left parenthesis 4 x minus 3 right parenthesis minus 1 over denominator left parenthesis 4 x minus 3 right parenthesis plus 5 end fraction

 

              = fraction numerator 8 x minus 6 minus 1 over denominator 4 x minus 3 plus 5 end fraction

 

              = undefined

Jadi (goh)(x) = fraction numerator 8 x minus 7 over denominator 4 x plus 2 end fraction comma space x equals space minus 1 half

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