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Diketahui fungsi f(x) = 3x + 4 dan g(x)=2x+14x−5​,x=−21​ invers (f o g) (x) adalah  ....

Pertanyaan

Diketahui fungsi f(x) = 3x + 4 dan begin mathsize 14px style g left parenthesis x right parenthesis equals fraction numerator 4 x minus 5 over denominator 2 x plus 1 end fraction comma space x not equal to negative 1 half end style invers (f o g) (x) adalah  ....

  1. begin mathsize 14px style left parenthesis f o g right parenthesis to the power of negative 1 end exponent equals fraction numerator x minus 14 over denominator negative 2 x plus 20 end fraction comma space x not equal to 10 end style

  2. begin mathsize 14px style left parenthesis f o g right parenthesis to the power of negative 1 end exponent equals fraction numerator x minus 11 over denominator negative 2 x plus 20 end fraction comma space x not equal to 10 end style

  3. begin mathsize 14px style left parenthesis f o g right parenthesis to the power of negative 1 end exponent equals fraction numerator x minus 16 over denominator negative 2 x plus 20 end fraction comma space x not equal to 10 end style

  4. begin mathsize 14px style left parenthesis f o g right parenthesis to the power of negative 1 end exponent equals fraction numerator x plus 11 over denominator negative 2 x plus 20 end fraction comma space x not equal to 10 end style

  5. begin mathsize 14px style left parenthesis f o g right parenthesis to the power of negative 1 end exponent equals fraction numerator x plus 14 over denominator negative 2 x plus 20 end fraction comma space x not equal to 10 end style

Pembahasan:

begin mathsize 14px style open parentheses straight f ring operator straight g close parentheses open parentheses straight x close parentheses equals straight f open parentheses straight g open parentheses straight x close parentheses close parentheses  straight f open parentheses fraction numerator 4 straight x minus 5 over denominator 2 straight x plus 1 end fraction close parentheses equals 3 straight space open parentheses fraction numerator 4 straight x minus 5 over denominator 2 straight x plus 1 end fraction close parentheses plus 4  straight f open parentheses fraction numerator 4 straight x minus 5 over denominator 2 straight x plus 1 end fraction close parentheses equals fraction numerator 12 straight x minus 15 over denominator 2 straight x plus 1 end fraction plus 4  straight f open parentheses fraction numerator 4 straight x minus 5 over denominator 2 straight x plus 1 end fraction close parentheses equals fraction numerator 12 straight x minus 15 plus 4 left parenthesis 2 straight x plus 1 right parenthesis over denominator 2 straight x plus 1 end fraction  straight f open parentheses fraction numerator 4 straight x minus 5 over denominator 2 straight x plus 1 end fraction close parentheses equals fraction numerator 12 straight x minus 15 plus 8 straight x plus 4 over denominator 2 straight x plus 1 end fraction  straight f open parentheses fraction numerator 4 straight x minus 5 over denominator 2 straight x plus 1 end fraction close parentheses equals fraction numerator 20 straight x minus 11 over denominator 2 straight x plus 1 end fraction    Karena space straight f open parentheses fraction numerator 4 straight x minus 5 over denominator 2 straight x plus 1 end fraction close parentheses equals open parentheses straight f ring operator straight g close parentheses open parentheses straight x close parentheses space maka  open parentheses straight f ring operator straight g close parentheses open parentheses straight x close parentheses equals fraction numerator 20 straight x minus 11 over denominator 2 straight x plus 1 end fraction    Untuk space mencari space inversnya comma space misalkan space left parenthesis straight f ring operator straight g right parenthesis left parenthesis straight x right parenthesis equals straight y comma space maka  straight y equals fraction numerator 20 straight x minus 11 over denominator 2 straight x plus 1 end fraction rightwards double arrow straight x equals fraction numerator straight y plus 11 over denominator negative 2 straight y plus 20 end fraction  Maka comma  open parentheses straight f ring operator straight g 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 negative 2 straight x plus 20 end fraction end style

Jawaban terverifikasi

Dijawab oleh:

S. Intan

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 04 Oktober 2021

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