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Diketahui f ( x ) = 2 x + 5 dan g − 1 ( x ) = x + 4 x − 1 ​ , x  = − 4 maka ( f o g ) − 1 ( x ) = ....

Diketahui dan , maka ....

  1. fraction numerator x minus 7 over denominator x plus 3 end fraction semicolon space x not equal to negative 3

  2. fraction numerator x minus 7 over denominator x minus 3 end fraction semicolon space x not equal to 3

  3. fraction numerator x plus 7 over denominator x plus 3 end fraction semicolon space x not equal to negative 3

  4. fraction numerator 7 minus x over denominator x plus 3 end fraction semicolon space x not equal to negative 3

  5. fraction numerator 7 minus x over denominator x minus 3 end fraction semicolon space x not equal to 3

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N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Jawaban

nilai .

nilai left parenthesis f o g right parenthesis to the power of negative 1 end exponent space left parenthesis x right parenthesis equals fraction numerator x minus 7 over denominator x plus 3 end fraction.

Pembahasan

Diketahui: Nilai sama dengan . Untuk itu, tentukan terlebih dahulu (x), yaitu: Memisalkan fungsi dengan y menjadi Maka nilai adalah Jadi, nilai .

Diketahui:

g to the power of negative 1 end exponent space left parenthesis x right parenthesis equals fraction numerator x minus 1 over denominator x plus 4 end fraction

f left parenthesis x right parenthesis equals 2 x plus 5

 

Nilai left parenthesis f space o space g right parenthesis to the power of negative 1 end exponent space left parenthesis x right parenthesis sama dengan left parenthesis g to the power of negative 1 end exponent space o space f to the power of negative 1 end exponent space right parenthesis left parenthesis x right parenthesis.

Untuk itu, tentukan terlebih dahulu f to the power of negative 1 end exponent(x), yaitu:

f left parenthesis x right parenthesis equals 2 x plus 5

Memisalkan fungsi dengan y menjadi

2 x plus 5 equals y

2 x equals y minus 5

x equals fraction numerator y minus 5 over denominator 2 end fraction

f to the power of negative 1 end exponent space left parenthesis x right parenthesis equals fraction numerator x minus 5 over denominator 2 end fraction

Maka nilai left parenthesis g to the power of negative 1 end exponent o space f to the power of negative 1 end exponent space right parenthesis left parenthesis x right parenthesis adalah

left parenthesis g to the power of negative 1 end exponent space o space f to the power of negative 1 end exponent right parenthesis space left parenthesis x right parenthesis equals g to the power of negative 1 end exponent space left parenthesis f to the power of negative 1 end exponent space left parenthesis x right parenthesis right parenthesis

equals g to the power of negative 1 end exponent space left parenthesis fraction numerator x minus 5 over denominator 2 end fraction right parenthesis

equals fraction numerator left parenthesis fraction numerator x minus 5 over denominator 2 end fraction right parenthesis minus 1 over denominator left parenthesis fraction numerator x minus 5 over denominator 2 end fraction right parenthesis space plus space 4 end fraction

equals fraction numerator fraction numerator x minus 5 minus 2 over denominator 2 end fraction over denominator fraction numerator x minus 5 plus 8 over denominator 2 end fraction end fraction

equals fraction numerator x minus 5 minus 2 over denominator x minus 5 plus 8 end fraction

equals fraction numerator x minus 7 over denominator x plus 3 end fraction

Jadi, nilai left parenthesis f o g right parenthesis to the power of negative 1 end exponent space left parenthesis x right parenthesis equals fraction numerator x minus 7 over denominator x plus 3 end fraction.

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