Pertanyaan

Diketahui fungsi f ( x ) = 3 x + 4 dan g ( x ) = 2 x + 1 4 x − 5 ; x  = − 2 1 . Invers ( f ∘ g ) ( x ) adalah ...

Diketahui fungsi $f (x) = 3 x + 4 dan g (x) = \frac{4 x - 5}{2 x + 1}; x \neq = - \frac{1}{2}$ . Invers $(f \circ g) (x)$ adalah ...

$open parentheses f ring operator g close parentheses to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x minus 16 over denominator negative 2 x plus 20 end fraction semicolon space x not equal to 10$
$open parentheses f ring operator g close parentheses to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x plus 11 over denominator negative 2 x plus 20 end fraction semicolon space x not equal to 10$
$open parentheses f ring operator g close parentheses to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x minus 11 over denominator negative 2 x plus 20 end fraction semicolon space x not equal to 10$
$open parentheses f ring operator g close parentheses to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x plus 14 over denominator negative 2 x plus 20 end fraction semicolon space x not equal to 10$
$open parentheses f ring operator g close parentheses to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x minus 14 over denominator negative 2 x plus 20 end fraction semicolon space x not equal to 10$

8 dari 10 siswa nilainya naik

dengan paket belajar pilihan

Habis dalam

Klaim

S. Yoga

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Diketahui: Mencari : Mencari invers dari : Jadi, . Dengan demikian, jawaban yang tepat adalah B.

Diketahui:

$f left parenthesis x right parenthesis equals 3 x plus 4 space dan space g left parenthesis x right parenthesis equals fraction numerator 4 x minus 5 over denominator 2 x plus 1 end fraction semicolon space x not equal to negative 1 half$

Mencari left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis :

$table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses ring operator g open parentheses x close parentheses end cell equals cell f left parenthesis g left parenthesis x right parenthesis right parenthesis end cell row blank equals cell f open parentheses fraction numerator 4 x minus 5 over denominator 2 x plus 1 end fraction close parentheses end cell row blank equals cell 3 open parentheses fraction numerator 4 x minus 5 over denominator 2 x plus 1 end fraction close parentheses plus 4 end cell row blank equals cell fraction numerator 12 x minus 15 over denominator 2 x plus 1 end fraction plus 4 end cell row blank equals cell fraction numerator 12 x minus 15 over denominator 2 x plus 1 end fraction plus fraction numerator 4 open parentheses 2 x plus 1 close parentheses over denominator 2 x plus 1 end fraction end cell row blank equals cell fraction numerator 12 x minus 15 plus 4 open parentheses 2 x plus 1 close parentheses over denominator 2 x plus 1 end fraction end cell row blank equals cell fraction numerator 20 x minus 11 over denominator 2 x plus 1 end fraction end cell end table$

Mencari invers dari left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis :

$table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis x right parenthesis end cell equals cell fraction numerator 20 x minus 11 over denominator 2 x plus 1 end fraction end cell row y equals cell fraction numerator 20 x minus 11 over denominator 2 x plus 1 end fraction end cell row x equals cell fraction numerator 20 y minus 11 over denominator 2 y plus 1 end fraction end cell row cell x open parentheses 2 y plus 1 close parentheses end cell equals cell 20 y minus 11 end cell row cell 2 y x plus x minus x end cell equals cell 20 y minus 11 minus x end cell row cell 2 y x minus 20 y end cell equals cell 20 y minus 11 minus x minus 20 y end cell row cell 2 y x minus 20 y end cell equals cell negative 11 minus x end cell row cell 2 y x minus 20 y end cell equals cell negative 11 minus x end cell row cell fraction numerator 2 y open parentheses x minus 10 close parentheses over denominator 2 open parentheses x minus 10 close parentheses end fraction end cell equals cell negative fraction numerator 11 over denominator 2 open parentheses x minus 10 close parentheses end fraction minus fraction numerator x over denominator 2 open parentheses x minus 10 close parentheses end fraction end cell row y equals cell fraction numerator negative 11 minus x over denominator 2 open parentheses negative 10 plus x close parentheses end fraction comma space x not equal to 10 end cell row cell open parentheses f ring operator g close parentheses left parenthesis x right parenthesis to the power of negative 1 end exponent end cell equals cell fraction numerator negative 11 minus x over denominator 2 open parentheses negative 10 plus x close parentheses end fraction comma space x not equal to 10 end cell row cell open parentheses f ring operator g close parentheses left parenthesis x right parenthesis to the power of negative 1 end exponent end cell equals cell fraction numerator x plus 11 over denominator negative 2 x plus 20 end fraction comma space x not equal to 10 end cell end table$

Jadi, $open parentheses f ring operator g close parentheses to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x plus 11 over denominator negative 2 x plus 20 end fraction semicolon space x not equal to 10$ .

Dengan demikian, jawaban yang tepat adalah B.