Pertanyaan

Diketahui dan Jika dan maka adalah....

Diketahui $f open parentheses x close parentheses equals fraction numerator x over denominator 2 x minus 3 end fraction$ dan $g open parentheses x close parentheses equals fraction numerator 2 minus x over denominator 3 x end fraction.$ Jika D subscript g equals open curly brackets x vertical line x not equal to 0 comma space x element of straight real numbers close curly brackets dan D subscript f equals open curly brackets x vertical line x not equal to 3 over 2 comma space x element of straight real numbers close curly brackets comma maka open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses adalah ....

$left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 4 x plus 2 over denominator 11 x plus 1 end fraction comma space x not equal to negative 1 over 11$
$left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 4 x minus 2 over denominator 11 x minus 1 end fraction comma space x not equal to 1 over 11$
$left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 4 x plus 1 over denominator 11 x plus 2 end fraction comma space x not equal to negative 2 over 11$
$left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 4 x minus 1 over denominator 11 x minus 2 end fraction comma space x not equal to 2 over 11$
$left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 4 over denominator 11 x end fraction plus 2 comma space x not equal to 0$

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Diketahui bahwa dan serta dan , maka dapat diperiksa bahwa terdefinisi dan memiliki invers. Ingat bahwa invers dari fungsi f ( x ) = c x + d a x + b adalah f ( x ) − 1 = c x − a − d x + b . Akan dicariinvers dari dan g ( x ) terlebih dahulu sebagai berikut. Selanjutnya, Ingat kembali bahwa ( f ∘ g ) − 1 ( x ) = ( g − 1 ∘ f − 1 ) ( x ) sehingga ( f ∘ g ) − 1 ( x ) dapat ditentukan sebagai berikut. Dengan demikian, . Jadi, jawaban yang tepat adalah B.

Diketahui bahwa $f open parentheses x close parentheses equals fraction numerator x over denominator 2 x minus 3 end fraction$ dan $g open parentheses x close parentheses equals fraction numerator 2 minus x over denominator 3 x end fraction$ serta D subscript g equals open curly brackets x vertical line x not equal to 0 comma space x element of straight real numbers close curly brackets dan D subscript f equals open curly brackets x vertical line x not equal to 3 over 2 comma x element of straight real numbers close curly brackets , maka dapat diperiksa bahwa open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses terdefinisi dan memiliki invers.

Ingat bahwa invers dari fungsi $f (x) = \frac{a x + b}{c x + d}$ adalah $f (x)^{- 1} = \frac{- d x + b}{c x - a}$ .

Akan dicari invers dari f open parentheses x close parentheses dan $g (x)$ terlebih dahulu sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell equals cell fraction numerator x over denominator 2 x minus 3 end fraction rightwards arrow f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator 3 x over denominator 2 x minus 1 end fraction end cell row blank blank blank row cell g left parenthesis x right parenthesis end cell equals cell fraction numerator 2 minus x over denominator 3 x end fraction equals fraction numerator negative x plus 2 over denominator 3 x end fraction rightwards arrow g to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 2 over denominator 3 x plus 1 end fraction end cell end table$

Selanjutnya, Ingat kembali bahwa $(f \circ g)^{- 1} (x) = (g^{- 1} \circ f^{- 1}) (x)$ sehingga $(f \circ g)^{- 1} (x)$ dapat ditentukan sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell open parentheses g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses open parentheses x close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses f to the power of negative 1 end exponent open parentheses x close parentheses close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses fraction numerator 3 x over denominator 2 x minus 1 end fraction close parentheses end cell row blank equals cell fraction numerator 2 over denominator 3 open parentheses begin display style fraction numerator 3 x over denominator 2 x minus 1 end fraction end style close parentheses plus 1 end fraction end cell row blank equals cell fraction numerator 2 over denominator begin display style fraction numerator 9 x over denominator 2 x minus 1 end fraction end style plus 1 end fraction end cell row blank equals cell fraction numerator 2 over denominator begin display style fraction numerator 9 x over denominator 2 x minus 1 end fraction end style plus begin display style fraction numerator 2 x minus 1 over denominator 2 x minus 1 end fraction end style end fraction end cell row blank equals cell fraction numerator 2 over denominator begin display style fraction numerator 11 x minus 1 over denominator 2 x minus 1 end fraction end style end fraction end cell row blank equals cell fraction numerator 2 open parentheses 2 x minus 1 close parentheses over denominator 11 x minus 1 end fraction end cell row blank equals cell fraction numerator 4 x minus 2 over denominator 11 x minus 1 end fraction end cell end table$

Dengan demikian, $left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 4 x minus 2 over denominator 11 x minus 1 end fraction comma space x not equal to 1 over 11$ .

Jadi, jawaban yang tepat adalah B.