Pertanyaan

Diketahui : ( g − 1 ∘ f − 1 ) ( x ) f − 1 ( x ) = = x − 2 x 2 x Tentukan : g ( x ) = ... ? Catatan : ( g − 1 ∘ f − 1 ) ( x ) = ( f ∘ g ) − 1 ( x )

Diketahui : $(g^{- 1} \circ f^{- 1}) (x) f^{- 1} (x) = = \frac{x}{x - 2} 2 x$

Tentukan : $g (x) = ... ?$

Catatan : $(g^{- 1} \circ f^{- 1}) (x) = (f \circ g)^{- 1} (x)$

K. Prameswari

Master Teacher

Mahasiswa/Alumni Universitas Gadjah Mada

Jawaban terverifikasi

Jawaban

diperoleh

diperoleh $g open parentheses x close parentheses equals fraction numerator 4 x over denominator x minus 1 end fraction$

Pembahasan

Perhatikan bahwa Misalkan . Maka dapat diperoleh Selanjutnyadimisalkan , maka didapat Jadi, diperoleh

Perhatikan bahwa

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

Misalkan y equals 2 x space left right arrow y over 2 equals x . Maka dapat diperoleh

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator x over denominator x minus 2 end fraction end cell equals cell g to the power of negative 1 end exponent open parentheses 2 x close parentheses end cell row cell fraction numerator begin display style y over 2 end style over denominator begin display style y over 2 end style minus 2 end fraction end cell equals cell g to the power of negative 1 end exponent open parentheses y close parentheses end cell row cell fraction numerator begin display style y over 2 end style over denominator begin display style fraction numerator y minus 4 over denominator 2 end fraction end style end fraction end cell equals cell g to the power of negative 1 end exponent open parentheses y close parentheses end cell row cell fraction numerator y over denominator y minus 4 end fraction end cell equals cell g to the power of negative 1 end exponent open parentheses y close parentheses space left right arrow space g to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x over denominator x minus 4 end fraction end cell end table$

Selanjutnya dimisalkan z equals g to the power of negative 1 end exponent open parentheses x close parentheses , maka didapat

$table attributes columnalign right center left columnspacing 0px end attributes row z equals cell g to the power of negative 1 end exponent open parentheses x close parentheses end cell row z equals cell fraction numerator x over denominator x minus 4 end fraction end cell row cell z open parentheses x minus 4 close parentheses end cell equals x row cell z x minus 4 z end cell equals x row cell z x minus x end cell equals cell 4 z end cell row cell x left parenthesis z minus 1 right parenthesis end cell equals cell 4 z end cell row x equals cell fraction numerator 4 z over denominator z minus 1 end fraction end cell row cell g open parentheses z close parentheses end cell equals cell fraction numerator 4 z over denominator z minus 1 end fraction space left right arrow space g open parentheses x close parentheses equals fraction numerator 4 x over denominator x minus 1 end fraction end cell end table$

Jadi, diperoleh $g open parentheses x close parentheses equals fraction numerator 4 x over denominator x minus 1 end fraction$

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