Pertanyaan

Diketahui fungsi f : R → R dan g : R → R , dengan f ( x ) = x − 6 x + 4 , x  = 6 dan g ( x ) = 2 x − 1 , tentukan fungsi ( f ∘ g ) − 1 ( x ) !

Diketahui fungsi $f : R \to R$ dan $g : R \to R$ , dengan $f (x) = \frac{x + 4}{x - 6}, x \neq = 6$ dan $g (x) = 2 x - 1$ , tentukan fungsi $(f \circ g)^{- 1} (x)$ !

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Habis dalam

Klaim

A. Fatta

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

Jawaban terverifikasi

Jawaban

fungsi adalah .

fungsi

open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses

adalah $fraction numerator 7 x plus 3 over denominator 2 open parentheses x minus 1 close parentheses end fraction$ .

Pembahasan

Dalam fungsi invers terdapat rumus khusus seperti berikut. Sehingga Dengan demikian, adalah sebagai berikut. Jadi, fungsi adalah .

Dalam fungsi invers terdapat rumus khusus seperti berikut.

$f open parentheses x close parentheses equals a x plus b space semicolon space a not equal to 0 rightwards double arrow f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x minus b over denominator a end fraction space semicolon space a not equal to 0 f open parentheses x close parentheses equals fraction numerator a x plus b over denominator c x plus d end fraction space semicolon space a not equal to fraction numerator negative d over denominator c end fraction rightwards double arrow f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative d x plus b over denominator c x minus a end fraction space semicolon space a not equal to a over c$

Sehingga

$f open parentheses x close parentheses equals fraction numerator x plus 4 over denominator x minus 6 end fraction comma space x not equal to 6 rightwards double arrow f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator 6 x plus 4 over denominator x minus 1 end fraction comma space x not equal to 1 g open parentheses x close parentheses equals 2 x minus 1 rightwards double arrow g to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x plus 1 over denominator 2 end fraction$

Dengan demikian, open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses adalah sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell open parentheses g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses open parentheses x close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses f to the power of negative 1 end exponent open parentheses x close parentheses close parentheses end cell row blank equals cell g to the power of negative 1 end exponent open parentheses fraction numerator 6 x plus 4 over denominator x minus 1 end fraction close parentheses end cell row blank equals cell fraction numerator fraction numerator 6 x plus 4 over denominator x minus 1 end fraction plus 1 over denominator 2 end fraction end cell row blank equals cell fraction numerator fraction numerator 6 x plus 4 over denominator x minus 1 end fraction plus fraction numerator x minus 1 over denominator x minus 1 end fraction over denominator 2 end fraction end cell row blank equals cell fraction numerator 7 x plus 3 over denominator 2 open parentheses x minus 1 close parentheses end fraction end cell end table$

Jadi, fungsi open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses adalah $fraction numerator 7 x plus 3 over denominator 2 open parentheses x minus 1 close parentheses end fraction$ .