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f ( x ) = x − 4 , g ( x ) = x − 5 11 x + 4 ​ , x  = 5 Tentukan ( f ∘ g ) − 1 ( x ) .

 

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L. Marlina

Master Teacher

Mahasiswa/Alumni Institut Teknologi Bandung

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.

 open parentheses f ring operator g close parentheses to the power of negative 1 end exponent equals fraction numerator 5 x plus 24 over denominator x minus 7 end fraction.

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sekarang kita akan mencari invers fungsinya, Jadi, .

open parentheses f ring operator g close parentheses open parentheses x close parentheses equals f open parentheses g open parentheses x close parentheses close parentheses space space space space space space space space space space space space space equals f open parentheses fraction numerator 11 x plus 4 over denominator x minus 5 end fraction close parentheses space space space space space space space space space space space space space equals fraction numerator 11 x plus 4 over denominator x minus 5 end fraction minus 4 space space space space space space space space space space space space space equals fraction numerator open parentheses 11 x plus 4 close parentheses minus 4 open parentheses x minus 5 close parentheses over denominator x minus 5 end fraction space space space space space space space space space space space space space equals fraction numerator 7 x plus 24 over denominator x minus 5 end fraction 

sekarang kita akan mencari invers fungsinya,

misal space fraction numerator 7 x plus 24 over denominator x minus 5 end fraction equals a space space space space space space space space space space 7 x plus 24 equals a open parentheses x minus 5 close parentheses space space space space space space space space space space 7 x plus 24 equals a x minus 5 a space space space space space space space space space space 7 x minus a x equals negative 5 a minus 24 space space space space space space space space space space open parentheses 7 minus a close parentheses x equals negative 5 a minus 24 space space space space space space space space space space space space space space space space space space space space x equals fraction numerator negative 5 a minus 24 over denominator 7 minus a end fraction space space space space space space space space space space space space space space space space space space space space x equals fraction numerator 5 a plus 24 over denominator a minus 7 end fraction space space space space space space space space space space space space rightwards arrow space space space space space open parentheses f ring operator g close parentheses to the power of negative 1 end exponent equals fraction numerator 5 x plus 24 over denominator x minus 7 end fraction 

Jadi, open parentheses f ring operator g close parentheses to the power of negative 1 end exponent equals fraction numerator 5 x plus 24 over denominator x minus 7 end fraction.

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