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Diketahui fungsi g ( x ) = x 2 + 4 x + 5 , x ∈ R . Agar g merupakan fungsi bijektif, tentukan: b.rumus fungsi g − 1 ,

Diketahui fungsi . Agar  merupakan fungsi bijektif, tentukan:
b. rumus fungsi ,

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D. Kamilia

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

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Rumus fungsi invers untuk fungsi kuadrat adalah: Karena sehingga diperoleh: Jadi, rumus adalah:

begin mathsize 14px style g open parentheses x close parentheses equals x squared plus 4 x plus 5 comma space x element of straight real numbers end style

Rumus fungsi invers untuk fungsi kuadrat adalah:

begin mathsize 14px style f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative b minus square root of b squared minus 4 a open parentheses c minus x close parentheses end root over denominator 2 a end fraction comma space jika space x less or equal than negative fraction numerator b over denominator 2 a end fraction f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative b plus square root of b squared minus 4 a open parentheses c minus x close parentheses end root over denominator 2 a end fraction comma space jika space x greater or equal than negative fraction numerator b over denominator 2 a end fraction end style

Karena begin mathsize 14px style a equals 1 comma space b equals 4 comma space dan space c equals 5 end style sehingga diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell g to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator negative open parentheses 4 close parentheses plus-or-minus square root of open parentheses 4 close parentheses squared minus 4 open parentheses 1 close parentheses open parentheses 5 minus x close parentheses end root over denominator 2 open parentheses 1 close parentheses end fraction end cell row blank equals cell fraction numerator negative 4 plus-or-minus square root of 16 minus 20 plus 4 x end root over denominator 2 end fraction end cell row blank equals cell fraction numerator negative 4 plus-or-minus square root of negative 4 plus 4 x end root over denominator 2 end fraction end cell row blank equals cell fraction numerator negative 4 plus-or-minus 2 square root of x minus 1 end root over denominator 2 end fraction end cell row blank equals cell negative 2 plus-or-minus square root of x minus 1 end root end cell end table end style

Jadi, rumus begin mathsize 14px style g to the power of negative 1 end exponent end style adalah:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell g to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell negative 2 minus square root of x minus 1 end root comma space jika space x less or equal than negative 2 end cell row cell g to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell negative 2 plus square root of x minus 1 end root comma space jika space x greater or equal than negative 2 end cell end table end style

 

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Diketahui fungsi f : R → R dan g : R → R dirumuskan dengan f ( x ) = x x − 1 ​ , untuk x  = 0 dan g ( x ) = x + 3 . Tentukanlah ( g ∘ f ( x ) ) − 1 .

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