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Diketahui g ( x ) = x − 2 x + 1 ​ , x  = 2 dan h ( x ) = x a x + 3 ​ , x  = 0 . Tentukan nilai a yang memenuhi ( h ∘ g − 1 ) ( 4 ) = 6 .

Diketahui  dan . Tentukan nilai  yang memenuhi 

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L. Rante

Master Teacher

Mahasiswa/Alumni Universitas Negeri Makassar

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nilai yang memenuhi adalah .

nilai begin mathsize 14px style straight a end style yang memenuhi adalah begin mathsize 14px style straight a equals 5 end style

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Misalkan , maka Untuk , . dan . Maka Jadi nilai yang memenuhi adalah .

begin mathsize 14px style g open parentheses x close parentheses equals fraction numerator x plus 1 over denominator x minus 2 end fraction comma space x not equal to 2 end style 

Misalkan begin mathsize 14px style y equals g open parentheses x close parentheses end style, maka 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator x plus 1 over denominator x minus 2 end fraction end cell row cell y open parentheses x minus 2 close parentheses end cell equals cell x plus 1 end cell row cell x y minus 2 y end cell equals cell x plus 1 end cell row cell x y minus x end cell equals cell 2 y plus 1 end cell row cell x open parentheses y minus 1 close parentheses end cell equals cell 2 y plus 1 end cell row x equals cell fraction numerator 2 y plus 1 over denominator y minus 1 end fraction end cell row cell g to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator 2 x plus 1 over denominator x minus 1 end fraction end cell end table end style  

Untuk begin mathsize 14px style x equals 4 comma space diperoleh space end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell g to the power of negative 1 end exponent open parentheses 4 close parentheses end cell equals cell fraction numerator 2 open parentheses 4 close parentheses plus 1 over denominator open parentheses 4 close parentheses minus 1 end fraction end cell row blank equals cell fraction numerator 8 plus 1 over denominator 3 end fraction end cell row blank equals cell 9 over 3 end cell row blank equals 3 end table end style 

begin mathsize 14px style h left parenthesis x right parenthesis equals fraction numerator straight a x plus 3 over denominator x end fraction comma space x not equal to 0 end stylebegin mathsize 14px style g to the power of negative 1 end exponent open parentheses 4 close parentheses equals 3 end style. dan begin mathsize 14px style open parentheses h ring operator g to the power of negative 1 end exponent close parentheses open parentheses 4 close parentheses equals 6 end style. Maka 

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis h ring operator g to the power of negative 1 end exponent right parenthesis left parenthesis 4 right parenthesis end cell equals 6 row cell h open parentheses g to the power of negative 1 end exponent open parentheses 4 close parentheses close parentheses end cell equals 6 row cell h open parentheses 3 close parentheses end cell equals 6 row cell fraction numerator straight a open parentheses 3 close parentheses plus 3 over denominator open parentheses 3 close parentheses end fraction end cell equals 6 row cell fraction numerator 3 straight a plus 3 over denominator 3 end fraction end cell equals 6 row cell 3 straight a plus 3 end cell equals 18 row cell 3 straight a end cell equals 15 row straight a equals cell 15 over 3 end cell row straight a equals 5 end table end style 

Jadi nilai begin mathsize 14px style straight a end style yang memenuhi adalah begin mathsize 14px style straight a equals 5 end style

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Jika f ( x ) = 3 x + 5 dan g ( x ) = 2 x + 3 x − 2 ​ , x  = − 2 3 ​ , hitunglah ( f ∘ g ) − 1 ( 6 ) + ( g ∘ f ) − 1 ( − 1 ) .

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