Pertanyaan

Diketahui g ( x ) = 2 x + 2 dan ( f ∘ g ) ( x ) = 2 x − 1 3 x + 1 , x  = 2 1 . Nilai dari f ( 2 ) = ...

Diketahui $g (x) = 2 x + 2$ dan $(f \circ g) (x) = \frac{3 x + 1}{2 x - 1}, x \neq = \frac{1}{2}$ . Nilai dari $f (2) = ...$

B. Hary

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah A

Pembahasan

Pertama kita menetukan fungsi : misal: maka: sehingga: Sehingga nilai adalah: Jadi, jawaban yang tepat adalah A

Pertama kita menetukan fungsi :

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g close parentheses open parentheses x close parentheses end cell equals cell fraction numerator 3 x plus 1 over denominator 2 x minus 1 end fraction end cell row cell f open parentheses g open parentheses x close parentheses close parentheses end cell equals cell fraction numerator 3 x plus 1 over denominator 2 x minus 1 end fraction end cell row cell f left parenthesis 2 x plus 2 right parenthesis end cell equals cell fraction numerator 3 x plus 1 over denominator 2 x minus 1 end fraction end cell end table end style$

misal: maka:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell 2 x plus 2 end cell row cell 2 x end cell equals cell y minus 2 end cell row x equals cell fraction numerator y minus 2 over denominator 2 end fraction end cell end table end style$

sehingga:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis 2 x plus 2 right parenthesis end cell equals cell fraction numerator 3 x plus 1 over denominator 2 x minus 1 end fraction end cell row cell f open parentheses y close parentheses end cell equals cell fraction numerator 3 open parentheses begin display style fraction numerator y minus 2 over denominator 2 end fraction end style close parentheses plus 1 over denominator 2 open parentheses fraction numerator y minus 2 over denominator 2 end fraction close parentheses minus 1 end fraction end cell row cell f left parenthesis y right parenthesis end cell equals cell fraction numerator begin display style fraction numerator 3 y minus 6 plus 2 over denominator 2 end fraction end style over denominator begin display style fraction numerator 2 y minus 4 minus 2 over denominator 2 end fraction end style end fraction end cell row cell f left parenthesis y right parenthesis end cell equals cell fraction numerator 3 y minus 4 over denominator 2 y minus 6 end fraction end cell row cell f open parentheses x close parentheses end cell equals cell fraction numerator 3 x minus 4 over denominator 2 x minus 6 end fraction end cell end table end style$

Sehingga nilai adalah:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis x right parenthesis end cell equals cell fraction numerator 3 x minus 4 over denominator 2 x minus 6 end fraction end cell row cell f open parentheses 2 close parentheses end cell equals cell fraction numerator 3 open parentheses 2 close parentheses minus 4 over denominator 2 open parentheses 2 close parentheses minus 6 end fraction end cell row cell f open parentheses 2 close parentheses end cell equals cell fraction numerator 6 minus 4 over denominator 4 minus 6 end fraction end cell row cell f open parentheses 2 close parentheses end cell equals cell fraction numerator 2 over denominator negative 2 end fraction end cell row cell f open parentheses 2 close parentheses end cell equals cell negative 1 end cell end table end style$

Jadi, jawaban yang tepat adalah A