Pertanyaan

Jika f ( x ) = x 1 dan g ( x ) = 2 x − 1 , tentukan ( f ∘ g ) − 1 ( x ) .

Jika $f (x) = \frac{1}{x}$ dan $g (x) = 2 x - 1$ , tentukan $(f \circ g)^{- 1} (x)$ .

F. Ayudhita

Master Teacher

Jawaban terverifikasi

Jawaban

$begin mathsize 14px style bold left parenthesis bold f bold ring operator bold g bold right parenthesis to the power of bold minus bold 1 end exponent begin bold style left parenthesis x right parenthesis end style bold equals fraction numerator bold x bold plus bold 1 over denominator bold 2 bold x end fraction end style$

Pembahasan

Ingat! artinya dimasukkan ke Berdasarkan konsep tersebut diperoleh Akan dicari invers dari Dengan demikian , atau Jadi,

Ingat!

artinya dimasukkan ke

Berdasarkan konsep tersebut diperoleh

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

Akan dicari invers dari

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator 1 over denominator 2 x minus 1 end fraction end cell row cell 2 x minus 1 end cell equals cell 1 over y end cell row cell 2 x end cell equals cell 1 over y plus 1 end cell row x equals cell fraction numerator 1 over denominator 2 y end fraction plus 1 half end cell row x equals cell fraction numerator 1 over denominator 2 y end fraction plus fraction numerator y over denominator 2 y end fraction end cell row x equals cell fraction numerator 1 plus y over denominator 2 y end fraction end cell row x equals cell fraction numerator y plus 1 over denominator 2 y end fraction end cell end table end style$

Dengan demikian $begin mathsize 14px style open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses y close parentheses equals fraction numerator y plus 1 over denominator 2 y end fraction end style$ , atau $begin mathsize 14px style open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x plus 1 over denominator 2 x end fraction end style$

Jadi, $begin mathsize 14px style bold left parenthesis bold f bold ring operator bold g bold right parenthesis to the power of bold minus bold 1 end exponent begin bold style left parenthesis x right parenthesis end style bold equals fraction numerator bold x bold plus bold 1 over denominator bold 2 bold x end fraction end style$