Diketahui fungsi dan . Jika fungsi dan , Nilai ...

Diketahui fungsi $begin mathsize 14px style f colon R rightwards arrow R end style$ dan $begin mathsize 14px style g colon R rightwards arrow R end style$ . Jika fungsi $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator 2 x plus 7 over denominator 3 x minus 5 end fraction comma space x not equal to 5 over 3 end style$ dan $begin mathsize 14px style g to the power of negative 1 end exponent left parenthesis x right parenthesis equals 4 x minus 1 end style$ , Nilai fungsi $begin mathsize 14px style left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent open parentheses 2 close parentheses end style$ adalah ...

$begin mathsize 14px style negative 8 end style$
$begin mathsize 14px style 8 end style$
$begin mathsize 14px style 10 end style$
$begin mathsize 14px style 12 end style$
$begin mathsize 14px style 16 end style$

Jawaban:

Pertama akan dicari invers dari fungsi $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator 2 x plus 7 over denominator 3 x minus 5 end fraction comma space x not equal to 5 over 3 end style$

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator 2 x plus 7 over denominator 3 x minus 5 end fraction end cell row cell 3 y x minus 5 y end cell equals cell 2 x plus 7 end cell row cell 3 y x minus 2 x end cell equals cell 5 y plus 7 end cell row cell x open parentheses 3 y minus 2 close parentheses end cell equals cell 5 y plus 7 end cell row x equals cell fraction numerator 5 y plus 7 over denominator 3 y minus 2 end fraction end cell row cell f to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator 5 x plus 7 over denominator 3 x minus 2 end fraction end cell end table end style$

Dengan menggunakan sifat invers akan ditentukan $begin mathsize 14px style left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent open parentheses x close parentheses end style$

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell open parentheses g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses open parentheses x close parentheses end cell row cell space space space space space space space space space space space space space space space space end cell equals cell g to the power of negative 1 end exponent open parentheses fraction numerator 5 x plus 7 over denominator 3 x minus 2 end fraction close parentheses end cell row cell space space space space space space space space space space space space space space space space end cell equals cell 4 open parentheses fraction numerator 5 x plus 7 over denominator 3 x minus 2 end fraction close parentheses minus 1 end cell row cell space space space space space space space space space space space space space space space space end cell equals cell fraction numerator 20 x plus 28 minus 3 x plus 2 over denominator 3 x minus 2 end fraction end cell row cell space space space space space space space space space space space space space space space space end cell equals cell fraction numerator 17 x plus 30 over denominator 3 x minus 2 end fraction end cell end table end style$

Nilai fugsi $begin mathsize 14px style left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent open parentheses 2 close parentheses end style$ adalah

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent open parentheses 2 close parentheses end cell equals cell fraction numerator 17 open parentheses 2 close parentheses plus 30 over denominator 3 open parentheses 2 close parentheses minus 2 end fraction end cell row cell space space space space space space space space space space space space space space space space space end cell equals cell fraction numerator 34 plus 30 over denominator 6 minus 2 end fraction end cell row cell space space space space space space space space space space space space space space space space space end cell equals cell 64 over 4 end cell row cell space space space space space space space space space space space space space space space space space end cell equals 16 end table end style$

Jadi, jawaban yang tepat adalah E

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