Pertanyaan

F u n g s i f : R → R d an g : R → R d i r u m u s kan d e n g an f ( x ) = x x − 1 , x  = 0 , g ( x ) = x + 3 , maka ( g ( f ( x ) ) − 1 ) a d a l ah ...

$F u n g s i f : R \to R d an g : R \to R d i r u m u s kan d e n g an f (x) = \frac{x - 1}{x}, x \neq = 0, g (x) = x + 3, maka (g (f (x))^{- 1}) a d a l ah ...$

$fraction numerator 2 minus 3 x over denominator x minus 1 end fraction$
$fraction numerator 2 plus 3 x over denominator x plus 1 end fraction$
$fraction numerator x minus 2 over denominator x end fraction$
$fraction numerator 4 x minus 1 over denominator x end fraction$
$fraction numerator 1 over denominator 4 x minus 1 end fraction$

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S. Nur

Master Teacher

Jawaban terverifikasi

Pembahasan

$f left parenthesis x right parenthesis equals fraction numerator x minus 1 over denominator x end fraction comma x not equal to 0 space d a n space g left parenthesis x right parenthesis equals x plus 3 g o f left parenthesis x right parenthesis equals fraction numerator x minus 1 over denominator x end fraction plus 3 g o f left parenthesis x right parenthesis equals fraction numerator x minus 1 plus 3 x over denominator x end fraction g o f left parenthesis x right parenthesis equals fraction numerator 4 x minus 1 over denominator x end fraction left parenthesis g left parenthesis f left parenthesis x right parenthesis right parenthesis to the power of negative 1 end exponent right parenthesis equals fraction numerator 4 x minus 1 over denominator x end fraction y equals fraction numerator 4 x minus 1 over denominator x end fraction x y equals 4 x minus 1 x y minus 4 x equals negative 1 x left parenthesis y minus 4 right parenthesis equals negative 1 x equals fraction numerator negative 1 over denominator y minus 4 end fraction y equals fraction numerator 1 over denominator 4 minus x end fraction$

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