Pertanyaan

I n v ers d a r i f u n g s i f ( x ) = x − 1 x + 2 , x  = − 1 , a d a l ah ...

$I n v ers d a r i f u n g s i f (x) = \frac{x + 2}{x - 1}, x \neq = - 1, a d a l ah ...$

$f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x plus 2 over denominator x minus 1 end fraction semicolon x not equal to 1$
$f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x minus 2 over denominator x minus 1 end fraction semicolon x not equal to 1$
$f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x plus 2 over denominator x plus 1 end fraction semicolon x not equal to negative 1$
$f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x plus 1 over denominator x plus 2 end fraction semicolon x not equal to negative 2$
$f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x minus 1 over denominator x plus 2 end fraction semicolon x not equal to negative 2$

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$f left parenthesis x right parenthesis equals fraction numerator a x plus b over denominator c x plus d end fraction rightwards arrow f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator negative d x plus b over denominator c x plus left parenthesis negative a right parenthesis end fraction p e r h a t i k a n space l e t a k space a space d a n space d space d i t u k a r space d a n space tan d a n y a space b e r u b a h space m e n j a d i space n e g a t i f space left parenthesis negative right parenthesis space comma space m a k a f left parenthesis x right parenthesis equals fraction numerator x plus 2 over denominator x minus 1 end fraction rightwards arrow f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator x plus 2 over denominator x minus 1 end fraction$