What is the inverse of y=(-3x^2-1)/(x^2-1) ?

The inverse of y=(-3x^2-1)/(x^2-1) is sqrt((-1+x)/(x+3)),-sqrt((-1+x)/(x+3))

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inverse of y=(-3x^2-1)/(x^2-1)

inverse

y = \frac{- 3 x ^{2} - 1}{x ^{2} - 1}

\frac{- 1 + x}{x + 3}, - \frac{- 1 + x}{x + 3}

Solution steps

y = \frac{- 3 x ^{2} - 1}{x ^{2} - 1}

Replace

x

with

y

x = \frac{- 3 y ^{2} - 1}{y ^{2} - 1}

Solve for

y, x = \frac{- 3 y ^{2} - 1}{y ^{2} - 1}

\frac{- 1 + x}{x + 3}, - \frac{- 1 + x}{x + 3}

f (x) = 7 x - 28

f (x) = \frac{x - 5}{11}

f (x) = \frac{5 x - 3}{x + 1}

f (x) = - x - 1 + 2

\frac{2 x - 5}{x + 7}

What is the inverse of y=(-3x^2-1)/(x^2-1) ?
The inverse of y=(-3x^2-1)/(x^2-1) is sqrt((-1+x)/(x+3)),-sqrt((-1+x)/(x+3))