What is the inverse of f(x)=((x^2-1))/((x^2-4)) ?

The inverse of f(x)=((x^2-1))/((x^2-4)) is sqrt((-1+4x)/(x-1)),-sqrt((-1+4x)/(x-1))

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inverse of f(x)=((x^2-1))/((x^2-4))

inverse

f (x) = \frac{( x ^{2} - 1 )}{( x ^{2} - 4 )}

\frac{- 1 + 4 x}{x - 1}, - \frac{- 1 + 4 x}{x - 1}

Solution steps

y = \frac{x ^{2} - 1}{x ^{2} - 4}

Replace

x

with

y

x = \frac{y ^{2} - 1}{y ^{2} - 4}

Solve for

y, x = \frac{y ^{2} - 1}{y ^{2} - 4}

\frac{- 1 + 4 x}{x - 1}, - \frac{- 1 + 4 x}{x - 1}

6 - lo g_{5} (2 x - 7)

f (x) = \frac{( 2 x + 3 )}{( x + 4 )}

1 - x^{2}, - 1 \leq x \leq 1

\frac{1}{( z - 1 ) ^{3}}

cos (2 θ)

What is the inverse of f(x)=((x^2-1))/((x^2-4)) ?
The inverse of f(x)=((x^2-1))/((x^2-4)) is sqrt((-1+4x)/(x-1)),-sqrt((-1+4x)/(x-1))