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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

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

f (x) = - 3 x^{2} + 3, x \geq 0

f (x) = \frac{1}{3} lo g_{10} (2 x)

f (x) = \frac{- 12 - 7 x}{4}

f (x) = 2^{x - 1} + 2

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