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}

x + 1 - 8

f (x) = \frac{( x + 4 )}{( x + 14 )}

- \frac{3}{x ^{2}}

f (x) = \frac{( 13 x + 2 )}{( 3 x - 7 )}

7^{x - 2}

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))