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

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

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

inverse

f (x) = \frac{3}{x ^{2} - 1}

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = 13 x + 3

f (x) = 13 x + 8

f (x) = \frac{1}{6} x - \frac{1}{3}

g (x) = \frac{x + 5}{x - 4}

f (x) = (7 x - 8)^{3}

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