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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

(3 x^{2} + 1) - (x + 4)

f (x) = 2^{x}

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

\frac{( e ^{x} - e ^{- x} )}{9}

f (x) = x^{6} - 3 x^{4}

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