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}

\frac{1}{4} (t^{4} + 8)

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

f (x) = a 3 - x

f (x) = f (x) = \frac{1}{x + 5} - 1

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

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)