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

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

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

inverse

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

\frac{2 - x}{x}, - \frac{2 - x}{x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{2 - x}{x}, - \frac{2 - x}{x}

f (x) = x^{4} + 12 x^{2} + 36

\frac{x - 1}{1 + x ^{2}}

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

2 - (x - 1)

f (x) = lo g_{6} (x + 5)

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