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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = 2 x^{2} + 32 x

f (x) = 2 + 4 - x^{2}

f (x) = \frac{2}{3 x + 4}

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