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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

5 \frac{x - 2}{2}

x^{2} + 3 x - 3

f (x) = 3^{2 x} - π

f (x) = 3 - 4 x - 10

f (x) = arccos (\frac{1}{2})

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