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

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

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

inverse

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

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

f (x) = 12 x^{2} - 14 x

f (x) = (x - 3)^{0.5}

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

f (y) = - x + 10

- λ ln (x)

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