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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = - \frac{1}{2} ∣ x - 6 ∣ + 1

f (x) = 4 x

f (x) = \frac{2 x + 5}{x - 3}

f (x) = \frac{5}{x ^{2}}

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