What is the inverse of x/(sqrt(4-x^2)) ?

The inverse of x/(sqrt(4-x^2)) is (2xsqrt(x^2+1))/(x^2+1),-(2xsqrt(x^2+1))/(x^2+1)

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inverse of x/(sqrt(4-x^2))

inverse

\frac{x}{4 - x ^{2}}

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

Solution steps

y = \frac{x}{4 - x ^{2}}

Replace

x

with

y

x = \frac{y}{4 - y ^{2}}

Solve for

y, x = \frac{y}{4 - y ^{2}}

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

g (x) = x^{3} - 5

- \frac{19}{( 3 + t ) ^{2}}

x - 2 y = 3

\frac{5 x}{6 x - 1}

\frac{x ^{3}}{1 - 2 x ^{3}}

What is the inverse of x/(sqrt(4-x^2)) ?
The inverse of x/(sqrt(4-x^2)) is (2xsqrt(x^2+1))/(x^2+1),-(2xsqrt(x^2+1))/(x^2+1)