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

The inverse of f(x)=(t^2)/(sqrt(4-t^2)) is

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inverse of f(x)=(t^2)/(sqrt(4-t^2))

inverse

f (x) = \frac{t ^{2}}{4 - t ^{2}}

\frac{- t ^{2} + t t ^{2} + 16}{2}, - \frac{- t ^{2} + t t ^{2} + 16}{2}, \frac{- t ^{2} - t t ^{2} + 16}{2}, - \frac{- t ^{2} - t t ^{2} + 16}{2}

Solution steps

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

Replace

t

with

x

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

Solve for

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

\frac{- t ^{2} + t t ^{2} + 16}{2}, - \frac{- t ^{2} + t t ^{2} + 16}{2}, \frac{- t ^{2} - t t ^{2} + 16}{2}, - \frac{- t ^{2} - t t ^{2} + 16}{2}

f (x) = 12 - 3 x

f (x) = 13 - x^{2}, x \geq 0

f (x) = \frac{x + 1}{5 x + 1}

f (x) = x^{2} + 4 x - 21

f (x) = \frac{x - 4}{2 x + 5}

What is the inverse of f(x)=(t^2)/(sqrt(4-t^2)) ?
The inverse of f(x)=(t^2)/(sqrt(4-t^2)) is