What is the inverse of \sqrt[3]{1-x^2} ?

The inverse of \sqrt[3]{1-x^2} is sqrt(-x^3+1),-sqrt(-x^3+1)

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inverse of \sqrt[3]{1-x^2}

inverse

3 1 - x^{2}

- x^{3} + 1, - - x^{3} + 1

Solution steps

y = 3 1 - x^{2}

Replace

x

with

y

x = 3 1 - y^{2}

Solve for

y, x = 3 1 - y^{2}

- x^{3} + 1, - - x^{3} + 1

\frac{2}{3} - 4 x - 9 + 9

2 + 4 - x, x \leq 4

\frac{1 + x}{1 - x}

f (x) = \frac{1}{4} - (\frac{x - 10}{x + 6})

f (x) = (x - 1)^{2}, [1, \infty]

What is the inverse of \sqrt[3]{1-x^2} ?
The inverse of \sqrt[3]{1-x^2} is sqrt(-x^3+1),-sqrt(-x^3+1)