What is the inverse of sqrt((-3x^2-2)/2) ?

The inverse of sqrt((-3x^2-2)/2) is sqrt(-(2x^2+2)/3),-sqrt(-(2x^2+2)/3)

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inverse of sqrt((-3x^2-2)/2)

inverse

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

- \frac{2 x ^{2} + 2}{3}, - - \frac{2 x ^{2} + 2}{3}

Solution steps

y = \frac{- 3 x ^{2} - 2}{2}

Replace

x

with

y

x = \frac{- 3 y ^{2} - 2}{2}

Solve for

y, x = \frac{- 3 y ^{2} - 2}{2}

- \frac{2 x ^{2} + 2}{3}, - - \frac{2 x ^{2} + 2}{3}

f (x) = \frac{x - 11}{x + 9}

e^{2 - x}

f (x) = - 6 + x

- 4 - j^{6}

f (x) = \frac{1}{3 x + 5}

What is the inverse of sqrt((-3x^2-2)/2) ?
The inverse of sqrt((-3x^2-2)/2) is sqrt(-(2x^2+2)/3),-sqrt(-(2x^2+2)/3)