What is the inverse of 1/3 (y^2+2)^{3/2} ?

The inverse of 1/3 (y^2+2)^{3/2} is sqrt(9^{1/3)y^{2/3}-2},-sqrt(9^{1/3)y^{2/3}-2}

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inverse of 1/3 (y^2+2)^{3/2}

inverse

\frac{1}{3} (y^{2} + 2)^{\frac{3}{2}}

9^{\frac{1}{3}} y^{\frac{2}{3}} - 2, - 9^{\frac{1}{3}} y^{\frac{2}{3}} - 2

Solution steps

y = \frac{1}{3} (y^{2} + 2)^{\frac{3}{2}}

Replace

y

with

y

y = \frac{1}{3} (y^{2} + 2)^{\frac{3}{2}}

Solve for

y, y = \frac{1}{3} (y^{2} + 2)^{\frac{3}{2}}

9^{\frac{1}{3}} y^{\frac{2}{3}} - 2, - 9^{\frac{1}{3}} y^{\frac{2}{3}} - 2

g (x) = \frac{8 x + 7}{5}

tan (x - \frac{3 π}{2})

f (x) = 2.17

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

What is the inverse of 1/3 (y^2+2)^{3/2} ?
The inverse of 1/3 (y^2+2)^{3/2} is sqrt(9^{1/3)y^{2/3}-2},-sqrt(9^{1/3)y^{2/3}-2}