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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

(0.01, 0.2), (0.025, 0.6)

1 - \frac{8 y + 6 x}{2} = 4

9 x + 48

f (x) = 3 tan (\frac{2}{3} x)

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