What is the inverse of f(x)=(3x^3+2)^{1/5} ?

The inverse of f(x)=(3x^3+2)^{1/5} is cube root of (x^5-2)/3

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inverse of f(x)=(3x^3+2)^{1/5}

inverse

f (x) = (3 x^{3} + 2)^{\frac{1}{5}}

Solution steps

y = (3 x^{3} + 2)^{\frac{1}{5}}

Replace

x

with

y

x = (3 y^{3} + 2)^{\frac{1}{5}}

Solve for

y, x = (3 y^{3} + 2)^{\frac{1}{5}}

f (x) = \frac{2 x + 1}{x - 2}

f (x) = \frac{( 3 - 2 x )}{3 x + 2}

f (x) = ln (x + x^{2} + 1)

y = x + 3

What is the inverse of f(x)=(3x^3+2)^{1/5} ?
The inverse of f(x)=(3x^3+2)^{1/5} is cube root of (x^5-2)/3