What is the inverse of f(x)=(x^3+4)^{1/5}-3 ?

The inverse of f(x)=(x^3+4)^{1/5}-3 is cube root of x^5+15x^4+90x^3+270x^2+405x+239

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inverse of f(x)=(x^3+4)^{1/5}-3

inverse

f (x) = (x^{3} + 4)^{\frac{1}{5}} - 3

3 x^{5} + 15 x^{4} + 90 x^{3} + 270 x^{2} + 405 x + 239

Solution steps

y = (x^{3} + 4)^{\frac{1}{5}} - 3

Replace

x

with

y

x = (y^{3} + 4)^{\frac{1}{5}} - 3

Solve for

y, x = (y^{3} + 4)^{\frac{1}{5}} - 3

3 x^{5} + 15 x^{4} + 90 x^{3} + 270 x^{2} + 405 x + 239

f (x) = \frac{( 2 x - 1 )}{( 6 - 5 x )}

\frac{d}{d} (\frac{x}{x - 2})

f (x) = 4^{2}

f (x) = 3 + 4 + 6 x

f (x) = \frac{3}{5 x - 10}

What is the inverse of f(x)=(x^3+4)^{1/5}-3 ?
The inverse of f(x)=(x^3+4)^{1/5}-3 is cube root of x^5+15x^4+90x^3+270x^2+405x+239