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

The inverse of f(x)=(x^3-5)^{1/3}-3 is cube root of x^3+9x^2+27x+32

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

inverse

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{x - 4}{3 x - x ^{2}}

- x^{4} + 11 x^{2} - 18

f (x) = sin (x)

y = x^{3} - 1

f (x) = 7 x^{3} - 4 x - 2

What is the inverse of f(x)=(x^3-5)^{1/3}-3 ?
The inverse of f(x)=(x^3-5)^{1/3}-3 is cube root of x^3+9x^2+27x+32