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

The inverse of f(x)=(x^3+3)^{1/3}+3 is cube root of x^3-9x^2+27x-30

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

inverse

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

3 x^{3} - 9 x^{2} + 27 x - 30

Solution steps

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

Replace

x

with

y

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

Solve for

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

3 x^{3} - 9 x^{2} + 27 x - 30

175 - 4 h

f (x) = \frac{1}{1 + 8 x ^{2}}

arccsc (e^{x}) - 2

f (x) = 5^{2} + 600 x - 70

g (x) = 2 + 2 - x

What is the inverse of f(x)=(x^3+3)^{1/3}+3 ?
The inverse of f(x)=(x^3+3)^{1/3}+3 is cube root of x^3-9x^2+27x-30