What is the inverse of f(x)=\sqrt[3]{x^2}-3 ?

The inverse of f(x)=\sqrt[3]{x^2}-3 is sqrt(x^3+9x^2+27x+27),-sqrt(x^3+9x^2+27x+27)

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inverse of f(x)=\sqrt[3]{x^2}-3

inverse

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

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

Solution steps

y = 3 x^{2} - 3

Replace

x

with

y

x = 3 y^{2} - 3

Solve for

y, x = 3 y^{2} - 3

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

\frac{\frac{4}{x} - 12}{x - \frac{1}{9}}

f (x) = 3 x^{2} - 1

y = \frac{x - 1}{x}

f (x) = 7^{\frac{x ^{\frac{1}{3}}}{7}}

f (x) = x^{2} - 3, x \geq 0

What is the inverse of f(x)=\sqrt[3]{x^2}-3 ?
The inverse of f(x)=\sqrt[3]{x^2}-3 is sqrt(x^3+9x^2+27x+27),-sqrt(x^3+9x^2+27x+27)