What is the inverse of f(x)=\sqrt[3]{x^2}-1 ?

The inverse of f(x)=\sqrt[3]{x^2}-1 is sqrt(x^3+3x^2+3x+1),-sqrt(x^3+3x^2+3x+1)

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inverse of f(x)=\sqrt[3]{x^2}-1

inverse

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

x^{3} + 3 x^{2} + 3 x + 1, - x^{3} + 3 x^{2} + 3 x + 1

Solution steps

y = 3 x^{2} - 1

Replace

x

with

y

x = 3 y^{2} - 1

Solve for

y, x = 3 y^{2} - 1

x^{3} + 3 x^{2} + 3 x + 1, - x^{3} + 3 x^{2} + 3 x + 1

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

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

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

f (x) = (2 x + 3) = 5 x - 2

f (x) = 3 - lo g_{\frac{1}{5}} (1 + x)

What is the inverse of f(x)=\sqrt[3]{x^2}-1 ?
The inverse of f(x)=\sqrt[3]{x^2}-1 is sqrt(x^3+3x^2+3x+1),-sqrt(x^3+3x^2+3x+1)