What is the inverse of f(x)=(x^3+2)^{1/3}-4 ?

The inverse of f(x)=(x^3+2)^{1/3}-4 is cube root of x^3+12x^2+48x+62

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inverse of f(x)=(x^3+2)^{1/3}-4

inverse

f (x) = (x^{3} + 2)^{\frac{1}{3}} - 4

3 x^{3} + 12 x^{2} + 48 x + 62

Solution steps

y = (x^{3} + 2)^{\frac{1}{3}} - 4

Replace

x

with

y

x = (y^{3} + 2)^{\frac{1}{3}} - 4

Solve for

y, x = (y^{3} + 2)^{\frac{1}{3}} - 4

3 x^{3} + 12 x^{2} + 48 x + 62

f (x) = e^{x + 4}

f (x) = \frac{5 ( x - 2 ) ^{2} + 1}{3}

f (x) = e^{x + 5}

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

f (x) = (x - 8)^{2} + 7

What is the inverse of f(x)=(x^3+2)^{1/3}-4 ?
The inverse of f(x)=(x^3+2)^{1/3}-4 is cube root of x^3+12x^2+48x+62