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

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

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

inverse

f (x) = 3 x + 1 g (x) = x^{3} + 1

3 x - 1

Solution steps

y = x^{3} + 1

Replace

x

with

y

x = y^{3} + 1

Solve for

y, x = y^{3} + 1

3 x - 1

f (x) = \frac{2 x + 5}{x - 11}

f (x) = - (x + 1)^{2} - 1, x < - \frac{π}{2}

y = ln (x)

f (x) = x^{3} + 25

- 2 x^{3} - 6

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