What is the inverse of f(x)=\sqrt[3]{x+8}+1 ?

The inverse of f(x)=\sqrt[3]{x+8}+1 is x^3-3x^2+3x-9

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inverse of f(x)=\sqrt[3]{x+8}+1

inverse

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

Solution steps

Replace

x

with

y

Solve for

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

f (x) = 2 cos (6 x + \frac{π}{2})

f (x) = \frac{x - 4}{x - 1}

f (x) = 18 - t^{2}

\frac{x ^{2} - 36}{x ^{3} - 2 x ^{2} - 24 x}

(0, 3), (3, 375)

What is the inverse of f(x)=\sqrt[3]{x+8}+1 ?
The inverse of f(x)=\sqrt[3]{x+8}+1 is x^3-3x^2+3x-9