What is the inverse of f(x)=(\sqrt[3]{x+1})^5-1 ?

The inverse of f(x)=(\sqrt[3]{x+1})^5-1 is (x+1)^{3/5}-1

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inverse of f(x)=(\sqrt[3]{x+1})^5-1

inverse

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

Solution steps

Replace

x

with

y

Solve for

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

f (x) = x^{2} - 2 x + 2, x \leq 1

f (x) = 18500 (0.25 - r^{2})

f (x) = ln (7 x + 4) - 2

f (x) = 9 x^{3} - 6

f (x) = - 7 cos (2 x)

What is the inverse of f(x)=(\sqrt[3]{x+1})^5-1 ?
The inverse of f(x)=(\sqrt[3]{x+1})^5-1 is (x+1)^{3/5}-1