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

The inverse of f(x)=(x^3-1)^{1/3} is cube root of x^3+1

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

inverse

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

f (x) = 8 (\frac{x ^{\frac{1}{3}}}{9} + 4)

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

y = a^{x}

lo g_{2} (k x - 6)

f (x) = 8 + 0.75 x

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