What is the inverse of f(x)=(x^3-8)^{1/3} ?

The inverse of f(x)=(x^3-8)^{1/3} is cube root of x^3+8

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inverse of f(x)=(x^3-8)^{1/3}

inverse

f (x) = (x^{3} - 8)^{\frac{1}{3}}

Solution steps

y = (x^{3} - 8)^{\frac{1}{3}}

Replace

x

with

y

x = (y^{3} - 8)^{\frac{1}{3}}

Solve for

y, x = (y^{3} - 8)^{\frac{1}{3}}

\frac{4 x - 9}{7}

f (x) = \frac{x \cdot 200}{( x + 200 ) + 1}

f (x) = \frac{( - 2 x - 8 )}{5}

What is the inverse of f(x)=(x^3-8)^{1/3} ?
The inverse of f(x)=(x^3-8)^{1/3} is cube root of x^3+8