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

The inverse of f(x)=sqrt(x-1)^{(1/3)} is x^6+1

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

inverse

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

x^{6} + 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

x^{6} + 1

f (x) = 3 x + 2 - 7

f (x) = \frac{( 1 + 5 x )}{3 x - 4}

100 (1 - \frac{t}{40})^{2}

f (x) = 3 x + (- 2)

\frac{1}{10 y + 9}

What is the inverse of f(x)=sqrt(x-1)^{(1/3)} ?
The inverse of f(x)=sqrt(x-1)^{(1/3)} is x^6+1