What is the inverse of f(x)=((x^3-2))/(x^3-1) ?

The inverse of f(x)=((x^3-2))/(x^3-1) is cube root of (-2+x)/(x-1)

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inverse of f(x)=((x^3-2))/(x^3-1)

inverse

f (x) = \frac{( x ^{3} - 2 )}{x ^{3} - 1}

Solution steps

y = \frac{x ^{3} - 2}{x ^{3} - 1}

Replace

x

with

y

x = \frac{y ^{3} - 2}{y ^{3} - 1}

Solve for

y, x = \frac{y ^{3} - 2}{y ^{3} - 1}

f (x) = \frac{4 t ^{2}}{4 + t ^{3}}

(4, - 1), (- 2, - 3)

2 x^{2} - 3 x + 6

y = x^{2} - 3 x - 54

What is the inverse of f(x)=((x^3-2))/(x^3-1) ?
The inverse of f(x)=((x^3-2))/(x^3-1) is cube root of (-2+x)/(x-1)