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

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

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

inverse

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

y + 8 x = 4 (3.4)

\frac{2 x - 3}{x + 4}

f (x) = \frac{2 x + 5}{x ^{3} + x ^{2} - 9 x - 9}

f (x) = - 4.9 (x + 3)^{2} + 45.8

f (x) = \frac{x ^{2} + 16}{x ( 3 x - 6 )}

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