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}

8 x - 1

x - 5

f (x) = 5 + (6 + x)^{\frac{1}{2}}

f (x) = - \frac{1}{x - 6}

\frac{2 x + 3}{x + 1}

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