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

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

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

inverse

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

- x^{2} + 2 x - 6

y = - \frac{2}{5} x + 8

y = x^{2} - 6 x + 7

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

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