What is the inverse of f(x)= 1/(x^3-4) ?

The inverse of f(x)= 1/(x^3-4) is cube root of (1+4x)/x

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inverse of f(x)= 1/(x^3-4)

inverse

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

Solution steps

y = \frac{1}{x ^{3} - 4}

Replace

x

with

y

x = \frac{1}{y ^{3} - 4}

Solve for

y, x = \frac{1}{y ^{3} - 4}

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

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

f (x) = 5 x^{\frac{1}{4}}

h (x) = \frac{2 x + 17}{3 x + 1}

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

What is the inverse of f(x)= 1/(x^3-4) ?
The inverse of f(x)= 1/(x^3-4) is cube root of (1+4x)/x