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/(x+4)

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

inverse

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

3 \frac{1}{x + 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

3 \frac{1}{x + 4}

f (x) = - 9 - x

f (x) = \frac{23 - 3 x}{6 x}

f (x) = 1.4

y = - \frac{1}{2} 0 + \frac{4}{5}

f (x) = 8 - 4 x^{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/(x+4)