What is the inverse of f(x)=(x^3-4)/(x^3-5) ?

The inverse of f(x)=(x^3-4)/(x^3-5) is cube root of (-4+5x)/(x-1)

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inverse of f(x)=(x^3-4)/(x^3-5)

inverse

f (x) = \frac{x ^{3} - 4}{x ^{3} - 5}

Solution steps

y = \frac{x ^{3} - 4}{x ^{3} - 5}

Replace

x

with

y

x = \frac{y ^{3} - 4}{y ^{3} - 5}

Solve for

y, x = \frac{y ^{3} - 4}{y ^{3} - 5}

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

f (x) = 7 x + 5 y = 2

f (x) = \frac{3}{x - 8}

f (x) = 7 x + 3

f (x) = 7 x + 6

What is the inverse of f(x)=(x^3-4)/(x^3-5) ?
The inverse of f(x)=(x^3-4)/(x^3-5) is cube root of (-4+5x)/(x-1)