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}

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

\frac{2 x + 3}{x - 4}

f (x) = x - 3 - 4

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

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)