What is the inverse of f(x)=((x^2-25))/(x^2-4x-5) ?

The inverse of f(x)=((x^2-25))/(x^2-4x-5) is (5-x)/(x-1)

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inverse of f(x)=((x^2-25))/(x^2-4x-5)

inverse

f (x) = \frac{( x ^{2} - 25 )}{x ^{2} - 4 x - 5}

\frac{5 - x}{x - 1}

Solution steps

y = \frac{x ^{2} - 25}{x ^{2} - 4 x - 5}

Replace

x

with

y

x = \frac{y ^{2} - 25}{y ^{2} - 4 y - 5}

Solve for

y, x = \frac{y ^{2} - 25}{y ^{2} - 4 y - 5}

\frac{5 - x}{x - 1}

f (x) = 7 x - 25

\frac{∣ 7 x + b ∣}{7 x + b}

12 x^{2} - 14 x

f (x) = e^{2}

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

What is the inverse of f(x)=((x^2-25))/(x^2-4x-5) ?
The inverse of f(x)=((x^2-25))/(x^2-4x-5) is (5-x)/(x-1)