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

The inverse of f(x)= 1/(x^2-4)+1 is sqrt((4x-3)/(x-1)),-sqrt((4x-3)/(x-1))

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inverse of f(x)= 1/(x^2-4)+1

inverse

f (x) = \frac{1}{x ^{2} - 4} + 1

\frac{4 x - 3}{x - 1}, - \frac{4 x - 3}{x - 1}

Solution steps

y = \frac{1}{x ^{2} - 4} + 1

Replace

x

with

y

x = \frac{1}{y ^{2} - 4} + 1

Solve for

y, x = \frac{1}{y ^{2} - 4} + 1

\frac{4 x - 3}{x - 1}, - \frac{4 x - 3}{x - 1}

f (x) = \frac{3 x - 2}{2 x - 5}

f (x) = 9.2452 e^{- 7.289 x}

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

f (x) = \frac{9 + 10 x}{8}

What is the inverse of f(x)= 1/(x^2-4)+1 ?
The inverse of f(x)= 1/(x^2-4)+1 is sqrt((4x-3)/(x-1)),-sqrt((4x-3)/(x-1))