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

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

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

inverse

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

\frac{1 - 2 x}{x - 1}, - \frac{1 - 2 x}{x - 1}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{1 - 2 x}{x - 1}, - \frac{1 - 2 x}{x - 1}

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

g (x) = x + 1

y = ln (x + x^{2} + 1)

5^{2 x + 7}

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