What is the inverse of f(x)=(x^2+1)/(x^2+3) ?

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

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inverse of f(x)=(x^2+1)/(x^2+3)

inverse

f (x) = \frac{x ^{2} + 1}{x ^{2} + 3}

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

Solution steps

y = \frac{x ^{2} + 1}{x ^{2} + 3}

Replace

x

with

y

x = \frac{y ^{2} + 1}{y ^{2} + 3}

Solve for

y, x = \frac{y ^{2} + 1}{y ^{2} + 3}

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

f (x) = 6 (x - 5)

f (x) = 2 (x + 3)^{2} - 5

f (x) = \frac{ln ( 3 x )}{8 + ln ( 3 x )}

f (x) = 9 sin (x - 5) + 8

f (x) = \frac{2 x - 6}{4}

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