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}

g (x) = \frac{1}{x}

\frac{2 x - 5}{3 x}

f (7) = 2 x + 1

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

f (x) = 1 + 5 + 7 x

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))