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

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

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

inverse

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

\frac{x + x ^{2} - 4 ( 3 - 3 x )}{2}, \frac{x - x ^{2} - 4 ( 3 - 3 x )}{2}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{x + x ^{2} - 4 ( 3 - 3 x )}{2}, \frac{x - x ^{2} - 4 ( 3 - 3 x )}{2}

f (x) = 32 + y \frac{9}{5}

y = \frac{2 - x}{2 x - 1}

f (x) = \frac{- 6 - 7 x}{2}

f (x) = 2 x - 38

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