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

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

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

inverse

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

\frac{1 + - x ( x + 3 ) + 1}{x}, \frac{1 - - x ( x + 3 ) + 1}{x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{1 + - x ( x + 3 ) + 1}{x}, \frac{1 - - x ( x + 3 ) + 1}{x}

f (x) = x^{2} + 9

f (x) = \frac{4 x + 5}{7}

\frac{x + 6}{7}

m = 0 b = 12

f (x) = 18 x^{4} - 108 x^{2}

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