What is the inverse of f(x)=(x^2+3)/(2x-5) ?

The inverse of f(x)=(x^2+3)/(2x-5) is x+sqrt(x^2-5x-3),x-sqrt(x^2-5x-3)

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inverse of f(x)=(x^2+3)/(2x-5)

inverse

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

x + x^{2} - 5 x - 3, x - x^{2} - 5 x - 3

Solution steps

y = \frac{x ^{2} + 3}{2 x - 5}

Replace

x

with

y

x = \frac{y ^{2} + 3}{2 y - 5}

Solve for

y, x = \frac{y ^{2} + 3}{2 y - 5}

x + x^{2} - 5 x - 3, x - x^{2} - 5 x - 3

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

f (x) = 41 - 1.25 x

f (x) = \frac{4}{7} x + 2

f (x) = (x^{3} + 1)^{5}

f (x) = (\frac{1}{7})^{x}

What is the inverse of f(x)=(x^2+3)/(2x-5) ?
The inverse of f(x)=(x^2+3)/(2x-5) is x+sqrt(x^2-5x-3),x-sqrt(x^2-5x-3)