What is the inverse of y=(x^2+3x+7)/(2x-1) ?

The inverse of y=(x^2+3x+7)/(2x-1) is (-3+2x+sqrt(4x^2-16x-19))/2 ,(-3+2x-sqrt(4x^2-16x-19))/2

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inverse of y=(x^2+3x+7)/(2x-1)

inverse

y = \frac{x ^{2} + 3 x + 7}{2 x - 1}

\frac{- 3 + 2 x + 4 x ^{2} - 16 x - 19}{2}, \frac{- 3 + 2 x - 4 x ^{2} - 16 x - 19}{2}

Solution steps

y = \frac{x ^{2} + 3 x + 7}{2 x - 1}

Replace

x

with

y

x = \frac{y ^{2} + 3 y + 7}{2 y - 1}

Solve for

y, x = \frac{y ^{2} + 3 y + 7}{2 y - 1}

\frac{- 3 + 2 x + 4 x ^{2} - 16 x - 19}{2}, \frac{- 3 + 2 x - 4 x ^{2} - 16 x - 19}{2}

cos^{x} (x)

x + 15

f (x) = x + 1

f (x) = 3 x^{3} + 7

y

What is the inverse of y=(x^2+3x+7)/(2x-1) ?
The inverse of y=(x^2+3x+7)/(2x-1) is (-3+2x+sqrt(4x^2-16x-19))/2 ,(-3+2x-sqrt(4x^2-16x-19))/2