What is the inverse of f(x)=((3x^2-2x+7))/(2x+5) ?

The inverse of f(x)=((3x^2-2x+7))/(2x+5) is (1+x+sqrt(x^2+17x-20))/3 ,(1+x-sqrt(x^2+17x-20))/3

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inverse of f(x)=((3x^2-2x+7))/(2x+5)

inverse

f (x) = \frac{( 3 x ^{2} - 2 x + 7 )}{2 x + 5}

\frac{1 + x + x ^{2} + 17 x - 20}{3}, \frac{1 + x - x ^{2} + 17 x - 20}{3}

Solution steps

y = \frac{3 x ^{2} - 2 x + 7}{2 x + 5}

Replace

x

with

y

x = \frac{3 y ^{2} - 2 y + 7}{2 y + 5}

Solve for

y, x = \frac{3 y ^{2} - 2 y + 7}{2 y + 5}

\frac{1 + x + x ^{2} + 17 x - 20}{3}, \frac{1 + x - x ^{2} + 17 x - 20}{3}

f (x) = \frac{9}{x + 2}

f (x) = 16 + x^{2} - x + 1

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

f (x) = \frac{x ^{2}}{x ^{2} - 9}

f (x) = x^{2} - 4 x + 1, x < 2

What is the inverse of f(x)=((3x^2-2x+7))/(2x+5) ?
The inverse of f(x)=((3x^2-2x+7))/(2x+5) is (1+x+sqrt(x^2+17x-20))/3 ,(1+x-sqrt(x^2+17x-20))/3