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

The inverse of (2x+3)/(3x^2) is (1+sqrt(9x+1))/(3x),(1-sqrt(9x+1))/(3x)

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

inverse

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

\frac{1 + 9 x + 1}{3 x}, \frac{1 - 9 x + 1}{3 x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{1 + 9 x + 1}{3 x}, \frac{1 - 9 x + 1}{3 x}

\frac{( x + 3 )}{( x + 8 )}

f (x) = \frac{( - 2 x + 9 )}{x - 3}

(x + 7) 2

f (x) = 1 6^{x}

y = 35000 - 6000 x

What is the inverse of (2x+3)/(3x^2) ?
The inverse of (2x+3)/(3x^2) is (1+sqrt(9x+1))/(3x),(1-sqrt(9x+1))/(3x)