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

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

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

inverse

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

\frac{- 1 - x ^{2}}{2 x ^{2} - 3}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{- 1 - x ^{2}}{2 x ^{2} - 3}

f (x) = ln (x^{2} - 4) + π

\frac{3 x - 4}{2 - x}

\frac{5 - 3 x}{2 x - 3}

f (x) = 0.2 x + 5.3

f (x) = (2 x + 17)^{2}

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