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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

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