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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

\frac{x + 3}{( x + 8 ) ( x - 2 )}

x = 2

x^{4} (x - 2) (x + 3)

\frac{5 x - y}{3} = \frac{5 x - 4}{6}

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

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