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

The inverse of f(x)=(y^2+1)/y is (y+sqrt(y^2-4))/2 ,(y-sqrt(y^2-4))/2

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

inverse

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

\frac{y + y ^{2} - 4}{2}, \frac{y - y ^{2} - 4}{2}

Solution steps

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

Replace

y

with

x

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

Solve for

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

\frac{y + y ^{2} - 4}{2}, \frac{y - y ^{2} - 4}{2}

\frac{2 x + 7}{2}

f (x) = \frac{x}{8 x - 5}

f (x) = (5 x + 10)^{\frac{1}{2}} + 14

f (x) = 3 x - 8 + 4

f (x) = x - 3 + 6

What is the inverse of f(x)=(y^2+1)/y ?
The inverse of f(x)=(y^2+1)/y is (y+sqrt(y^2-4))/2 ,(y-sqrt(y^2-4))/2