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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = - 7 x + 7

f (x) = - 7 x + 5

f (x) = - 5 - 4 x

f (x) = x^{\frac{2}{7}}

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

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