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

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

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

inverse

y = \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) = \frac{7 ^{x + 6}}{7}

10 - 3 - x

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

y = \frac{4}{x ^{2}}

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