What is the inverse of y= 5/(x^2+1) ?

The inverse of y= 5/(x^2+1) is sqrt((5-x)/x),-sqrt((5-x)/x)

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inverse of y= 5/(x^2+1)

inverse

y = \frac{5}{x ^{2} + 1}

\frac{5 - x}{x}, - \frac{5 - x}{x}

Solution steps

y = \frac{5}{x ^{2} + 1}

Replace

x

with

y

x = \frac{5}{y ^{2} + 1}

Solve for

y, x = \frac{5}{y ^{2} + 1}

\frac{5 - x}{x}, - \frac{5 - x}{x}

f (x) = 9 x - 7 + 14

f (x) = \frac{x}{5} + 6

f (x) = - 1273 x^{2} + 148 x + 13393

\frac{x ^{2} - 9}{7 x ^{2}}

f (x) = \frac{( - 3 - 4 x )}{2 + 2 x}

What is the inverse of y= 5/(x^2+1) ?
The inverse of y= 5/(x^2+1) is sqrt((5-x)/x),-sqrt((5-x)/x)