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

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

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

inverse

f (x) = \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}

y = x^{2} y = - 2 x + 8

f (x) = 27 x^{3} - 9 x + 9

f (x) = (9 - x^{2})^{\frac{3}{2}}

f (x) = x^{3} - 27 x + 10

(53) (5 - 8)

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