What is the inverse of f(x)= 8/(1+x^2) ?

The inverse of f(x)= 8/(1+x^2) is sqrt((8-x)/x),-sqrt((8-x)/x)

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inverse of f(x)= 8/(1+x^2)

inverse

f (x) = \frac{8}{1 + x ^{2}}

\frac{8 - x}{x}, - \frac{8 - x}{x}

Solution steps

y = \frac{8}{1 + x ^{2}}

Replace

x

with

y

x = \frac{8}{1 + y ^{2}}

Solve for

y, x = \frac{8}{1 + y ^{2}}

\frac{8 - x}{x}, - \frac{8 - x}{x}

\frac{e ^{x}}{9 + 4 e ^{x}}

f (x) = - 4 x - 9

20

2 x^{3} - 10

What is the inverse of f(x)= 8/(1+x^2) ?
The inverse of f(x)= 8/(1+x^2) is sqrt((8-x)/x),-sqrt((8-x)/x)