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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

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

f (x) = t + 6

- \frac{36 ( x - 1 )}{( x + 3 ) ^{3}}

f (x) = 7 x - 6

(1) (0.2)

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