What is the inverse of f(x)= 9/(x^2+1) ?

The inverse of f(x)= 9/(x^2+1) is sqrt((9-x)/x),-sqrt((9-x)/x)

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inverse of f(x)= 9/(x^2+1)

inverse

f (x) = \frac{9}{x ^{2} + 1}

\frac{9 - x}{x}, - \frac{9 - x}{x}

Solution steps

y = \frac{9}{x ^{2} + 1}

Replace

x

with

y

x = \frac{9}{y ^{2} + 1}

Solve for

y, x = \frac{9}{y ^{2} + 1}

\frac{9 - x}{x}, - \frac{9 - x}{x}

\frac{2}{x - 5}

x^{2} - 6 x + 5

x + 4 + 2

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

5 \cdot 2^{1 - x} - 1

What is the inverse of f(x)= 9/(x^2+1) ?
The inverse of f(x)= 9/(x^2+1) is sqrt((9-x)/x),-sqrt((9-x)/x)