What is the inverse of f(s)= 4/(s^2-9) ?

The inverse of f(s)= 4/(s^2-9) is sqrt((4+9s)/s),-sqrt((4+9s)/s)

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inverse of f(s)= 4/(s^2-9)

inverse

f (s) = \frac{4}{s ^{2} - 9}

\frac{4 + 9 s}{s}, - \frac{4 + 9 s}{s}

Solution steps

y = \frac{4}{s ^{2} - 9}

Replace

s

with

y

s = \frac{4}{y ^{2} - 9}

Solve for

y, s = \frac{4}{y ^{2} - 9}

\frac{4 + 9 s}{s}, - \frac{4 + 9 s}{s}

3^{x} - 1

x^{3} - 27 x + 9

5 y - 4 x = - 15

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

f (x) = \frac{- 5 x + 2}{6 x + 3}

What is the inverse of f(s)= 4/(s^2-9) ?
The inverse of f(s)= 4/(s^2-9) is sqrt((4+9s)/s),-sqrt((4+9s)/s)