What is the inverse of 1/(s(s+3)) ?

The inverse of 1/(s(s+3)) is (-3s+sqrt(9s^2+4s))/(2s),(-3s-sqrt(9s^2+4s))/(2s)

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inverse of 1/(s(s+3))

inverse

\frac{1}{s ( s + 3 )}

\frac{- 3 s + 9 s ^{2} + 4 s}{2 s}, \frac{- 3 s - 9 s ^{2} + 4 s}{2 s}

Solution steps

y = \frac{1}{s ( s + 3 )}

Replace

s

with

y

s = \frac{1}{y ( y + 3 )}

Solve for

y, s = \frac{1}{y ( y + 3 )}

\frac{- 3 s + 9 s ^{2} + 4 s}{2 s}, \frac{- 3 s - 9 s ^{2} + 4 s}{2 s}

f (x) = \frac{( - 3 x + 18 )}{5}

f (x) = 12 x^{3} - 4

9 + x^{\frac{1}{3}}

f (x) = x^{2} + 5 x + 3

\frac{6}{x + 8}

What is the inverse of 1/(s(s+3)) ?
The inverse of 1/(s(s+3)) is (-3s+sqrt(9s^2+4s))/(2s),(-3s-sqrt(9s^2+4s))/(2s)