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

The inverse of (3s)/((s-1)(s+3)) is (-2s+3+sqrt(16s^2-12s+9))/(2s),(-2s+3-sqrt(16s^2-12s+9))/(2s)

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

inverse

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

\frac{- 2 s + 3 + 16 s ^{2} - 12 s + 9}{2 s}, \frac{- 2 s + 3 - 16 s ^{2} - 12 s + 9}{2 s}

Solution steps

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

Replace

s

with

y

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

Solve for

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

\frac{- 2 s + 3 + 16 s ^{2} - 12 s + 9}{2 s}, \frac{- 2 s + 3 - 16 s ^{2} - 12 s + 9}{2 s}

f (x) = x + [x]

f (x) = \frac{1}{5} (5 x - 25) + 5

f (x) = 4 x + 5 = 17

f (x) = x + e^{1}

cos (1.18)

What is the inverse of (3s)/((s-1)(s+3)) ?
The inverse of (3s)/((s-1)(s+3)) is (-2s+3+sqrt(16s^2-12s+9))/(2s),(-2s+3-sqrt(16s^2-12s+9))/(2s)