What is the inverse of (3s)/(s^2+3s+2) ?

The inverse of (3s)/(s^2+3s+2) is (-3s+3+sqrt(s^2-18s+9))/(2s),(-3s+3-sqrt(s^2-18s+9))/(2s)

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inverse of (3s)/(s^2+3s+2)

inverse

\frac{3 s}{s ^{2} + 3 s + 2}

\frac{- 3 s + 3 + s ^{2} - 18 s + 9}{2 s}, \frac{- 3 s + 3 - s ^{2} - 18 s + 9}{2 s}

Solution steps

y = \frac{3 s}{s ^{2} + 3 s + 2}

Replace

s

with

y

s = \frac{3 y}{y ^{2} + 3 y + 2}

Solve for

y, s = \frac{3 y}{y ^{2} + 3 y + 2}

\frac{- 3 s + 3 + s ^{2} - 18 s + 9}{2 s}, \frac{- 3 s + 3 - s ^{2} - 18 s + 9}{2 s}

f (x) = ∣ x + 3 ∣

y = (x - 5)^{2}

f (x) = \frac{x ^{2} - 2 x - 24}{x ^{2} - 64}

f (x) = \frac{1}{x} - x

What is the inverse of (3s)/(s^2+3s+2) ?
The inverse of (3s)/(s^2+3s+2) is (-3s+3+sqrt(s^2-18s+9))/(2s),(-3s+3-sqrt(s^2-18s+9))/(2s)