What is the inverse of (3s)/(s^2-s-1) ?

The inverse of (3s)/(s^2-s-1) is (s+3+sqrt(5s^2+6s+9))/(2s),(s+3-sqrt(5s^2+6s+9))/(2s)

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inverse of (3s)/(s^2-s-1)

inverse

\frac{3 s}{s ^{2} - s - 1}

\frac{s + 3 + 5 s ^{2} + 6 s + 9}{2 s}, \frac{s + 3 - 5 s ^{2} + 6 s + 9}{2 s}

Solution steps

y = \frac{3 s}{s ^{2} - s - 1}

Replace

s

with

y

s = \frac{3 y}{y ^{2} - y - 1}

Solve for

y, s = \frac{3 y}{y ^{2} - y - 1}

\frac{s + 3 + 5 s ^{2} + 6 s + 9}{2 s}, \frac{s + 3 - 5 s ^{2} + 6 s + 9}{2 s}

f (x) = \frac{x + 8}{x ^{2} - 64}

f (x) = (\frac{- 1}{2}) \cdot (x - 2)^{3}

f (x) = 3 x - 6

y = 2 (x - 6) + 3

f (x) = 4^{\frac{1}{x - 2}}

What is the inverse of (3s)/(s^2-s-1) ?
The inverse of (3s)/(s^2-s-1) is (s+3+sqrt(5s^2+6s+9))/(2s),(s+3-sqrt(5s^2+6s+9))/(2s)