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

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

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inverse of y= 1/(s^2-s-2)

inverse

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

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

Solution steps

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

Replace

s

with

y

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

Solve for

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

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

g (x) = \frac{x}{3} + 6

7 x^{4} - 6 x^{2} - 1

f (x) = - 2^{3}

\frac{1}{x - 2} + 1

f (x) = 10 - 3 (13 - 9 x)

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