What is the inverse of (2s-8)/(s^2-4s-4) ?

The inverse of (2s-8)/(s^2-4s-4) is (2s+1+sqrt(8s^2-4s+1))/s ,(2s+1-sqrt(8s^2-4s+1))/s

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inverse of (2s-8)/(s^2-4s-4)

inverse

\frac{2 s - 8}{s ^{2} - 4 s - 4}

\frac{2 s + 1 + 8 s ^{2} - 4 s + 1}{s}, \frac{2 s + 1 - 8 s ^{2} - 4 s + 1}{s}

Solution steps

y = \frac{2 s - 8}{s ^{2} - 4 s - 4}

Replace

s

with

y

s = \frac{2 y - 8}{y ^{2} - 4 y - 4}

Solve for

y, s = \frac{2 y - 8}{y ^{2} - 4 y - 4}

\frac{2 s + 1 + 8 s ^{2} - 4 s + 1}{s}, \frac{2 s + 1 - 8 s ^{2} - 4 s + 1}{s}

y = - 4.9 (x + 3)^{2} + 45.8

f (x) = \frac{x + 6}{2}

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

f (x) = \frac{25}{x ^{2}}, x > 0

10 cos (\frac{2}{5} x)

What is the inverse of (2s-8)/(s^2-4s-4) ?
The inverse of (2s-8)/(s^2-4s-4) is (2s+1+sqrt(8s^2-4s+1))/s ,(2s+1-sqrt(8s^2-4s+1))/s