What is the inverse of s/(s^2+2s+5) ?

The inverse of s/(s^2+2s+5) is (-2s+1+sqrt(-16s^2-4s+1))/(2s),(-2s+1-sqrt(-16s^2-4s+1))/(2s)

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inverse of s/(s^2+2s+5)

inverse

\frac{s}{s ^{2} + 2 s + 5}

\frac{- 2 s + 1 + - 16 s ^{2} - 4 s + 1}{2 s}, \frac{- 2 s + 1 - - 16 s ^{2} - 4 s + 1}{2 s}

Solution steps

y = \frac{s}{s ^{2} + 2 s + 5}

Replace

s

with

y

s = \frac{y}{y ^{2} + 2 y + 5}

Solve for

y, s = \frac{y}{y ^{2} + 2 y + 5}

\frac{- 2 s + 1 + - 16 s ^{2} - 4 s + 1}{2 s}, \frac{- 2 s + 1 - - 16 s ^{2} - 4 s + 1}{2 s}

\frac{( 2 x - 5 )}{x - 7}

f (x) = \frac{4}{3 x - 2}

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

f (x) = \frac{4 - x ^{2}}{3}

f (x) = \frac{9 - 8 x}{10 x - 9}

What is the inverse of s/(s^2+2s+5) ?
The inverse of s/(s^2+2s+5) is (-2s+1+sqrt(-16s^2-4s+1))/(2s),(-2s+1-sqrt(-16s^2-4s+1))/(2s)