What is the inverse of (3s+5)/(s^2+4s+13) ?

The inverse of (3s+5)/(s^2+4s+13) is (-4s+3+sqrt(-36s^2-4s+9))/(2s),(-4s+3-sqrt(-36s^2-4s+9))/(2s)

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inverse of (3s+5)/(s^2+4s+13)

inverse

\frac{3 s + 5}{s ^{2} + 4 s + 13}

\frac{- 4 s + 3 + - 36 s ^{2} - 4 s + 9}{2 s}, \frac{- 4 s + 3 - - 36 s ^{2} - 4 s + 9}{2 s}

Solution steps

y = \frac{3 s + 5}{s ^{2} + 4 s + 13}

Replace

s

with

y

s = \frac{3 y + 5}{y ^{2} + 4 y + 13}

Solve for

y, s = \frac{3 y + 5}{y ^{2} + 4 y + 13}

\frac{- 4 s + 3 + - 36 s ^{2} - 4 s + 9}{2 s}, \frac{- 4 s + 3 - - 36 s ^{2} - 4 s + 9}{2 s}

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

ln (4 - 2 x)

f (x) = f (x) = 49 - x^{2}

f (x) = f (x) = 6 x - 5

\frac{- 4 x}{7}

What is the inverse of (3s+5)/(s^2+4s+13) ?
The inverse of (3s+5)/(s^2+4s+13) is (-4s+3+sqrt(-36s^2-4s+9))/(2s),(-4s+3-sqrt(-36s^2-4s+9))/(2s)