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

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

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inverse of (s+1)/(s^2+2)

inverse

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

\frac{1 + 1 - 4 s ( 2 s - 1 )}{2 s}, \frac{1 - 1 - 4 s ( 2 s - 1 )}{2 s}

Solution steps

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

Replace

s

with

y

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

Solve for

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

\frac{1 + 1 - 4 s ( 2 s - 1 )}{2 s}, \frac{1 - 1 - 4 s ( 2 s - 1 )}{2 s}

f (x) = 2 x^{2} + 12 x - 14

f (x) = x + c

f (x) = e^{4 x + 3} + 2

f (x) = \frac{5 x}{5}

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

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