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

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

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

inverse

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

- \frac{s}{s - 1}, - - \frac{s}{s - 1}

Solution steps

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

Replace

s

with

y

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

Solve for

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

- \frac{s}{s - 1}, - - \frac{s}{s - 1}

f (x) = \frac{- 5 x + 4}{3 x - 1}

f (x) = \frac{5}{2} x - \frac{15}{2}

f (x) = 2 x [x + 3]

h (x) = [(x)]^{2} + 1

f (x) = \frac{- 5 x + 4}{3 x - 2}

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