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

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

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inverse of (1e^-(2s))/((s^2-s))

inverse

\frac{1 e ^{-} ( 2 s )}{( s ^{2} - s )}

\frac{2 e}{s} + 1

Solution steps

y = \frac{1 \cdot e ^{-} \cdot 2 s}{s ^{2} - s}

Replace

s

with

y

s = \frac{1 \cdot e ^{-} \cdot 2 y}{y ^{2} - y}

Solve for

y, s = \frac{1 e ^{-} 2 y}{y ^{2} - y}

\frac{2 e}{s} + 1

7 x^{2} - 2

f (x) = - 5 x - 5, x \geq 5

\frac{8 x - 3}{x}

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

\frac{2}{s ^{2} + 2 s - 13}

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