What is the inverse of s/(s^2+2s-3) ?

The inverse of s/(s^2+2s-3) is (-2s+1+sqrt(16s^2-4s+1))/(2s),(-2s+1-sqrt(16s^2-4s+1))/(2s)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of s/(s^2+2s-3)

inverse

\frac{s}{s ^{2} + 2 s - 3}

\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 - 3}

Replace

s

with

y

s = \frac{y}{y ^{2} + 2 y - 3}

Solve for

y, s = \frac{y}{y ^{2} + 2 y - 3}

\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{d}{d} tan (x)

- ln (x - 2) + 3

f (x) = 2 x + b

\frac{x ^{2} - 2 x - 15}{x ^{2} - 4 x - 21}

g (x) = \frac{x}{x + 1}

What is the inverse of s/(s^2+2s-3) ?
The inverse of s/(s^2+2s-3) is (-2s+1+sqrt(16s^2-4s+1))/(2s),(-2s+1-sqrt(16s^2-4s+1))/(2s)