What is the inverse of F(s)=(s+1)/(s^3+1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of F(s)=(s+1)/(s^3+1)

inverse

F (s) = \frac{s + 1}{s ^{3} + 1}

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

Solution steps

y = \frac{s + 1}{s ^{3} + 1}

Replace

s

with

y

s = \frac{y + 1}{y ^{3} + 1}

Solve for

y, s = \frac{y + 1}{y ^{3} + 1}

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

f (x) = 2 x

f (x) = 500 - 16 t^{2}

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

9 s^{9} s^{2} + 1

f (x) = 6 x^{7} - 3

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