求反函数 (s+1)/(s^2-4s)

解答

求反函数

\frac{s + 1}{s ^{2} - 4 s}

\frac{4 s + 1 + 16 s ^{2} + 12 s + 1}{2 s}, \frac{4 s + 1 - 16 s ^{2} + 12 s + 1}{2 s}

求解步骤

y = \frac{s + 1}{s ^{2} - 4 s}

用

y

替代

s

s = \frac{y + 1}{y ^{2} - 4 y}

求解

y, s = \frac{y + 1}{y ^{2} - 4 y}

\frac{4 s + 1 + 16 s ^{2} + 12 s + 1}{2 s}, \frac{4 s + 1 - 16 s ^{2} + 12 s + 1}{2 s}

in v erse f (x) = \frac{6 x - 7}{3 x + 5}

in v erse f (x) = 3 4 x - 11

in v erse f (x) = - 3 - \frac{8}{5} x

in v erse f (x) = \frac{( x - 2 )}{2 x + 1}

in v erse f (x) = (x + 1)^{3} + 4