What is the inverse of (s+3)/(s^2-2s+10) ?

The inverse of (s+3)/(s^2-2s+10) is (2s+1+sqrt(-36s^2+16s+1))/(2s),(2s+1-sqrt(-36s^2+16s+1))/(2s)

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inverse of (s+3)/(s^2-2s+10)

inverse

\frac{s + 3}{s ^{2} - 2 s + 10}

\frac{2 s + 1 + - 36 s ^{2} + 16 s + 1}{2 s}, \frac{2 s + 1 - - 36 s ^{2} + 16 s + 1}{2 s}

Solution steps

y = \frac{s + 3}{s ^{2} - 2 s + 10}

Replace

s

with

y

s = \frac{y + 3}{y ^{2} - 2 y + 10}

Solve for

y, s = \frac{y + 3}{y ^{2} - 2 y + 10}

\frac{2 s + 1 + - 36 s ^{2} + 16 s + 1}{2 s}, \frac{2 s + 1 - - 36 s ^{2} + 16 s + 1}{2 s}

f (x) = \frac{3 - 2 ^{a r c c o s (x)}}{5}

\frac{5 s + 3}{s ^{2} - 1}

f (x) = \frac{1}{2} x - \frac{3}{7}

1 + \frac{4 x}{x + 3}

f (x) = - 3 + 3 2 x + 3

What is the inverse of (s+3)/(s^2-2s+10) ?
The inverse of (s+3)/(s^2-2s+10) is (2s+1+sqrt(-36s^2+16s+1))/(2s),(2s+1-sqrt(-36s^2+16s+1))/(2s)