What is the inverse of (2s+3)/(s^2+9) ?

The inverse of (2s+3)/(s^2+9) is (1+sqrt(-3s(3s-1)+1))/s ,(1-sqrt(-3s(3s-1)+1))/s

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inverse of (2s+3)/(s^2+9)

inverse

\frac{2 s + 3}{s ^{2} + 9}

\frac{1 + - 3 s ( 3 s - 1 ) + 1}{s}, \frac{1 - - 3 s ( 3 s - 1 ) + 1}{s}

Solution steps

y = \frac{2 s + 3}{s ^{2} + 9}

Replace

s

with

y

s = \frac{2 y + 3}{y ^{2} + 9}

Solve for

y, s = \frac{2 y + 3}{y ^{2} + 9}

\frac{1 + - 3 s ( 3 s - 1 ) + 1}{s}, \frac{1 - - 3 s ( 3 s - 1 ) + 1}{s}

h (t) = - 16 t^{2} + 30 t + 5

f (x) = 1.25 (38 - x)

y = 9 x

arccos (- 4)

f (x) = lo g_{10} (x - 5) - 4

What is the inverse of (2s+3)/(s^2+9) ?
The inverse of (2s+3)/(s^2+9) is (1+sqrt(-3s(3s-1)+1))/s ,(1-sqrt(-3s(3s-1)+1))/s