What is the inverse of (3s+4)/(s^2+9) ?

The inverse of (3s+4)/(s^2+9) is (3+sqrt(9-4s(9s-4)))/(2s),(3-sqrt(9-4s(9s-4)))/(2s)

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inverse of (3s+4)/(s^2+9)

inverse

\frac{3 s + 4}{s ^{2} + 9}

\frac{3 + 9 - 4 s ( 9 s - 4 )}{2 s}, \frac{3 - 9 - 4 s ( 9 s - 4 )}{2 s}

Solution steps

y = \frac{3 s + 4}{s ^{2} + 9}

Replace

s

with

y

s = \frac{3 y + 4}{y ^{2} + 9}

Solve for

y, s = \frac{3 y + 4}{y ^{2} + 9}

\frac{3 + 9 - 4 s ( 9 s - 4 )}{2 s}, \frac{3 - 9 - 4 s ( 9 s - 4 )}{2 s}

f (x) = (x^{3} - 4)^{\frac{1}{5}} + 4

4 x^{3} - 7

- 5 x^{2} + x

6 x + 7 - 9

- 219 \cdot (- 4)

What is the inverse of (3s+4)/(s^2+9) ?
The inverse of (3s+4)/(s^2+9) is (3+sqrt(9-4s(9s-4)))/(2s),(3-sqrt(9-4s(9s-4)))/(2s)