What is the inverse of (4s+1)/((s+2)(s+3)) ?

The inverse of (4s+1)/((s+2)(s+3)) is (-5s+4+sqrt(s^2-36s+16))/(2s),(-5s+4-sqrt(s^2-36s+16))/(2s)

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inverse of (4s+1)/((s+2)(s+3))

inverse

\frac{4 s + 1}{( s + 2 ) ( s + 3 )}

\frac{- 5 s + 4 + s ^{2} - 36 s + 16}{2 s}, \frac{- 5 s + 4 - s ^{2} - 36 s + 16}{2 s}

Solution steps

y = \frac{4 s + 1}{( s + 2 ) ( s + 3 )}

Replace

s

with

y

s = \frac{4 y + 1}{( y + 2 ) ( y + 3 )}

Solve for

y, s = \frac{4 y + 1}{( y + 2 ) ( y + 3 )}

\frac{- 5 s + 4 + s ^{2} - 36 s + 16}{2 s}, \frac{- 5 s + 4 - s ^{2} - 36 s + 16}{2 s}

f (x) = \frac{9 x - 4}{8 x + 7}

lo g_{10} (6^{x - 1})

f (x) = 15.5 - 5 x

150000. 6^{x}

f (x) = \frac{2 x + 42}{x + 6}

What is the inverse of (4s+1)/((s+2)(s+3)) ?
The inverse of (4s+1)/((s+2)(s+3)) is (-5s+4+sqrt(s^2-36s+16))/(2s),(-5s+4-sqrt(s^2-36s+16))/(2s)