What is the inverse of (3s-14)/(s^2-4s+8) ?

The inverse of (3s-14)/(s^2-4s+8) is (4s+3+sqrt(-16s^2-32s+9))/(2s),(4s+3-sqrt(-16s^2-32s+9))/(2s)

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inverse of (3s-14)/(s^2-4s+8)

inverse

\frac{3 s - 14}{s ^{2} - 4 s + 8}

\frac{4 s + 3 + - 16 s ^{2} - 32 s + 9}{2 s}, \frac{4 s + 3 - - 16 s ^{2} - 32 s + 9}{2 s}

Solution steps

y = \frac{3 s - 14}{s ^{2} - 4 s + 8}

Replace

s

with

y

s = \frac{3 y - 14}{y ^{2} - 4 y + 8}

Solve for

y, s = \frac{3 y - 14}{y ^{2} - 4 y + 8}

\frac{4 s + 3 + - 16 s ^{2} - 32 s + 9}{2 s}, \frac{4 s + 3 - - 16 s ^{2} - 32 s + 9}{2 s}

f (x) = - \frac{x - 17}{12}

f (x) = \frac{4 x + 1}{5 x - 2}

- \frac{3}{4} x + 12

72 x^{\frac{1}{2}}

\frac{( 2 x + 9 )}{9 x - 4}

What is the inverse of (3s-14)/(s^2-4s+8) ?
The inverse of (3s-14)/(s^2-4s+8) is (4s+3+sqrt(-16s^2-32s+9))/(2s),(4s+3-sqrt(-16s^2-32s+9))/(2s)