What is the inverse of (3x^2-18x+24)/(x^2-4x) ?

The inverse of (3x^2-18x+24)/(x^2-4x) is -6/(x-3)

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inverse of (3x^2-18x+24)/(x^2-4x)

inverse

\frac{3 x ^{2} - 18 x + 24}{x ^{2} - 4 x}

- \frac{6}{x - 3}

Solution steps

y = \frac{3 x ^{2} - 18 x + 24}{x ^{2} - 4 x}

Replace

x

with

y

x = \frac{3 y ^{2} - 18 y + 24}{y ^{2} - 4 y}

Solve for

y, x = \frac{3 y ^{2} - 18 y + 24}{y ^{2} - 4 y}

- \frac{6}{x - 3}

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

f (x) = 6 - \frac{1}{2} x

f (x) = \frac{1}{sin ( x )}

f (x) = (x - 3) (x - 5)

f (x) = 1 0^{x + 1}

What is the inverse of (3x^2-18x+24)/(x^2-4x) ?
The inverse of (3x^2-18x+24)/(x^2-4x) is -6/(x-3)