What is the inverse of (x^2-9)/(x^2+x-12) ?

The inverse of (x^2-9)/(x^2+x-12) is (3-4x)/(x-1)

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inverse of (x^2-9)/(x^2+x-12)

inverse

\frac{x ^{2} - 9}{x ^{2} + x - 12}

\frac{3 - 4 x}{x - 1}

Solution steps

y = \frac{x ^{2} - 9}{x ^{2} + x - 12}

Replace

x

with

y

x = \frac{y ^{2} - 9}{y ^{2} + y - 12}

Solve for

y, x = \frac{y ^{2} - 9}{y ^{2} + y - 12}

\frac{3 - 4 x}{x - 1}

f (x) = x^{3}

f (x) = - 3 x - 4 + 9

+ tan^{2} (x)

f (x) = 3 \frac{x + 3}{5}

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

What is the inverse of (x^2-9)/(x^2+x-12) ?
The inverse of (x^2-9)/(x^2+x-12) is (3-4x)/(x-1)