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

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

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

inverse

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

\frac{2 - 4 x}{x - 2}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{2 - 4 x}{x - 2}

f (x) = 2 (5 x - 7) + 3

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

f (x) = 5^{x} + 7

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

f (x) = \frac{3 - x}{x}

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