What is the inverse of (x^2-9)/(4x^2) ?

The inverse of (x^2-9)/(4x^2) is sqrt(-9/(4x-1)),-sqrt(-9/(4x-1))

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inverse of (x^2-9)/(4x^2)

inverse

\frac{x ^{2} - 9}{4 x ^{2}}

- \frac{9}{4 x - 1}, - - \frac{9}{4 x - 1}

Solution steps

y = \frac{x ^{2} - 9}{4 x ^{2}}

Replace

x

with

y

x = \frac{y ^{2} - 9}{4 y ^{2}}

Solve for

y, x = \frac{y ^{2} - 9}{4 y ^{2}}

- \frac{9}{4 x - 1}, - - \frac{9}{4 x - 1}

f (x) = \frac{2 x ^{2} - x - 9}{x ^{2} - 1}

lo g_{4} (x)

ln (5 - x)

y = (2 - x)^{2}

f (x) = 2 x - 3

What is the inverse of (x^2-9)/(4x^2) ?
The inverse of (x^2-9)/(4x^2) is sqrt(-9/(4x-1)),-sqrt(-9/(4x-1))