What is the inverse of f(x)=(x^2-9)/(4x^2),x>0 ?

The inverse of f(x)=(x^2-9)/(4x^2),x>0 is sqrt(-9/(4x-1)),-sqrt(-9/(4x-1))

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inverse of f(x)=(x^2-9)/(4x^2),x>0

inverse

f (x) = \frac{x ^{2} - 9}{4 x ^{2}}, x > 0

- \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}

2 x = 10 - 3 y

f (x) = 16 x^{2}

2 x + 5 y = 4

f (x) = \frac{2 + ( x + 6 ) ^{2}}{3 - x}

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

What is the inverse of f(x)=(x^2-9)/(4x^2),x>0 ?
The inverse of f(x)=(x^2-9)/(4x^2),x>0 is sqrt(-9/(4x-1)),-sqrt(-9/(4x-1))