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

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

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inverse of f(x)=(x^2-9)/(2x^2)

inverse

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

- \frac{9}{2 x - 1}, - - \frac{9}{2 x - 1}

Solution steps

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

Replace

x

with

y

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

Solve for

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

- \frac{9}{2 x - 1}, - - \frac{9}{2 x - 1}

f (x) = 9 x^{2} - 13

\frac{5 - 2 X}{4 X - 2}

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

(6)^{\frac{1}{2}}

g (x) = - \frac{3}{4} x^{2} - \frac{1}{2}

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