what is the inverse of function (fx) = ( x^2-4) / 2x^2?

Question

what is the inverse of function (fx) = ( x^2-4) / 2x^2?

Answer 1

Solve
x = ( y^2-4) / 2y^2,

but it will be easier if we first separate out the two terms:

(y^2)/(2y^2) - 4/(2y^2)
= 1/2 - 4/(2y^2)
= 1/2 - 2/(y^2)

Now rearrange the equation as
2/y^2 = 1/2 - x

2/y^2 = (1 - 2x)/2
Hence
(y^2)/2 = 2/(1-2x)

y^2 = 4/(1-2x)
and so, provided x < 1/2,

y = 2/sqrt(1-2x) or -2/sqrt(1-2x)

When I saw you already had two answers, I almost didn't bother to look, but they are both wrong; and the three who followed me also are not quite right, as they didn't put in both possibilities and incorrectly gave it as an inverse function. The inverse of f(x) in this case is not a function, because it's not a single-valued relation. If we impose the condition y > 0, we get a function, the one the three following answers give; but if we impose the condition y < 0, we get another function, using the second expression I found above.

Answer 2

Set the function equal to another variable, say "y":

f(x) = y
y = (x^2 - 4) / (2x^2)

Now solve this for x in terms of y. You'll get a new function in terms of y. This function will be the inverse of f(x). The easiest way to do this is to probably just multiply both sides by 2x^2 and combine like terms. You should eventually get:

(2y - 1)*x^2 = 4
x^2 = 4 / (2y-1)
x = sqrt(4 / (2y-1)) etc.

So the inverse function of f(x), let's call it g(y), is:

g(y) = (2 * sqrt(2y-1)) / (2y-1)

Answer 3

Let us write y=f(x), then :
y=( x^2-4) / (2x^2)
=(1/2) - 2/(x^2), so:
2/(x^2)=(1/2) - y or
x^2 = 2/((1/2)-y) or
x=Sqrt(4/(1-2y))
or x = 2/Sqrt(1-2y). Interchanging x & y, we get the inverse function desired:
y=2/Sqrt(1-2x)

Answer 4

Inverse Of X

Answer 5

y = ( x^2 - 4) / 2x^2
2x^2y = (x^2 - 4)
x^2(2y - 1) = - 4
x^2 = -4/(2y - 1)
x = ±2/√(1 - 2y)

f(x) = ( x^2-4) / 2x^2, x ≠ 0

f^-1(x) = ±2/√(1-2x), x < 0.5

Answer 6

1/f(x)

Answer 7

chnge x to y and y(f(x)) to x
at the end we have
f(x)=2*sqrt(1/2x-1)

Answer 8

2x^2/(x^2-4)