Confused as to how the range of this function was calculated

[laughablehaha]

I was trying to find the range of the function shown below

1/(4x²-4x-3) = y

Domain: x != -1/2, x != 3/2

Range is where I had the most trouble. I got
Range: y != -1/4, y != 1/4

However, checking my answer on symbolab the range ended up being (-inf, -1/4] u [0, inf).

I tried multiple methods, one of which was isolating x and then finding the domain of that.
This is where I'm stuck. I can't isolate x(and I still don't understand how the range was calculated).

 

[Dr.Peterson]

Range is not easy, in general. What you mentioned trying is one of the better ways for this particular problem. Please show the details of your work, so we can see where you went wrong. (The key idea, which you haven't mentioned explicitly, is to find when x can be isolated, which depends on a discriminant.)

 

[HallsofIvy]

The basic concept here is that a fraction only 0 when the numerator is 0, irrespective of the denominator. Here the numerator is 1 so y is never 0. Yes, it is true that there are vertical asymptotes at x= -1/2 and at x= 3/2 and that y goes to both \(\displaystyle \infty\) and \(\displaystyle -\infty\) as x approaches those as well as approaching 0 as x goes to \(\displaystyle \infty\) and \(\displaystyle -\infty\). So the range is "all real numbers except 0",

 

[Steven G]

The number k is in the range if you can find an x value such that f(x) =k.

Here is how I usually find the range. I set f(x) = k and then solve for x.

Now if I get something like x= 1/(k-2) then I know k=2 is NOT in the range. Why? Because there is no x value when k=2, that is f of no value = 2.

So set 1/(4x²-4x-3) = k, then 4x²-4x-3 = 1/k, then 4x²-4x-(3 + 1/k) =0. Now use quadratic formula to solve for x.

Post back showing us your work.

 

[Dr.Peterson]

The basic concept here is that a fraction only 0 when the numerator is 0, irrespective of the denominator. Here the numerator is 1 so y is never 0. Yes, it is true that there are vertical asymptotes at x= -1/2 and at x= 3/2 and that y goes to both \(\displaystyle \infty\) and \(\displaystyle -\infty\) as x approaches those as well as approaching 0 as x goes to \(\displaystyle \infty\) and \(\displaystyle -\infty\). So the range is "all real numbers except 0",

No, it's more subtle than that. The existence of asymptotes doesn't determine all of the range. Here is a graph:

The provided answer, which excludes an interval, not just zero, is correct, and Jomo's approach, which I am hoping is the OP's and which I suggested how to complete, is good.