Genereally, understanding what it is you are looking at is most helpful. My whole point, here, is to suggest it may not be a trivial task. Sorry.
[math]f(x) = \dfrac{x^{2}-9}{4x^{2}+x} = \dfrac{(x+3)(x-3)}{x(4x+1)}[/math]
Just looking a the original expression, one can see immediately that there is an horizontal asymptote at f(x) = 1/4, so that is unlikely to be in the Range, but it may behave oddly close to x = 0, so don't rule it out just yet.
Looking at the second expression, one sees vertical asymptotes at x= 0 and x = -1/4. Given vertical asymptotes, it may seem likely that all real nunnbrs should be included in the Range. It's also important to note the DEGREE of each vertical asymptote causing factor in the denominator. These are both degree 1, so the extreme behavior is known if we know a little more about it.
What is f(-4) = 0.117
What is f(-3) = 0 -- That wasn't a surprise. Most notably, it is decreasing as x increases.
What is f(-2) = -0.357 -- Still decreasing
What is f(-1) = -2.667 -- Still decreasing
Okay, so we know the function decreases (increases without bound in the negative direction) as x approaches -1/4 from the negative side.
Looking at "x" in the denominator, being of degree 1, we know that the asymptote must come back from the top, since it exited from the bottom.
What is f(-1/8) = 143.75 -- Okay, so it IS coming down from the top.
What is f(-1/16) = 191.917 -- What? That may have been a surprise. It turns abound and goes back up as we continue to approach x = 0 on our trek in the positive x-direction.
Knowing it turns around and is unbounded in the positive y direction, what does that tell us of the other side of x = 0? It had better come back into the picture from the bottom.
What is f(1/4) = -17.875 -- Yup.
That's almost all we need to know. At the extremes, we are increasing to y = 1/4 as we increase x in either direction. f(x) crosses the x-axis at x = 3 and x = -3. We know everything EXCEPT where and how does it turn around between x = -1/4 and x = 0? Well, without the calculus, this is a much harder problem. You can just guess and check.
We already have:
f(-1/8) = 143.75
f(-1/16) = 191.917 -- Thus, it must have turned around somewhere between x = -1/8 and x = -1/16. Bisection can be instructive. Half way between -1/8 and -1/16 is -3/32 = -0.094.
f(-3/32) = 153.45 -- Half way between -1/8 and -3/32 is -7/64. = -0.109
f(-7/64) = 146.091 -- We could keep doing this for a VERY LONG TIME. The calculus immediately tells us that the minimum is at
[math]x = 3\cdot\sqrt{136} - 36 = -0.1252177707 [/math], so, we WERE getting there with our little bisection thing, but we were getting there rather slowly.
BTW: f(-0.1252177707) = 143.75 -- It really looks like we almost stumbled over it when we arbitrarily picked x = -1/8 for earlier testing.
One thing I cannot answer is why you were getting [math]\sqrt{136}[/math] and I was getting [math]\sqrt{143}[/math]. Someone made a typo along the way would be my first guess.
Like I said, without the calculus, this can be a much more difficult task. Even with the calculus, it can be rather tedious.
