Expert: Chanda Walker - 2/28/2009

Question
QUESTION: The following was an exam question I happened to get right. I have no idea how, but it happened lol.

F(x) defined by F(x)= -x^2/x^2-4

A) Sketch the graph

Done.

B) State the equations of the vertical asymptotes

I see two asymptotes that touch that graph at (-2,2)
X=-2
X=2

Done.

C) State the domain and ranges of the function

The Domain in this case would where the graph cannot touch X right?

D: X cannot equal-2, 2

How would I get the Range?

I went to 2nd+Graph and noticed that Y was moving towards -1.

So Y < -1 is one range, how would I go about getting the other range?

I know the answer is Y > 0 but for the life of me can't remember how I got that answer. Does the 0 come from when the whole equation cancels out and equals? I know that if input 0 into the original which is:

x^2
----
x^2-4

(0)^2=0 (0)^2-4
0/4=0

Can you provide me with an easy way to attain the functions? Thank you in advance!

ANSWER: Warning: Math humor ahead.  No it isn't really funny but eh, it is better than the normal fare.

The domain is all the numbers that x can be without making the function do something nasty and terrible--as it looks like you know already.  An example of nasty and terrible is that the denominator of a fraction tries to equal zero.  That is very bad.  Mathematicians have a love/hate relationship with infinite.

So in your example, the denominator (if I'm reading it correctly) would be:

x^2 - 4

You can see pretty easily that if x = 2, the world explodes.  BOOM!  That is bad.  Also, x = -2. BANG.  So the domain is any number except those two numbers.  That is often written like this:

x = (-infinite, -2), (-2, 2), (-2, infinite)

Notice the brackets.  All brackets are not created equal.  Really.  Bracketism is alive and well in mathematics.  When you have a square bracket it means the number standing beside it is INCLUDED.  When you have round, it means the number is excluded.  Infinite is ALWAYS excluded.  Don't EVER put a square bracket next to infinite.  We are not sure what happens but my theory is that you will instantaneously become a drama queen or king.

-2 and 2 are also EXCLUDED from the domain.  The stuff in between the -2 and 2 is included which is why we bother writing out (-2, 2).  Just the number WRITTEN next to the parenthesis (or round bracket) is out.

We don't have ANY square brackets YET in this problem but when we do, you'll see why we write it in this funny way.  Just keep reading . . .




The range is all the numbers that f(x) can be if we let x be any number that x can be.  Always figure out the domain first.

So let's check the extremes . . .

x = - infinite means that f(x) = -1.  The same true of x = infinite since our ^2 makes it so.  Hmmm.  -1 to -1.  Sounds boring and our equation doesn't look like the equation of a straight line.  More investigation is needed.  (Technically speaking if I write x = infinite the gang of nerds I call colleagues are supposed to break down my door and hang me.  I should type -> which means approaches but I'm not sure you know that notation so forgive me for my lax notation.  Ah, there is the knock.  I'd better hurry.)

Before I go, let's finish this.  What happens between -2 and 2?  Let's look at 0, say.  Well that goes to 0.  F(0) = 0.  Now let's look at x = +/- 1, whichever you want, just because integers are easy to work with.  You could look at 3/7's if you want but, ugh, we already have enough fractions no?  Here f(1) = -1/-3 or 1/3.  So from zero we traipse up to 1/3 (in both directions) and keep going.  Up and up and up!  As we approach 2 (or -2), f(x) gets big, fast.  So here f(x) has started at 0 and went to positive infinite.  The range, thus far, is [0, infinite).  Look!  A square bracket!

But we know this thing has to end up at -1, right?  So there is more to learn.

Let's start at some number on the other side of 2 (or -2) that isn't far from 2 and see what is going on there.  Let's look at 3.  Again, I picked an small integer.  I'm lazy.  You can pick Sqrt[64.3] if you'd like.  F(3) = -9/5.  Hmm that is smaller than -1.  It turns out, the closer you look to 2, the smaller the number gets, and does so quickly.  So on that side of 2, f(x) approaches negative infinite.

Aha!  A discontinuity!  Mathematicians LOATHE discontinuities.  There is the REAL reason we have to exclude 2 and -2.  We can't decide whether f(2) is -infinite or +infinite and that is a problem.  So we throw 2 and -2 out and just ignore them.  If we applied this rule in real life, many women I know would just disappear.  Drama queens.  

We have an asymptote where we have the discontinuity.   The equation of the asymptote is x = +/- 2.  To derive that equation, just investigate the denominator.  x^2 - 4 = 0.  You can solve that equation without much effort.

BUT now we know that the range is (-infinite,-1), [0, infinite).  Note, round brackets again AND we have a square bracket!  Zero CAN be included.  f(0) = 0.  We do not include -1 because the function only APPROACHES -1.  (You know if a function only approaches a value if you can't find a real number for x that will make it happen--without doing any rounding.  You'll get to have lots of fun thinking about those things in pre-calc.  I'll be here if you need more silly babble.)

So, in summary: The first and last part of the domain yields a range f(x) < -1.  The middle part of the range yields f(x) > 0.

In general, to figure out the domain and range, you have to be creative.  There is not a set of rules to figure this stuff out.  You look for "problem" areas and then investigate what happens when you are near them.  Don't forget, funny things can happen on both sides of "problem" areas.

Let me mention two other things to know that are helpful.  

1)  There is another nasty and terrible thing to watch out for.  Remember the first was zero in the denominator.  Well, the second is a negative number in a radical.  We don't like negative numbers in our radicals.  Imaginary numbers are not our friends until we move to physics.  In physics, imaginary numbers save the day.

If the negative number were in a cube root though, that would be ok. It doesn't give us an imaginary number.  You have to pay attention.

2)  There is one more kind of bracket.  The squiggledy one: {}.  That is what you use to talk about a specific point--not a continuum of points.  Sometimes you'll need that.  It is too early in the morning for me to think of an example.  If I do I'll come back and let you know.

You asked for an easy way to attain the functions.  If I understand your request, the answer is that there isn't an easy way.  It takes creativity and experience, which is why you have homework.  If you get good at it, you get paid big bucks.  Or not.  We don't get paid here.  Please don't forget to rate me.  

And you're welcome in advance.  I just hope my humor wasn't too annoying and I hope that I was helpful.  If I wasn't helpful, please feel free to ask more questions.  I enjoy trying to help.

---------- FOLLOW-UP ----------

QUESTION: I think I'm with you so far. Can you give me a few examples to try on my own?

ANSWER: Sure here's one:  Try to answer all the same questions.


This one is much easier . . .

f(x) = 1 /( x - 1)

Here's one that is substantially harder.

f(x) = 1/(sqrt(8 - 2x) - 4)

---------- FOLLOW-UP ----------

QUESTION: For f(x)=1/(x-1)

The Vertical Asymptote is located at (1,0)

X cannot equal 1, XER
Y > 0, YER

f(x)=1(sqrt(8-2x)-4)

I have no idea where to start with that, I don't think we'd have anything as hard as that shwo up on any of our exams our testbook work yet. How would I go about tackling that?

Thanks in advance.  

Answer
First, you don't want your sqrt to be negative.  So solve for values that will allow that argument to be less than zero and those values are out.

8 - 2x < 0
8 < 2x
x > 4

When x is greater than 4, you'll get imaginary numbers.

Next, the denominator cannot equal zero.

Sqrt(8 - 2x) - 4 = 0
Sqrt(8 - 2x) = 4
8 - 2x = 16
-8 = 2x
x = -4

So x cannot equal -4 either.  Here is the domain.

x = (-infinity, -4), (-4, 4]

The range . . .

f(x) = 1/(sqrt(8 - 2x) - 4)

when x -> -infinity, f(x) -> 0
when x = -5, f(-5) = 1/(Sqrt(33) - 4) > 0

So the function goes from zero up to + infinity in that portion of the domain.

when x = -1/2, f(-1/2) = 1/(Sqrt(9) - 4) = 1/-1 = -1
when x = 4, f(4) = 1/4

So the function goes from negative infinity up to 1/4 in the second portion of the domain.

y = (-infinity, + infinity)