Int Algebra: Satisfying the Domain of a Function

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Relations and Functions
Domain and Range
Satisfying the Domain of a Function Page [1 of 2]
Now I really want to take a look at looking at a whole bunch of functions and finding their domains. That is finding all
the allowable inputs for x values that would then produce some sort of allowable output. Let’s just start right in and
take a look at f(x). That just means that this is a function that depends upon x and here’s there recipe. Here’s the
little machine, x2 - x =1. There we go.
Now the question is, “What are the allowable x values?” Those x values that I could actually put in here, where this
would actually produce a legal, honest to goodness real number. Answer?. Well, let’s see. What x values or what
restrictions would I have to place on it? If I have any x value at all, any number at all, think of your favorite number or
even you least favorite number, you could always take that number and you can square it. That gives you another
number. Then you could always subtract the original number away from that and add 1. So, the domain of this is
everything. All the allowable x’s would be the real. So the domain here would be all the real numbers. Again, I’m
using this little notation for reals. Because any x value is allowable.
That wasn’t too bad, how about this one. This next function I’ll call g(x). Remember that’s just the name for it. I could
have called it Sam(x) or Mary(x), it doesn’t make a difference, I’m calling it g. 3 - 4x and I take the square root of the
whole show, everything. What’s the domain of this function? What are the allowable x values that I could plug in?
Let’s think about it for a second. I’m taking the positive square root of this quantity. So the only allowable x’s are
those x’s for which this entire thing under the radical is going to be positive. What I have to actually do is actually
write down an inequality and then solve an inequality. So, the ability to solve an inequality now comes in handy here.
I’d want 3 - 4x to be greater than or equal to 0. Now I want to solve that. How would I solve that? One thing I could
do is bring the 3 over to the other side, and it would become a -3. So, I subtract 3 from both sides and I would have a
-4x greater than or equal to -3. Even though I’m subtracting 3 from both sides, the sign doesn’t flip in this case,
because I’m just adding or subtracting something from both sides. It doesn’t change the inequality. Now, I want to
divide both sides by -4. What do I see? I see that x, I’ll have x on this side and here I’ll have - 3/4. What’s the sign to
put here? We’ll, you may be tempted just to keep the sign, but that is a classic mistake. That’s right, it’s a classic
mistake number 7. It’s the multiplying the inequality mistake. It’s a classic mistake. It definitely makes my top 10 list.
You have to remember that when you multiply an inequality or divide an inequality by a negative sign, that inequality
symbol flips over. So, here, it should look like this. Classic mistake and I hope that you will never do it.
So, the domain for this function g would be all the values for x that are less than or equal to -3/4. If you look back on
that you see that--oh hold on there, I made a little typo here. Maybe you saw it. If I’m dividing both sides by negative.
Look at this, major typo. In fact, I would call that a little typo. If I were grading this, I would be very harsh on myself.
This should be a minus sign here. If that’s a minus sign then a minus and a minus makes this a plus. Sorry about
that. I have to be honest I didn’t catch that one myself. Anyway, that’s a positive 3/4 so x has to be less than or equal
to positive 3/4, because I have to divide both sides by -4. Not only, the left-hand side, but the right-hand side.
That means that the only allowable values that I could put in for x in this function g are any value for x that’s less than
or equal to 3/4. Look what happens if I put in a value bigger than 3/4. For example, let’s say I put in the value 1, which
is bigger than 3/4. If I put in the value 1 for x, look what happens. I see 3 - 4, which is -1 and I’m taking the square
root of -1. That’s not going to happen because that would not give me a real number. That’s an imaginary number.
So, the only values for x, which give me real numbers are the values of x that are less than are equal to 3/4.
Let’s take a look at another example. I’ll call this function h. I’m just trying to show you that you can call these things
anything you want. This is going to be called h(x), x - 1 over x2 - 5x + 6. There’s the function. I want to find the
domain. I want to find all the allowable x’s. Take any number x, if you subtract 1 that’s a completely fine number.
Take any number x, square it, subtract 5 times itself and add 6, that’s a completely fine number. But what about when
you take that quotient? Well, you could take the quotient of any two numbers as long as the denominator isn’t 0. So,
the domain of this will be any value at all for x, except those values that make this equal to 0. So, let’s find out those
restrictions by taking the bottom and setting it equal to 0, equating it to 0.
You can see now all the stuff that you thought about in terms of factoring, solving inequalities and so forth, comes into
play in just finding the domain of a function. Let’s see, the signs are going to be the same and they’re going to be
negative. The product has to be 6 and have to combine to be -5, so 2 and 3 seem to work really well. So the
Relations and Functions
Domain and Range
Satisfying the Domain of a Function Page [2 of 2]
denominator can be factored in this way. When is the denominator 0? Either that is 0, which means x = 2, or this
term is 0, which means x = 3. So, therefore, what I see is the domain of this function, h, are all the values for x except
two, the number 2 and the number 3. So, here the domain would be everyone except 2 and 3. Now, let’s take a look
and see if we’re in good shape here.
If I look at this, you factor, you see an x time x is x2. Here’s a 2x and here’s a +3x. Which gives a -5x. Then I see a -
2 times -3 gives me 6. So, I think I’m in good shape. Looks good to me. The point is the domain is everywhere.
Every single person except x = 2, 3. You could write it the following way. That’s sort of a funny way to write
something, domain is everything one except that. You could say the domain is all the real numbers except, minus the
numbers 2 and 3. This notation just means all the people that are real except for, but remove the number 2 and the
number 3. So there’s the domain there.
Let’s do one last on together. k(x) equals x-2 over x2 + 3x - 10. So, again, I see it’s a fraction, so all I care about is
when the bottom is zero and I have to avoid those points. So, I see x2 + 3x - 10 = 0. If I factor that, I’d have an x and
an x. This negative sign means I’m going to have opposite signs here, so plus and a minus. They have to combine to
give 10. I have to multiply to get 10 and combine to give 3 when I subtract them. So, let’s see, that looks like a 5 and
a 2 are going to work well. I should put the 5 by the positive and the 2 here and they combine to give me plus 3.
So, if this product equals 0, then either this term equals 0, which means x = -5, or this term equals 0, which means x =
2. So those are the places we have to avoid. Other than that we can take any other value, any other real number at
all and plug into the top and plug into the bottom and we’ll get a real number. So, the domain here would be all the
real numbers except for the numbers -5 and 2. Everybody except for these guys here. That’s the domain. Basically,
bottom line, if you have no denominators, no square roots, you’re probably cool. If you’ve got square roots, make
sure the things in the square roots are always positive or 0. Otherwise you’re not in the domain and if you’ve got a
fraction, always make sure that bottom is not 0.
Now, one last point about this one. Suppose you are really on top of things and you’re trying to be slick. What could
you have done? You could have said, “Hey, if I would have factored early on in life, something interesting would have
come to pass.” Here’s the factorization of the bottom. So, I could say that k = x - 2 divided by and then I’d factor the
bottom like we did before, x + 5 and x - 2. If you’re really slick you would have said, “Hey, I could cancel the factor of
x - 2 on the top with the bottom.” Which is completely legal, by the way. If you want to do that, you can do that. Then
you’d say, “Oh, okay, well that means that k equals 1 over x + 5.” So when is the bottom there equal to 0? Only when
x = -5. So, maybe I’m allowed to plug in 2 now. I’m allowed to plug in 2 now, because it doesn’t make the bottom 0.
So, what went wrong? What went wrong is, whenever you cancel you can only cancel if you’re not dividing by 0. So
the moment you canceled, you made a promise, you made a covenant with the problem. That covenant was that
you’re not dividing by 0. So, by canceling, you’re promising me that x does not equal 2. So, if you’re going to do any
kind of algebraic gymnastics here, which is great, you have to remember to play by the rules. Which means that if
you’re going to cancel a common factor, you must record that x can’t equal 2 and therefore it would appear right here.
If you didn’t do that you might have gotten the answer of just all the real numbers except for -5 and that wouldn’t have
been quite right. So really be careful. When you cancel things away, those things you’re canceling can’t be 0.
Otherwise, when you do this you’ll see the correct answer, everybody except -5 and 2. Enjoy finding the domain of
these functions.