HELLO again. PreCalc lesson number four of the day or something like that. So let's talk
about determining the domain of a function without the use of a graphing calculator.
Now domain is the x's that a graph can take on. Well, equations makes those graphs. If
you want to find domain without the, let's call it a crutch of a graphing calculator,
you need to look at the equation itself. There is really only two problems or issues when
you are looking for the domain of an algebraic equation or function. Now unless of course
you are talking about trig functions which are a whole different story. That is chapters
from now, but sine, tangent, are a little bit different. But when you are looking at
a pure algebra equation at this level of math, precalculus we are looking for the fact that
you cannot divide by zero and you cannot take the even root of a negative number. So let's
run through quite a few examples here in the next fifteen minutes, so I have got about
15 minutes. Let's see how many I can get through before our time runs out. Is there an issue
with the domain of f of x equals three x squared plus four x minus seven? Well, the answer
to this question is no. There is not a problem with domain. The domain is going to be your
x's from negative infinity to positive infinity and beyond. The reason why is because...ok.
Let's talk about... We are not going to break out a graphing calculator, but when you have
one power of two is that not a parabola. And, when x is squared your parabola will open
up or down, and therefore it will move left and right forever. So, the domain is all real
numbers. Let's take a look at another function. You know what I like about chalk boards so
much even though they make me a mess? A box with 12 pieces of chalk costs a dollar which
is way less than dry board markers. You know what else I like about chalk boards for? Because
you can make the kids cringe at the slightest drop of a hat. I love that chalk board sound:)
Anyway. Let's take a look at the next problem. We have f of x is equal to two x squared plus
six x plus four over x plus one. Now, what is the problem here? Right away we can look
at this and see that we have x in the denominator. X in the denominator? If x is in the denominator,
that is a problem. You cannot divide by zero. So let's see if that denominator can be made
to equal zero. Well, clearly if I just subtract both sides by one, x cannot equal negative
one. So my domain is going to be all my x's from negative infinity to negative one, I
can skip over that, and I can pick back up on the other side of negative one and go to
positive infinity. OR BEYOND! So why am I leaving you this extra space here? Well, actually
I don't think that I left enough. Let's see if I can fit this in. The top, I want to simplify
this function here and show you sometimes the domain issues, or the domain restrictions,
kind of disappear but you still need to pay attention to them. A lot of times when the
denominator equals zero, or can be equal to zero, you have some form of asymptote, whether
it is a vertical or a slanted asymptote. With this though, we are not going to be left with
this denominator. If I factor the numerator, which is going to be... Well, you might want
to see that reviewed. I just did this in another video, I think I used this exact same function
but let's see. I told you that you could take the first and last coefficient/term, multiply
them together. This is my scratch work on the side. Look for factors of that product
that add to the middle term. Well, the middle term is 6x. So if I use two and four. This
is the same example if you just watched the other lesson. Two x squared, I am going to
use these and put x's next to them. I am using the two and four because they add up to six.
So, plus 2x from there plus 4x from there plus four all over x plus one. Now with these
four terms, the way that I have split up the six in the middle, I can do factoring by grouping.
These two terms both share a 2x. 2 x squared divided by 2x is x. 2x divided by 2x is one.
Then these both have a common factor of four, so we factor that out. Now look. We have two
big terms which also both have a factor of x plus one. I can take that out and write
the x plus once instead of twice, leaving me with 2x plus four all over x plus one.
Guess what? Our denominator cancelled out. Now if you think 2x plus four is the answer,
you would right. But, the 2x+4 implies that the domain is all real numbers because there
is no division left. That undefined value of negative one, actually if you looked at
the graph would be a hole at negative one and not a vertical asymptote. But that is
beyond the scope today's lesson, other than we just want to make sure that x cannot equal
negative one for this domain. So, let's take a look at the next problem, shall we??? The
next problem I would like to go over with you today is f of x is equal to the square
root of x squared plus three. Well now you go, woooh. you just told me a second ago I
cannot take the even root of a negative number. So what is inside has to remain positive.
X squared plus three has to remain positive. Well how good are you with your math? You
take a number and you square it, it stays positive. Then you add three and it is still
positive. So there is no way that this will come out to be less than zero so the domain
is all real numbers. But if you don't catch that and you are actually trying to solve
it you are going to bring three over to the other side. And, if you still don't catch
it you will try and square root both sides and haha! At some point you are going to realize
that you cannot square root a negative number unless you are working with imaginary numbers.
Right now, we are not. So, there is no problem with the domain. This will never be negative,
so I can plug in anything that I want. If I can plug in anything that I want, then the
domain are our x's from negative infinity to positive infinity AND BEYOND!!! Alright,
so moving on to the next example. Let's take a look at a square root that will have a limit
on the domain. So, the square root of x square minus four. Ok, nice four. That is what I
get for looking forward while I am writing. Once again x is in the square root symbol.
So, what is inside the square root symbol, the even root, has to remain positive. So,
x squared minus four is greater than or equal to zero. That means that x squared must be
greater than or equal to four after you add the four over to the other side. Then finally
we got that power of two. We need to undo that. I am going to square root both sides
and don't forget you have both a positive and negative answer. So, x has to be greater
than or equal to two or x has to be, I am changing the direction of the sign for the
negative answer. Don't forget that with even powers and you are working with inequalities.
x has to be less than or equal to negative two. So there is my domain. X has to be greater
than or equal to two or less than or equal to negative two. Negative two? uh. So what
does that mean? Just means what it says. From negative infinity up until negative two, and
it can be equal to because I can square root zero, or you can use values that start at
two and go to infinity. Of course we all know that is ...AND BEYOND!!! So if you plug in
say three, three squared is nine, and nine minus four is positive. You can take the square
root of it. If you plug in negative three. Negative three squared is positive nine, not
negative nine, and positive nine minus four is also going to be positive so BOOOM! There
you go, these values will work in this equation. Domain. Alright. I don't hear my warning bell
going off, so that means that i have time for more examples. Oh, let's see what else.
F of x is the square root of x plus five over x plus two. Well, now this equation gives
us both problems. We can't divide by zero and we can't take an even root of a negative
number. We gotta check both of them if you have both issues in your equation. X cannot
equal negative two. Now the numerator. Let's highlight that. X plus five must remain greater
than or equal to zero. So, when you solve that inequality x has to remain greater than
or equal to negative five. Now x can be greater than or equal to negative five, but negative
two is also bigger than negative five and you cannot use that value. So, our final domain
is your x's such that it is from negative infinity up until negative five. OOOHH, no
it is not. NO, NO, NO.... Negative five is the smallest x I can use, sorry about that.
So, negative five, I got infinity stuck in my head, negative five up until we get to
a value of negative two. And then I can skip over that two because I can't use exactly
negative two. Pick back up on the other side of negative two and go to infinity....and
beyond...yes I know. Alright, so there is my domain for this function. You have to check
the radical if it has an even root and you have to check that you do not divide by zero.
There you go. These numbers I can plug into this function and not make it undefined. Alrighty!
I think I have time for one more example and I won't scratch my nails down the board. I
will just finish with a big ol' BAM!!! That is f of x is equal to the square root of x
minus seven plus the square root of negative x plus two. Let's see how this works. Again,
even roots. Let's put a four in here. I am tired of writing square roots. Even roots.
You can't even root a negative number. So, once again this has to remain greater than
or equal to zero. So, x minus seven has to be greater than or equal to zero and negative
x plus two has to remain greater than or equal to zero. What do we have? This says that x
has to be, after you add both sides by seven, greater than or equal to seven. Here we are
going to need to subtract by two. Negative x is greater than or equal to negative two.
We are going to divide by negative one. I am going to show that step because when you
divide by a negative number, please remember to change the direction of your inequality.
We get x is less than or equal to two. Ok, so I have to meet both of those restrictions
if I am going to be find the domain, or excuse me, work out this equation. Both of these
radicals are in the same equation, so I have to meet both requirements. Whoo.. Well, I
can't!