BAM!!! Mr. Tarrou In this video we going to do two examples of testing for either you
know think of it either, you know continuity or discontinuity, whatever. Is a graph continuous
and smooth at any given point or through the entire function itself. so, Limits and continuity.
A function "f" is continuous at "a" when 3 conditions are satisfied. Is the function
itself defined at a particular value "a", and that means is a in the function of the
domain, or is f(a) equal to a real number? The limit as x approaches a of f(x), does
it exist? are the left and right hand limits, when you come together, and you come towards
a from the left, and you come towards a from the right, are you getting the same limit?
Or is it approaching the same y value from the left and the right? Does it exist? If
their equal, if the left and the right limits are equal, well then yes, the two sided limit,
that's when you don't make this little exponent notation here, does it exist? And finally,
if the function is defined at a, and if there is a limit, are they equal? If they're equal,
than the function at that particular point is continuous. Now what the heck does that
mean. Well first of all, most functions are smooth and continuous everywhere within their
domain. Think of a parabola. Parabola, like f(x) is equal to x^2, I would like to sketch
a graph quickly that's correct. So let's just say it's really simple, like y equals, or
f(x) is equal to x^2. Parabolas, we know what those look like. They are smooth and continuous
everywhere, I can draw that graph without ever having to pick up my pencil. So, that
would be smooth and continuous everywhere, no matter what value of a I choose. If we
look at a square root function, f(x) is equal to the square root of x, now there's a restriction
on the domain here, I can't square root a negative number. But I can square root 0,
and I can square root all positive real numbers. So, when a textbook gives you, you know, some
kind of radical symbol. If there is no indication on the beginning of that symbol, then you
are only going to be expecting, or accepting, the positive answers. So, y equals, or f(x)
equals the square root of x, looks like this. That function is, once you get in the domain
where x is greater than or equal to 0, it is smooth and continuous everywhere. So, most
functions at most points are smooth and continuous. Really their only issues are. Uh, if we have
a quotient, we can't divide by 0, so if your domain is, this domain is x where, you know,
let's say with interval notation, it's negative infinity to positive infinity. But if your
domain is, well, x can be anything except for 4, where you can see where there is like
only one particular value, that's omitted from the domain. That is the place where you
want to check for continuity, unless, in my first example I'm going to tell you where
to check. Um, and then in the next example, we're going to figure out if a function is
continuous at any point that we think it could be. So, if your domain is like all real numbers
except x equals 4, where that value maybe makes the function undefined, then we need
to check for continuity there, or piece wise functions. That's where we are really have
our biggest issues, where we need to spend some time and check for continuity at a particular
point. So that's what continuous is. I can take my pencil, lay it down and make a smooth
continuous curve and never have to pick my pencil up. That is is a smooth and continuous
function. A discontinuous function. Maybe we have a line with a hole in it. To draw
that line, did you see I had to go back and erase it? I had to go through and maybe at
this value of a I had to pick up my pencil and draw a circle and then continue on. Or
it is a piecewise function where at the value of a I come up, I stop for some reason, and
then there is some other kind of function that takes over after I get through the value
of a and I am on the right side. My left and right hand limits are not equal, and I had
to pick up my pencil. Graphically it is very easy to see continuity. Is there a break?
Did you have to pick up your pen or pencil? Algebraically there are three steps. Let's
see what those look like with two examples. BAM!!! For our first example we have f of
x is equal to x plus 2 over 4 x squared plus 7 x minus 2. We are going to check for continuity
at the value of negative two and we are going to find if it is continuous at the value of
one. I am verbally just going to talk to you about if i just gave you this function and
said, "Check for discontinuities." Remember what I just said for the first check. The
function must be defined at a particular value of a. Well that little y equals the square
root of x, that is not defined at all real numbers but it was a continuous curve once
you started at zero and moved on to the right. So if I wanted you to check for places where
this was discontinuous, I would be taking a look at that denominator because you cannot
divide by zero. Setting a denominator equal to zero and and seeing if there are any values
of x that must be specifically omitted from the domain. At those points, double check...and
it is probably going to be discontinuous at those points...but double check for discontinuity
there through these steps I am about to do right now. However we are not given a function
and asked to find where it is discontinuous we are just supposed to check. So, is it continuous
at a equals negative two. I have not left much room here so might have an idea of what
is going to happen. Then we are going to check at one. Step one, is the function defined
at that value of a? So f of negative 2 is equal to negative 2 plus 2 all over 4 times
negative 2 squared plus 7 times negative 2 minus 2. We have negative plus 2 is equal
to zero. You cannot stop there because yes when the numerator of a fraction is equal
to zero generally the entire fraction or quotient is equal to zero, but you cannot divide by
zero. So if this denominator ends up being zero, then this function is undefined at the
a value of negative two and we fail our first check and we can just stop and say no it is
not continuous at this a value. Ok, so we have negative 2 squared. Negative 2 squared,
negative two times negative two is positive four. So four times four is 16. Seven times
negative two is negative 14 and then another negative two. So we have zero over zero, which
is undefined at that point. So this function is not even defined at negative two so it
cannot be continuous at that point. Ok, so at an a value of one. Let's go and check that
one. Step one again, is the function defined at that value. So f of one, one plus two all
over four times one squared plus seven times one minus two. One plus two is equal to three.
One squared is one, so four times one is four. Seven times one is seven and minus two. We
get four and seven is eleven minus two is nine. So yes...let me double check my notes
here...excuse me three over nine is one third. I know that! So the function does exist at
one and it is defined and it's value is one third. So now we are going to find the limit.
The limit as x approaches one of x plus two all over four x squared plus seven x minus
two... There is a little bit of construction so hopefully you can't hear that drill running,
that is what the hum is. So the limit of this quotient. Well if you want to find the limit
of a quotient, you find the limit of the numerator and the denominator separately. So the limit
as x approaches 1 of x plus two over the limit as x approaches 1 of four s squared plus seven
x minus two. We are going to plug one into the numerator, inside this limit function
now is just a polynomial so I can just plug in 1. Again we are letting x approach infinitely
close to 1 when we find limits, but this is a polynomial so we can just plug in 1. Plug
one into the denominator and we get four parenthesis...man this is sounding familiar, didn't I just write
this? Oh, yes I did. So that limit is going to be 1/3. Since the limit as x approaches
1 of f of x is equal to f of 1, this function is continuous at the value of 1. That means
again the graph goes continuously through that a value of 1. You would not have to pick
up your pencil and have a break in the graph and pick it up somewhere else. That is the
end of my first example. Let's do one more checking for continuity. I will leave that
blooper in there...OH...I will leave two in there. I am talking, thinking, writing, sometimes
the brain goes all wacko. Let's try that again. Continuous at the a value of 1. F of 1. Less
than negative one, use f(x) equals negative two. x is greater than or equal to negative
one and less than one, use the second piece 3x plus 1. When x is exactly equal to one
or greater than we are supposed to use the last piece which is x squared. So f of 1 is
equal to one squared which is equal to one. And yes from my picture I can see that was
supposed to happen. Left and right hand limits. That is step one. Step two. The limit as x
approaches one from the left... approaching one from the left, that means that I am using
values that are less than one...that means that I am using the second piece. So if I
am finding the limit as x approaches one from the left which means we are plugging into
3x+1 so we get 3 times 1 plus 1 which is equal to 4. Now we are going to find the limit as
x approaches one from the right. If we are approaching one from the right, if we are
coming in here at one...if we are coming in from the right again my x values are bigger
so when I am using x values that are greater than one, I am using the last piece of the
piecewise function. We have x squared. Take that value of 1 and plug it in. And our left
and right hand limits are not equal, so the limit as x approaches one of f(x) does not
exist. The left and right hand limits are not equal. Now again just one more time before
this video is over with, you see me actually using the x value of one but we are approaching
one. We are getting infinity close to one from the left hand side, so as x is becoming
.9999999999 but I don't want to write that out so I am just writing one. My limit notation,
you are supposed to understand that we are using..technically using..values that are
infinitely close to one from the left so they are just a tiny bit smaller. Here approaching
one from the right, even though I am using one because we understand the idea of limits,
it is like I am using 1.00000000001 but we are just rounding it off to 1 and using the
appropriate piece of the function to match that x is less than or x is greater than one.
I am Mr. Tarrou. Hopefully that helps you check for continuity of functions. BAM!!!
Go Do Your Homework!