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[Music] in this video we learn about continuous functions

we begin with what continuity is and then explain the intermediate value theorem

stick around to the end where i work through some fun exercises [Music]

okay welcome back this is the second episode in the

calculus one explore discover learn series and so today we’re going to

be talking about continuity continuous functions work through lots of examples

we’re going to start off with what continuous and discontinuous

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is we’re going to talk about limits of composition of functions

and we’re going to talk about left and right continuity

and then at the end we’re going to talk about the intermediate value theorem

and then followed by some examples you got to see the examples at the end stick

around to the end i hope you do all right so let’s get started [Music]

okay so um first we’re going to start off with a function is continuous

so what does that mean we’re going to be talking about continuity and

discontinuity and we’re going to be working a bunch of examples

and so you know i want to emphasize in this video here not only how to work

out a problem like brainstorming it but also how to write it up and how to make

good graphs and things like that so let’s start off here with uh the

definition of continuity here and now you’ve probably seen

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something like this when you’re in precalculus because in pre-calculus

although you don’t study limits usually so you don’t have a formal definition

and so your your your unrigorous definition or your intuitive definition

your intuitive meaning of continuity was something probably like this if i

can draw it without picking up my pencil right so that’s continuous that’s

continuous but what are we talking about here formally in terms of calculus

and the last video we studied limits so we’re going to use limits of course

to define what continuity is so a function is continuous at a number c

means that we can plug c into our function in other words f of c is defined um

and the two-sided limit exists we talked about two side limits in the last video

um and then the third condition is um we need that two-sided limit not only to

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exist but we need it to be equal to the functional value the output

that you get from plugging in c so we need all three conditions to hold

and if all three conditions hold then you say a function is continuous

at that c so um we know there’s some ways you can reword

that and so it’s important i think to see the

the way it can be reworded so i think number one also says so

you know when you say f of c exists you can also say

c is in the domain of f that would be another way of saying that

number two another way of saying number two

that the two-sided limit exists is to say that

both the one-sided limits exist and they’re equal to each other

right we talked last time about how if two if if a two-sided limit exists if and

only if the one-sided limits exist and they’re equal to each other

and the third condition well that’s just an equation there

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and so there’s there’s another way of writing the

continuity definition for continuity at a point or

number c so we have three things and the emphasis is there is on all

three conditions if one of the conditions doesn’t hold then

it’s not continuous and if it’s not continuous then we’re going to call it

discontinuous so a function is discontinuous

if one of those three conditions doesn’t hold and we’re going to say has a

discontinuity at c and the next thing is well

we talked about continuity at c what about continuity

on an interval so we’re continuous on an interval

precisely when it’s continuous at every number

in the interval and you might say how can i check every number in an interval

aren’t there infinitely many numbers in the interval yes well

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there are um but we can talk about that and also the thing to notice about this

is that it says an open interval so we never say something is continuous

on the boundary on the left endpoint or the right endpoint

it’s always in the interior of the interval on the open interval

all right so we’ll talk about that a little bit more later but right now

let’s talk about what happens if we have the functions

and we have the operations on the functions

how does continuity um compatible with our operations right

so if we start with two continuous functions we may ask

is the sum of those functions also continuous

and what about the difference and what about the product and what about the

quotient well they’re actually all continuous even the composition

they’re all continuous at c so if you start with two continuous functions

then all those combinations of functions with those operations

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are also continuous um you know provided you stay within the domain of the

function so if you look at that quotient f over g then you have to obviously be

careful um at the domain of g and make sure that that function is

defined and if it is then we’re going to be continuous so

if a function is a polynomial right so here are some examples of some

type of functions and you know what’s the domain of a polynomial polynomial is

has domain all real numbers and in fact the theorem is that the polynomial

function is continuous on its domain which is all real numbers

so every polynomial is continuous along the whole real line that makes it

very useful rational functions are also continuous

of course you have to stay away from the isotopes so you have to stay within the

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domain of the rational function trigonometric functions like sine and

cosine their domain is all real numbers and the others four trigonometric

functions you know tangent cotangent secant and cosecant

they’re also continuous as long as you stay within their domain

and same thing for the inverse trig functions and so if f is continuous

for any of those functions f any of those types of functions

as long as it’s defined it’s continuous now this is an extension

of the limit laws that we learned last time

in other words we can apply the limit laws and obtain this theorem here um so

let’s look at some examples so we’re going to talk about some examples of

some continuous functions um so here’s the polynomial 2x squared

minus 2x plus 5. that’s a polynomial that’s continuous everywhere

along the whole real line all real numbers there’s a rational function it’s

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continuous everywhere except at zero at zero we’re going to have an

uh vertical isotope so it’s not going to be continuous there

and then c we have a trigonometric function

cosecant so that also has um vertical isotopes

so as long as you’re not on the vertical isotope you’re at a

number c that’s in the domain of cosecant you will be continuous

at that number same thing for the inverse trigonometric function whatever secant

inverse looks like it is continuous on its domain

so those are continuous functions on their domain for example

and we just saw that theorem where we said we can add and subtract and

multiply and divide them right so this is also a continuous function

this takes the function in a and the function in b and the function c

and combines them together b times c plus the function in a right

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so if you combine all them together because each of the individual functions

are continuous and they’re the sum and product of two continuous functions

so that function is also continuous so is this function

this is the composition we took the function in b

and substituted into the function in d and that’s a continuous function

and here’s another continuous function we took a

times b and a and b are both continuous functions

so the product is also continuous honest on the on their domains of course

so it’s very important when you’re talking about continuity what the domain

of a function is when you’re talking about continuity of you know

more than just a point all right so what about discontinuity if something is

discontinuous one of the three conditions in the

continuity definition doesn’t hold right so what are some ways

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so if we look at this function right here it’s discontinuous at one

well we talked about that that’s a rational function

we have a polynomial over a polynomial so it’s going to be continuous everywhere

on its domain but you know one is not in the domain

right so f of one is not defined so that would be condition one

of the in the definition of continuity and it fails

so that function right there would have a discontinuity

it has a whole that’s one type of discontinuity

another type is well let’s look at this function here

we have two parabolas x squared plus one and negative x squared minus two and

they’re pieced together they’re pieced together at zero so you might be

suspicious on how they’re pieced together you know

if you look at each polynomial each parabola by itself

it’s going to be continuous everywhere x squared plus one is continuous

everywhere and negative x squared minus two is continuous everywhere

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because it’s just a polynomial so how is it pieced together at zero

so if we look at the limit from the left of this function right so that’s less

than zero so we look at the second piece

and we plug in zero we get out minus two and we look at the right piece

we approach from the right so that’s when we use the top piece the x squared

plus one we substitute in zero we can find those

limits those one-sided limits there and so then we can say the two-sided

limit doesn’t exist so this is an example of a function that does not satisfy

rule number two in the continuity definition

all right so we have a discontinuity zero this is a different type of

constant discontinuity it’s called a jump so we have one more type of

discontinuity and it’s an asymptote so this function

right here is a rational function it’s continuous on its domain and so if

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you look at something that’s not in its domain for example two

then it’s discontinuous and this type of discontinuity is called a pole

and that’s because we have unbounded behavior or

we have an isotope so we’re going to approach infinity

as we approach two from the right so there’s the walkthrough of three

different types a whole a jump and a pole and so here’s uh here’s what they look

like graphically right so there’s the first type of

discontinuity and i use software to to look at that graph there now looking

using software it’s kind of hard to tell that there’s a hole there

right so you know you have this line and it’s missing something

at some number here has a hole in it and the way we usually do this is we

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draw a little circle or a little put a little hole there

to represent that it’s missing right it’s not a circle it’s just

there’s a hole missing there um so right there i believe it was zero right

or or at one yeah f of one is not defined

so we’re looking at that first function right there

so we have a hole right there at one so back to this there’s one right there

and a two and so on so at one there there’s a hole there now for the second type

we have a jump so again using software to graph it now

if we want to use our hands then what are we doing here

so it’s at one and two here so we’re looking at one

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we’re looking at two and we’re asking yourself

because both of them cannot be filled in because it’s a function right so one of

them has to be open and perhaps both of them are open but we

certainly cannot have both of them so if we look back

we’re gonna have a jump here if they’re if this one’s filled in we still have a

jump or if they’re both open we still have a

jump no matter how you look at it we still have a jump right here

so in other words intuitively you have to pick up your pencil to keep drawing

this right but formally we look at the limit from the left and

we’re approaching this height and we look at the limit from the right

and we’re approaching this height and those heights are different so the

two-sided limit here doesn’t exist so this is called a jump this is a jump

and then we have one more type of example so this is a pole

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so here i have an isotope a vertical isotope at x equals two um

a vertical isotope of x equals two and as you can tell as we approach

two from the right the function is just growing unbounded

and if we approach two from the left that the function is just falling down

right so the limit from both sides to two doesn’t exist and and it doesn’t exist

because of the limit from the left is not equal to limit from the right

uh and there’s unbounded growth so we’re

going to call this type of discontinuity a pole and so if you look at these two

slides together here this one with um all the explanation and this one

right here with all three graphs of those functions right there

you get a you know good picture of just discontinuous discontinuity and how how

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it might arise a hole a jump or a pull all right so let’s go on to a different

type of problem now let’s talk about how do we come up with a function that

looks like different the different parts of it look like something that we want

for example the first piece there looks like a parabola and the third piece

looks like a parabola and i’m trying to piece them together

right so how do we do that well one way is to put some parameters

into our problem and find those parameters in other words

i want to take two parabolas you know just just intuitively

what are we thinking about here right so i have a coordinate system

and then i have a parabola and then i might have another parabola

and obviously they can’t overlap so i’ll just chop this off

and maybe i’m interested in a point here like or a number here like two

and obviously there’s a jump there right so i want to curve this parabola down a

little bit harder and i want to piece them together in a continuous way

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so i take this one parabola and say another parabola

and how do you piece them together so you kind of parameterize the

quadratic the parabola using some parameters

and we want to try to piece it together to get a continuous function so how do

we do that so we don’t know what a and b are in

fact we don’t even know if we can find any a and b

and we’re piecing it together at 2. so let’s see how to piece them together

to be continuous right so the first thing i want to notice is that

if there are an a and a b that exists to make this continuous

well the function is going to be continuous f everywhere because it’s made up of

polynomials as long as you can piece it together at

2 in other words the only place where this function

if a and b exists the only place where this function

might be discontinuous is away from 2. uh sorry the only place this function

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might be discontinuous is at two everywhere else it’s already continuous

as long as we can find a and b right so in other words f is continuous for all

x not equal to two as long as we can find a

and b and actually we will find an a and b we’ll see

so for any a and b that we choose so let’s find the a and b i claim i

claim we can find it right so the first thing is you know actually

before we look at that let’s look at our board here so i like to do this here

so you know what is our definition of continuity

so we have three three parts to it right

so we need f of two to exist so the c in our problem is a two

everywhere else we’re already continuous as long as we can find a and the b

so we’re just curious about the two what is f of two

and then we’re also looking at the limit as we approach two

and then the third condition is the the numbers that you get in one and two

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have to agree so this is a two here so we need to check those

those three things there right so what is f of two for our problem

here right so for our problem the 2 f of 2 is the 4.

when you plug in 2 you get out the 4. right so we have a 4 there now what is the

two-sided limit here well our function is pieced together so what we really

need to do is to look at the limit from the left of two

look at the limit as we’re approaching two from the left

and look at the limit as we’re approaching two from the right

and so what are those limits and we need to find those limits but those

limits need to be equal to four those limits need to be equal to four

because we don’t want to jump so the two-sided limit if it’s going to exist

it has to be four and if the two-sided limit is four

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that means the one-sided limits also have to be four

but we can actually find these limits right here using the pieces that we have

so we approach 2 from the left of the function

what does the function say when we’re looking at 2 on the left hand side we’re

going to use that third piece so we’re going to use the x squared minus a

x plus b so if we you know approach 2 from the left we’re going to be looking at

4 minus 2a plus b so i’m writing that here 4a 4 minus 2a plus b

and then if we’re looking at the limit from the right there

now we use the top piece a x squared plus b

so when as we’re approaching 2 that’s going to approach

4a plus b so that’s we have here we have 4a plus b here

so what we have here are these equations here

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which are missing a and b and that will allow us to find the a and the b so

you know this gives us zero right subtract four minus two a plus b

and then we have four equals four a plus b and if we just subtract these to

eliminate the b so i’ll say zero minus four then i’ll say minus two a

minus four a and then the b subtract off and so from this right here we can see

that a is just simply two-thirds right so that’s what the a is and then we can

come back over here or here and we can find the b so the b will be

four minus four a and a was two thirds so this would be twelve thirds which is

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four minus eight thirds right so b is just four thirds so we found the a

and we found the b by looking at the limit from the left and the limit from

the right and coming up with equations for them and knowing that

those equations we get the 4 because that’s the height of the function

and we needed the two-sided limit to exist all right so we need to look at the

limit from the right which we did and we got 4 for that which is the same

thing as plugging in two to make sure there’s no jump and then we

look at from the limit from the left and we get the system four a plus b

equals four and that was the other equation we got there

so solving that system of equations solving that we get a and b

and we get the two-sided limit to agree because it’s

both the limit from the left is 4 and the limit to the right is 4.

we’re guaranteed that to happen for that a and b

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so there’s our function if i substitute in the a and the b everywhere

there’s our function right there there’s our two parabolas they’re pieced

together and if we were to go graph those two parabolas and

we see the shape there it almost looks like one continuous parabola

but it’s actually two parabolas pieced together in a continuous way

all right so let’s look at another example

so in this example here we’re going to find constants a and b

and we want this function to be continuous and in this example we’re only

worried about continuity of one so we don’t need to argue anywhere else

so again i’m going to write down my definition of continuity

so here we go so condition one is f of c exists

now what are we using in our problem here we’re using

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we’re using f and we’re interested at one so this will be f of one exists

condition two is the limit from both sides of one exists and condition three

that those values have to be the same so let’s start working on that this f of

one exists so let’s go back to our function and our function says

f of one is three so that exists that’s good and we need to look at the

limit so f f one is three that exists so check

and we need to look at the two sided limit now to look at the two-sided limit

because again we’re looking at a piecewise function

if we weren’t looking at a piecewise function we don’t need to look at these

separate limits but because our function is a piecewise

function we got three pieces to it there right piecewise function

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so let’s look at the limit from the left and the limit from the right

so let’s find these limits here now whatever these limits are we have to

also look we have we have to also make sure three holds right so the limit

has to also be f of one which was three so

the two-sided limit has to be equal to three so the one-sided limits also have

to be equal to three so these limits have to be equal to

three but on the other hand i can try to find these limits because we’re given

the function right what does the function say the function says so if we’re

approaching one from the left which piece are we going to use

one from the left so we’re going to use that bottom piece there

so this is the limit as we approach one one from the left of that bottom piece

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there which is x squared minus four x plus b plus three

right so that’s what that’s what i’m getting right here

so the the function right here when we’re approaching from the left

we’re just looking less than zero so there’s the piece when we’re less than zero

and that has to be equal to three but what do we get for this limit right here

so we’re going to use a one minus four plus b plus three

so that’s the one-sided limit right there x squared minus 4x plus b

plus 3 which we get by looking at the function

where it’s less than 1. and now we find this limit right here

we substitute in one and one and b we get b plus three

and so what does this come out to be so this is minus three so that just comes

out to be b right so we get minus three plus three so that’s just b

so we know b is 3 now so what about this limit here now

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so now this limit from the right what’s the piece what’s the function

look like when we’re greater than one let’s go take a look like uh what the

function is um when we’re greater than one here so when we’re greater than one

it looks like ax plus b all right very good so we have ax plus b

here and now when we try to find this limit here

we’re just going to get a plus b and so now we know a plus b is three

so a plus b is three all this is equal to this

which has to be three because that’s what the two-sided limit has to be

so that tells me that a is zero so very good now we have our function

all right so let’s go write this up so to have continuity at x equals one

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we must have the limit from the left is equal to the output

which is three and we must have the limit from the right

which is the output which is equal to three

and that tells us by looking at those equations that a is zero and

b is three and that will make sure that we have the two-sided limit

which is three which is equal to f of one that will guarantee that

we will have continuity at x equals one all right

all right so now we’re going to talk about

limit of composition of functions that’s what’s up next here

limit of composition and so we’re going to have this wonderful theorem here

um now perhaps composition of function isn’t your

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favorite operation with functions right you have addition subtraction

multiplication and division which you’re used to working with with

numbers and so naturally those are straightforward or much more familiar to you

um but composition of functions right so let’s look at that real quick so here’s

what i’m talking about when i’m talking about

composition of functions so let’s say we have a function

f is say x squared and here’s the function g of x which is say sine x

so we could write this function here and this function right here will be f

composed g which will be f and then here’s the g of x which is just sine

so this function means place g inside of f so if i place sine x inside of f

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i get sine x and sine takes whatever you input it and you square it

of course this is usually written as sine squared x

and so we also have g composed of f here’s another way to compose

so this says f goes inside of g so g and then we have f of x first so f goes

first and what is f of x it’s x squared now x squared goes into g so g takes

something and it takes sine of whatever you give it so if i give it x

squared now we’re looking at sine of x squared

so as you can see these two things are not always equal to each other

we have the composition of two functions and we in whichever order you do it

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matters whichever way you do this matters so just a brief refresher there

so here’s what the limit says the limit of a composition so

we need to know that the limit of the function g

is exist is equal to l and we need to know that f is continuous at

l and then we have the composition this is a limit law here

but it uses continuous so that’s why we’re seeing it

this time as compared in the last video so this says that we can [Music]

pull the limit inside the function f as long as f is continuous

that’s an intuitive way to think about it so it’s equal to f of l

so if you know what the limit of g is which is l then to find the limit of the

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composition you just do f of l you just plug the l into f

and that will help you find the limit of the composition

so you know let’s do some examples to make sure we’re clear on this

so on a we have the limit as we approach three and our

function is the composition of two functions it’s

x squared plus three and that function is squared

so we have a function sitting inside of a function

so let’s try to identify um what everything is here

um in terms of the f and g and look back at the theorem so the theorem says

that the limit as we approach three in this case of f composed of g of x

right so that’s what the theorem says here

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the limit as we approach 3 of f composed of g of x that’s equal to f

of the limit as we approach 3 of the g of x here

so i find this limit first whatever it is let’s call it l that’s just f of l

all right so let’s see that here with this example here

so what is the g that’s going inside of the f so i’m going to say

that g is the inside function here so what do you think g should be what do

you think f should be right so here’s the problem right here

we’re approaching three of this function right here

so what’s the inside function what’s the outside function here

so let’s say g is the x squared plus 3 and the f

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is just squaring something x squared so this will be the right so if i do f

composed of g with these two right here we’re putting g into f

so this is f of g of x so this is g right here so i’m gonna replace

g right here it’s x squared plus three and then i’m gonna take that and put it

into f and f just squares something so i’m gonna square that

so that’s what we have right there so i’m trying to find the limit of x

squared plus 3 squared trying to find the limit of that but

it’s the same thing as the limit of this all right now by using the theorem i

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just need to find the limit of the g function and then substitute that into f

right so how do we find the limit of g as we approach three right so as we

approach three we’re going to get nine plus three so

that’s twelve so i’m gonna do f of twelve and then that means i need to plug 12

into the f which is 144. so that’s the understanding behind the

limit of the composition now you might just say hey wait a minute

i’m trying to take this limit why don’t i just plug in three

i plug in 3 so i get 9 plus 3 squared which is you know

12 squared 144. you know you get the same thing either way

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but how do you know you can plug in three how can you go straight to the x

there’s a two hanging out here you know you gotta square that

so how do you know you can go right into that x well the reason why that you can

take the shortcut is because well we’ve worked out all the

detail it works we know that this works so we’ve worked out all this detail and

we worked it out and we proved it in the theorem

and so once we have this theorem the composition limit theorem now now you

just go and use it you don’t need to work out these details all the time

you don’t need to work out all these details you can just go straight and use it

but how do you know how can you justify this step right here

so this step is justified by that limit theorem

so yeah just plug in the three and you’re done and you can do that

because of that limit law there all right so

so we’re going to let g be the x squared plus 3

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and we find the composition using the theorem and it just simply means

plug in the 12. plug in the three all right so what about the second one

so now we have sine to the fourth so the inside function the g is sine x

and we’re just going to take the limit of the sine x

and then that is square root of 2 over 2 and then we’re going to raise that to

the fourth so it’s 1 4. so that’s that way is a lot shorter

right we don’t need to go through the um [Music]

all this work every time i just want to go through that for you to make sure

that you understand how this is justified right here how can

you just go and plug it in why why is that true and so you can see

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this example right here of the of the theorem all right so

okay so now we’re going to talk about left and right

uh continuity and so this is just an extension of one-sided uh limits

where we’re looking at limit from the left and limit from the right

so we’re going to say a function is continuous

from the right and that’s just simply going to mean that the

one-sided limit is equal to the functional value the output so

and you know normal continuity continuity from both sides

means the two-sided limit is equal to f of c

the output value so if you just restrict yourself to looking at one of the sides

left or the right then you can talk about continuity from the left to the right

so continuity from the left and continuity from the right

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so what’s an example right so what’s an example of some function that’s

continuous from the right at zero so how about the square root of function

right it just looks like it’s going up there and we look at the

limit from the right and the square root of zero is zero so

that function right there is continuous from the right

so you know you may talk about continuous from the right or continuous

from the left because your domain may require you to only talk about one side

okay so now we’re going to talk about the intermediate value theorem

um and the intermediate value theorem is an important theorem that you may have

actually seen before you saw calculus and the reason why i say that is because

when you’re talking about intermediate value theorem you need to talk you need

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to know about what continuous means however in pre-calculus and in other courses

you may have just seen an intuitive definition of what continuity means

i can draw a function without picking up my pencil for example

right so if that’s your idea of what continuity means

then you can talk about the intermediate value theorem however here

in calculus we have a very formal rigorous definition of what continuity

means we have part one part two and part three of our definition of continuity

so this here has a more rigorous meaning here

but in either case the intermediate value theorem relies

a great deal upon continuity and so you know let’s refresh our memory of

what the intermediate value theorem says

so we have a continuous function we have a closed bounded interval

and we know that l is some number strictly between

f a and f b now remember those are outputs so we’re looking along the y

axis so we have f of a is some number along the

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y axis and f of b is some number along the y axis

and l is some number in between those two

so if you have some number between those two then what you can say is there

exists a number c between a and b that gives us that l

so you can see the diagram right there so hey let’s stop draw that redraw that

diagram as i’m reading the intermediate value

theorem there right so we can kind of get a

a feel for that so let’s go over here and do that

so the intermediate value theorem starts off with

a closed interval so let’s say we have a and b here and

it’s very important we have a continuous function it doesn’t matter what the

continuous function looks like but it’s continuous on a and b so it

could be discontinuous out here maybe there’s a hole maybe there’s a jump

00:42

but between a and b it’s continuous right so there’s a nice continuous

function there now at f of a we’re right here that’s f of a the height

and at b we’re right here and now we come over here and get a another

height in fact let me make the function a little bit more exciting make it come

up and come down like that so now here’s the b here f of b there’s the height

and here’s l right here so all that is in the hypothesis

of the intermediate value theorem for ivt the ivt is an ext in existence theorem

existence theorem what that means is it doesn’t tell us what something is

00:43

it just tells us that it exists and so here’s the way the intermediate value

theorem works if i’m given the a and the b and the continuous function

and i’m given some l in between the a and the b

then what the intermediate value theorem guarantees

is that there’s some c in here somewhere where i’m equal to this height so where

would that height come across here so there’s the c we can kind of eyeball

it right there there’s the c this intermediate value theorem says c has to exist

no matter what this function is as long as it’s continuous

if i choose some l in between here i have to be able to find a c

in between these two right there that’s the intermediate value theorem

so the intermediate value theorem tells us that something must happen

00:44

so given these conditions something must happen so

we’re given the condition a and b and and

that automatically tells us what f of a and f b is

but we’re given a and b and we’re given some l in between

it tells us that amongst all these values in here infinitely many values in here

one of them c has to exist that may happen that there’s more than one c

for example what if i draw this function here what if this function is even more

exciting comes up and down and then up and then down right

so now i’m getting lots of c’s here right equals

where where the uh you know let’s call this here c1 and c2 and c3 right

so the intermediate value theorem tells us that at least one of them has to

exist where if you go plug it in and you get that out you hit that l

as the height there has to be at least one of these c’s that has to exist

[Music] okay so that’s the intermediate value

00:45

theorem there and let’s look at a real world example to see

how that could be used in every day so we’re going to talk about bacteria

you know nothing in politics so we’re going to talk about

a colony of bacteria and we have this piecewise

function here and so someone poisoned the the bacteria and

now we know the population is given by this function here

and that population function was found by doing lots of

observations and modeling and once you get this function here then

you can use calculus to make some predictions

and to me intermediate value theorem is a calculus theorem

because it involves continuity so to prove it you know you need to rely

upon continuity all right so that’s the bacteria colony

00:46

and the first question is when does the colony die out

there’s no more no more bacteria and b we’re going to show that at some time t

between two and seven the population must have hit nine thousand

um now let’s see here um the pop okay uh the population is given

in thousands right so we need to keep that in mind

and the time is in minutes so two minutes between two minutes and seven minutes

the population was the output was equal to nine somewhere between two and seven

all right so p of five why am i looking at p of five

well in order to unplug in in order to use intermediate value theorem

we need to know the function is continuous right so they ask us between

two and seven so i need to know that this function is continuous

00:47

all the way in between two and seven now the only place where this function may

be discontinuous is at five right because the first piece the t square plus one

that’s a polynomial that’s continuous everywhere

and the second one is linear it’s it’s a

polynomial so it’s continuous everywhere so the only place where p may not be

continuous is where those continuous functions are pieced together

so i’m checking continuity at five so the first thing i do is i plug in

five and i use the second piece because it says t is greater than

or equal to five right so we plug in t is five

and we’re going to get minus 40 plus 66 so we get to 26

and now to be continuous i need to check the two-sided limit at five

so i’m looking at the limit from the right first now when we’re looking at

the limit from the right which piece we’re going to use we’re going to use

this bottom piece that’s when we’re greater than five and

now we find that limit and it’s 26. now if we didn’t find 26 there

00:48

knowing that the output is 26 the p of 5 is 26 then we would already have this

continuous function and so then you wouldn’t be able to use

intermediate value theorem so the intermediate value theorem

requires a continuous function on that domain now that the function

doesn’t need to be continuous everywhere but certainly it for this problem needs

to be continuous between two and seven all right so let’s check the limit from

the left now use the piece to the right of of a five

and so we get 26 also so those agree and they agree with the output 26 so p

is continuous at 5. so now we know we can use the intermediate value theorem

so but to answer question a when does the colony die out

that is when the [ __ ] um when the population reaches zero

00:49

so we’re going to look at minus eight t plus 66 when is that equal to zero

and we solve that for t and then we get 33 over four

so the colony dies out in about eight minutes and 15 seconds so

0.25 a quarter of a minute all right so part b um we’re going to

use intermediate value theorem and that was the the work that we did

making sure it was continuous so p is continuous on 2 7 the open

interval between 2 and 7 and so we’re going to use the

intermediate value theorem and so what the intermediate value

theorem says is there has to be some number c between right

now why is that because when we look at p of 2 we’re equal to 5

when we’re p of 7 we’re equal to 10 so we’re asking ourselves

00:50

so the the l in this example is the nine is the nine between the f of a and f of

b and the answer is yes five is not five is

nine is between the five and ten so once we know that the intermediate value

theorem tells us there has to be some c between such that the population is not

all right so next one all right now it’s time to do some exercises

and we’re going to be doing these three exercises here

the first one is going to require us to sketch a graph and

look at some some information from the graph

talking about continuity and limits and stuff and then the second one is

going to find constants so that the function will be continuous

and then the last one will be an exciting example where

00:51

we break one of the theorems that we just talked about guys stick to the end

to see that last example there our last exercise all right so for the

first exercise here we have to sketch this function here and

then answer eight questions the eight questions aren’t uh hard

um in fact the hardest part of this is probably just sketching that function

there which isn’t hard but you know we need to

make sure that we get a good sketch because we got to answer all these questions

but we should be able to read off all the answers from this question

just by looking at the function there all right so here we go let’s um

sketch this graph so let’s get some axes set up here [Music]

all right so let’s write the function over here

so that we can see it and the first branch is

00:52

x squared minus one and this is between minus one and zero all right

so between minus one and zero we look like x squared minus one

now what does x squared minus one look like so that just looks like a parabola

shifted down one so it looks like that it hits one

it when it’s one it hits zero so we’re looking at it between minus one

and zero so between minus one zero we just have this little branch here

so i’m just gonna put this little branch right here so this is minus one

and then just have this curve down here now this is equal here

so we’re going to fill this in and this is strict here so i’m going to

put a hole here all right so far so good just erase that now the next branch was

a 2x and that’s between zero and one strictly between zero and one

all right so between zero and one what does this look like it’s just like y

00:53

equals two 2x right and when when x is one then the output’s a two

right so it’s just a line going through the origin

so between zero and one we’ll just look at this piece of it here

and it’s strict so both of these will be

open right so let’s go over here and say okay that’s about a one right there

and we’re looking like we have a hole there we’re just going up and this height

should be a two all right good and now the next piece is

one one one if x equals one so x equals one so we’re filled in right about

here at one one all right and the next branch is minus two x

plus four and that’s between one and two strictly so what does this line look

00:54

like so we’re looking at one to two so at one we’re going to be at two

one we’re two and then at two were equal to zero

so that would just look like a line going just line straight through there

but we’re chopped off between one and two so we’ll cut all that off

and they’re both strict so they’ll both be holes there

let’s go put that in here so we have a hole there

one two and then we’re just coming down to two so let’s say two is about right

there and we’re coming down here and it’s the hole right there at two

all right so so far so good and then greater than two what happens

nothing happens at two so between two and three

00:55

were equal to zero so greater than two and three so three there we’re going to

end there’s nothing happening greater than three

and then here we’ll just hit zero right there all right so let’s uh

make that a little bit better try a little bit better

all right there’s our function all right there’s our function there’s our graph

and i just shade this in really thick just to make sure

um oh and then we’re open right there x y alright so i think we’re ready to

answer some questions now so the first question is does f of -1 exists

question so i’m looking over here of minus one

and we’re equals right there and we can see from the graph yes

00:56

it’s zero yes f of minus one is zero so question two does

um the limit as we approach minus one from the right does this exist

so looking at minus one from the right so here’s minus one

and we’re looking from the right of minus one and where’s the graph the

graph is right down here and the limit is so as we’re approaching

minus one from the right the graph is approaching zero so yes

this limit right here is equal to zero so it certainly exists

all right so there’s one and two so far so good and so 3 says

3 says how about the limit as we approach minus 1 from the right

00:57

does that equal to f of minus one and so now that’s just looking at one

and two so they’re both equal to zero and say yes c one and two all right so 4

this f of 1 exists let’s go over to 1 and yes

f of 1 is just one we could go here or we could look at the graph

so yes f of one exists f of one is actually just equal to one

all right so number five um is f defined at two [Music]

this f of two exists oh sorry now it says is f continuous at two

00:58

number five is f continuous at x equals two

question so to to do that we need three steps so

this f of two defined so f of two is oh f of two is not defined so

no no f of two is not defined so certainly cannot be continuous at two

there’s a there’s a hole there so we would have to pick up our pencil

if we were drawing it we were drawing it there’s the hole and then now continue

so of course we cannot talk about pencil and our answers calculus right

we just say f of 2 is not defined so no f of 2 is not defined

all right what about 6 let’s try 6 down here

00:59

6 says and what values is f continuous so f is continuous

on and so now i’m going to write some intervals

um so if we look at the graph here between minus 1 and 0 it’s continuous

it’s not continuous at 0. there’s a jump so i’m going to say -1 to 0

that open interval and then now 0 to 1. it’s not continuous at 1 there’s a jump

there so between 0 and 1 is continuous so i’m going to say union zero to one

union zero to one and now let’s look at the graph again zero to one is

continuous now from one to two it’s also continuous

it just looks like this line right here so one to two it’s continuous union

one to two union one to two and now um what about two to three it’s continuous

01:00

it’s not continuous at two so now i would say union two to three

and then the function didn’t go past three so you know make sure when you say

continuous on you don’t close any of those

in other words none of them should look like a square bracket because that means

you’re including the end point and you never include the end point when

you’re talking about continuity unless you’re talking about continuity

from the left or continuity from the right so this example there’s six for this

example here all right so let’s move that up all right so there’s six

part six there and let’s get rid of seven and two here

or one and two there and let’s go to 7. so 7 here

is and what value should be assigned to f of 2

to make the function continuous so you know here’s the function right here

and it’s missing a 2 right there so it’s saying what should we how should

01:01

we fix this function so you know f 2 should be equal to so assign f the value

of so when we’re looking at 2 right here to be continuous we need that point

filled in so two we should be zero so assign f the value of zero at

x equals two then f is continuous then f is continuous at two

or continuous at two so [Music] yeah we can see that from the graph we

want that to we don’t want there to be that hole there

so we would put an equals here and if we if we change the function and

put an equals here then that function would be continuous

at two so that’s what that’s saying right there

01:02

the sign f the value of zero we could put the equals here too right

but not both places we could put it either here or here

either way we would get out the zero there all right so part eight so

to what new value should f of one be assigned to

remove the discontinuity so assign f the value of something

at x equals one so now we’re interested in x equals one then f is continuous

at one right so i have the exact same sentence here

the difference is we need to look at the graph and we need to look at one

so one here it’s already equal to a one but we need to change the function so

that it’s continuous so it would be continuous if that hole was

01:03

there so we need to assign one to be a value of two if we can move

that point up here now it’s continuous right there at one

right so we need to change that to a two so i’m going to say

assign f the value of two at x equals one then f is continuous at one

all right so there’s my part eight there and that finishes the problem there

so sometimes people will start with a function a bunch of pieces because

they’re interested in certain shapes modeling of some real world real world

problem and then they’ll you know try to have the function continuous

so you can use certain theorems like the intermediate value theorem

all right so now let’s look at exercise two so here’s exercise two um

01:04

don’t take a peek at three yet three is our last problem

exercise two exercise two has two parts to it

and so we’re gonna find constants a and b so we work some examples earlier um

like this so let’s look at a first so our function is

square root of ax plus b plus a minus 1 and then and then a one

oh that’s all over x and then a one and so we’re looking at not zero and

equal to zero so we’re looking at continuous for all x

in the domain so we need to talk about all the x’s possible not just at zero

so if you look at a the question is is it continuous at zero right because

that’s where it feeds together but we need to talk about all x and the

domain right so here we go so there’s the function right there

01:05

square root of x plus b minus one over x and we got a one here

now obviously we have to look at zero well actually let’s just look at zero

first right so what’s happening at zero so first of all to be to be continuous

at x equals zero right so what do we need we need three things

we need f of zero to exist we need the two-sided limit to exist and we need

them to agree because they agree with each other

right so there’s just my aside there there’s my definition

so first off i’m going to check at f of zero so f of zero is one that exists

so check so the next thing is i look at this two-sided limit

01:06

now we’re not looking at both sides um you know this is pieced together just

by a single point here so this two-sided limit we don’t need to

go look at one-sided limits we can look at both sides by looking at

this part right here the question is though can we find the limit

so if we look at both sides here and if we try to find this limit here we

try to substitute in zero right so but we’re going to get zero down here

and we’re going to get square root of b minus one over zero

and we don’t know what the b is yet so we don’t know

what this limit is yet um i’m actually going to try to

use the conjugate here and see if that helps us

so i’m looking at the denominator the numerator

and i’m going to say square root of x plus b

01:07

plus 1 over square root of ax plus b plus 1. so

i’m going to get ax plus b out of this equals limit now i’m going to get square

root of square root so i’m going to get ax plus b

and i’m going to get a minus 1. and then i’m going to get

x and i’m going to get ax plus b plus 1 and then make sure that’s x is

times that whole thing there so let’s put parentheses in there

all right very good so how can we get this to cancel and

have this limit to exist so if b is 1 and a is one then this limit right here

if i choose these values right here then

i’m guaranteed to have an x over x and i can cancel the x’s

01:08

so then this limit right here will be ax plus b the a is one the b is one

so that’s x plus one and then plus one so that’s

um a nice thought there if we have this a is one

and this b is one then this goes away and this is ax this is one x

this is x over x they cancel and then the a and the b

are still one right and so this limit comes out to be

what so that’s square root of two so one over one plus square root of two

oh sorry nope my bad this is going to zero right

so we’re getting square root of one um plus once we’re getting one half there

so that’s not equal to the one there which we need here if this was the one

half here we’d be in business so it doesn’t look like we’re gonna be

01:09

able to find any a and b to work here well what if a is just

an a and not choose a is one there what if we just leave that as an a

right so we definitely need the b is equal to one so we can cancel

so let’s just not assume that a is one there

we’ll say b is one and then now what do we find the limit to b

we still have ax over x the x’s will still cancel

so then we’ll still have the limit as we approach

zero we’ll have an a left on top and we have an a here and the b is one

and we need this limit to be equal to one can we find an

a so what is the limit this is going to zero

so this is a over square root of two right so x is going to zero so that’s

01:10

zero square root of one is one oops so that’s a two all right so

therefore a is two oh sorry so x is going to zero here

so this is zero square root of one is one so one plus two is two we’re getting a

over two now so i jumped the gun there on the a should be equal to one

a can be anything at this point we’re going to cancel the x’s by making b

one b is one so that goes away then we can cancel the a so we get an a and

see if we don’t cancel those x’s we’ll never get we’ll never get to anything

down here because we have a zero right there so anyways a is two and we have

the two-sided limit equal to one we have that

and we have these equal to each other for this a and the b

so let’s just say a is two and b is one and then we will have

01:11

this one here accomplished also they’re both equal to one

so we’ve tackled continuity at zero so when a a is two and b is one

um how we know this is continuous so if we look at this piece here

a is 2 so we have 2x plus 1. so 2x plus 1 and then -1

all over x right so this is continuous on its domain

so this is continuous on its domain here and we have to worry about zero so it’s

continuous um everywhere except zero so let’s just say that

because this domain is all real numbers except zero right well actually

01:12

we have to um worry about when the x is 2x plus 1 is negative right there right

so 2x plus 1 has to be greater than zero so x has to be greater than minus one

half so knowing that now for that a and that b um

we’re never going to find a solution um for this a and the b there will be no

value of a and b where this function is continuous everywhere

so the problem here is let’s go back and read the problem

so for part a here if you’re asking for continuity

at zero only the a will be two and the b will be one

however if you’re asking for part a that function to be continuous on its domain

right so its domain is x is greater than or equal to minus one half

01:13

that will be the domain and the a will be two and the b will be one

and so that’s part a there okay so for part b now let’s look at part b

so part b we’re looking at the function f of x and it looks like ax plus 3 5

and x squared plus b and we’re looking less than 1 equal to 1

and greater than one all right so again we’re looking for continuous for

all x in the domain and this time though we’re broken up

to the left to the right of one so i’m going to be looking at the one-sided

limits right so but again we need f of one exists

it’s number one we need the two-sided limit to exist

as we’re approaching one from both sides we need that to exist

01:14

and then part three we need them to agree all right

so let’s get started first off what is f

of one f of one is five f of one is five got that part check um the two-sided

limit needs to exist at one so i need to look at limit from the left

and the limit from the right so let’s do that over here

so if we look at the limit from the left sorry one from the left of f of x

so which piece will we use we will use this piece right here ax plus three

and that limit if we use a one here that’s going to be a plus three

and that by part three here means we need to be equal to five

this has to be equal to five so a is 2. now let’s look at the limit from the

01:15

right so if we look at we needed the two-sided limit to exist so left and right

so when we look at the limit from the right which piece we use

so greater than one we’re looking like x squared plus b

so now we use a one here so we have b plus one

and that must be equal to five so b equals four so

we have this function here uh f of x equals two x plus three

five and x squared plus four we’re taking this line and this parabola

and we’re piecing them together in a way that the function is continuous

so this function right here is continuous

on its domain now what is the domain of this function right here

now if we look at 2x plus 3 the domain is all real numbers

01:16

and then that’s just a point if we look at this right here it’s just a parabola

the domain is all real numbers so the domain of this

function right here will be all real numbers

except possibly at one house it pieced together well one were five

so the domain of f is going to be all real numbers

and the only place where it could be discontinuous

because that function right here this piece is continuous everywhere this

piece is continuous everywhere so the only way the function as a

piecewise function would be discontinuous would be at one

but we already showed it’s continuous at one

right so we can say level from the left equals the limit from the right

so we can say this limit right here at 1 from both sides

of f of x this a and this b is 5. so we have all three of these checked

those therefore f is continuous at zero so therefore f is continuous

01:17

on its domain which is all real numbers so this function right here for this a

and this b is continuous for all real numbers

we verified it at zero and we argued at least i you know spoke it verbally

this function is continuous this is the linear function this is the quadratic

function right all right so there’s two uh exercise two part a and b

and now let’s do exercise three and no matter which way i go i cover up

parts of it all right so let’s look at exercise three now exercise three

is pointing out something in the theorem that we talked about earlier

01:18

which was the limit of a composition so let’s find that

limit theorem here and just briefly recall what that limit theorem was

so if the limit of g is l and f is continuous then the limit of the composition

is just f of l and we’re going to see that that breaks here for this problem

here exercise three [Music] so this right here is very important

right here that f is continuous at l and if you don’t have that

then you may not have equals here so let’s look at exercise three here okay so

the limit theorem says that if this limit is

01:19

l the limit of the g the inside function here

and if it’s continuous the the limit of the composition function

i can just say that’s f of l i could just slip the limit inside of the

function l and take the limit directly of that and then find that limit

and then find f of that now if you don’t have f is continuous

then you may not have equality here and that’s what this

exercise is going to show us right so exercise three here

what do we have so the function f of x is zero and one

it’s zero when we’re not zero and we’re a zero

at zero we’re one and the function u of x is equal to x

so u is just the identity function and then f is the zero or one function

01:20

so here we go so you know this is just u is just

the line y equals x right if we just graph u and if we were to just graph

f when we’re not when we’re not zero we’re at zero

when we’re equal to zero we’re one so this function is zero everywhere

except at zero we’re one right so clearly this function f here

notice f is notice f is not continuous not continuous at x equals zero

right clearly now how would you show that rigorously right so we got three steps

to our definition of continuous and which one would hold

01:21

so f of zero exists yes it’s one does the two-sided limit exist so the

two-sided limit as we approach zero from both sides the limit from the left

is zero so i’ll write that down here the limit from the left is zero

why is the limit from the left zero whereas we’re approaching zero from the left

the height is approaching zero it doesn’t matter what the height is

it matters what the height is approaching so as we’re approaching zero

the heights are zero the height is staying zero the hydra staying zero

so the height is approaching zero and the same thing from the limit from the

ride as we’re approaching zero from the right the height is zero there

so the two-sided limit does exist also so that’s number one in the definition

there’s number two in the definition but number three and the definition says

these have to be equal and you can see that they’re not

01:22

this is zero this is one so this one fails so that’s why f is not continuous at

zero you can clearly see it from the graph but you should be able to explain it

using calculus as well there’s the graph of of f

all right so let’s look at example three now now that we’ve understood our two

functions u and x u is basically just whatever you

plug in that’s what you get out very basic function there all right so

let’s look at the exercise the exercise says for us to um

find the limit of f of u and then find f of the limit u of x

and so let’s find those two limits now we’ve got to find the limit

01:23

for the one on the left and the limit for the one on the right [Music]

all right so here we go so we need to find the limit as we approach

zero of f of u of x and this will be the limit as we approach zero

of f of what is u of x it’s just an x and so what’s the limit as we approach

zero of f of x well we’ve already found that that was zero

and the way that we found that is we looked at the limit from the left and

the limit from the right they were both zero so this two-sided limit is zero

we could see it from the graph as we’re approaching zero from both sides

the height is approaching zero so this limit right here is zero

now um what about the right-hand side that we need to find

so we need to find f of the limit as we approach zero of u of x

01:24

we need to find f of this so this is the right-hand side so if we look back real

quick the right hand side there is f of the limit

of u of x so let’s find this limit uh let’s find this limit

here on the inside so f of what’s the limit as we approach

zero right so u of x is just an x so this would be

the limit as we approach zero of x and that’s just zero so this is f of zero

and what is f of zero when you plug in zero into f

you get out of one and you see these are not the same here

that’s a one so you see these are not the same here

so this right here is not equal to this right here

these two things are different this one’s equal to zero this one’s equal to one

so now we’ve done the not equals there and so this is an illustration of if you

01:25

try to find the limit of a composition of functions

and you try to pull that limit inside of the function f you may not be equals

if f is continuous you will be equal that’s what the theorem says

if f is continuous you will be but if you’re not equals

i mean if you’re not continuous if f is not continuous then it may not be equal

and there’s an example which shows you which shows that

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now i want to turn it over to you math can be difficult

01:26

because it requires time and energy to become skills

i want you to tell everyone what you do to succeed in your studies either way

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