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Kind: captions
Language: en
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The format of the exam i think its going to be 8 questions
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what will be on the exam will be limits
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definition of the derivative
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you dont have to write this down but you should know limits
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you should know definition of derivative
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you should be able to sketch curve very basic sketching
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you should be able to look at a curve and
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tell whats going on with the cure or the derivative
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will not be concavity on the test because we have not covered that yet
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so if you see the word concavity
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you could just ignore it, but it has to do with the curvature
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whether its concave up or concave down
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what else is on the test.. it will be similar too
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some of the midterms but one of the midterms had a bunch of pre calc on it
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i think thats the 07 one
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so you can ignore a lot of the 07 exam
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someone spent way to much time on that.. what else do you need to know
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thats about it.. some of you have been asking can you
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do the derivative short hand
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for those of you who had privilege of taking calculus before
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i think the answer is going to be yes and no
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if we say
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do it the hard way you have to do it the hard way
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and if we dont you can do it the easy way so there you go
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gonna be a whole bunch of limits
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i think the test is very straight forward
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youve been preparing you should do niceley on the test
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theres a couple of questions that would sort of stretch you
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but you know we have to kind of do that
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thats what they pay us for to ruin your lives
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make sure you never get to be a doctor
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no no no im not a doctor why should you be a doctor
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so im gonna go through the october 09 exam
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oh one last thing that i was telling some people if you go to your blackboard page
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and you dont have things like documents etc
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its gonna be a lot of emails but
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well figure out what to do but we dont understand why some of that is missing
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it shouldnt be
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im gonna pull this up so if you do not have it you can see it
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so this is the new term professor sutherland gave
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in october of 09
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and well do some of these
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feel free to ask questions
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raise your hands dont be shy
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some of you are not wearing red, remember your seawolves
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so were gonna do 1a
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the limit as x approaches 2
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x suqared minus 4
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over 5x times x minus 2
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so we look at that and say uh oh
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were gonna have 0 on the bottom 0 on the bottom is bad
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your gonna plug in 2 on top and your gonna get 4-4 is 0
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and youre gonna plug in 2 on the bottom and get 0 and thats a bad sign
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its not really a bad sign it just means youll get 0/0 which is undefined
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so how do you worka round that
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well i told my class before
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if you do when x goes to a number
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you get 0/0
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that means youll be able to factor out x minus that number
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so if you do the limit as x goes to 10
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and you get 0/0
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that means the top and the bottom both contain a term x-10
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it can contain more than 1 of those but at least 1
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thats what making it become 0/0
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that make sense
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when you plug in the 10 you get 0/0 so here when you plug in 2
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you get a 0 here, so you gotta go over here and take an x minus 2 out of that
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so its not a very hard expression to factor
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its the difference of 2 squares
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cancel the x minus 2
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opps that should be plus
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now you plug in 2
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cause you do the limit as x goes to 2 and you dont have that problem anymore
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youll get 4/10
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which you could leave that way you dont have to reduce to 2/5
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or turn it to a decimal or anything else
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there you go 3 points your on your way to becoming a doctor
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next one
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limit x goes to infinity
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2cosine
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pi over x
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gotta love trigonometry
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you know when youre gonna need trigonometry.. well for this exam
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but you guys know the general answer
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well if youre gonna be a physics major youre gonna need trigonometry
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physics majors yea applied math
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engineering
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the rest of you youre gonna run out of trigonometry at some point
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now x goes to infinity what happens with pu/x
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well x is going to infinity
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so pi over infinity is going to be 0
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whats the cosine of 0
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you sure its 1
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you should be sure. make sure you know your unit circle
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okay more points this is so much fun
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that wasnt very hard
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dont be paranoid some of these limits are very straight forward
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some of them are not but some of the are
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im obviously not gonna tell you which ones
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that would be cheating alright
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limit
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as x goes to 4
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of x squared
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over x minus 4
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squared
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now when you plug in 4 youre gonna 0 in the ottom but youre not gonna get 0 on top
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so when you hav a limit 0 on the bottom denominator
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but you dont have 0 in the numerator
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thats gonna be even for the limit
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infinity or minus infinity or something, we might get limit does not exist
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so you have to figure out whats going on when you approach four from the left sid
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from the right side and see if they agree
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when we do the limit as x approaches 4
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minus thats just less than 4
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the top would be 16
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and the bottom would be 0
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would the bottom be positive or negative
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positive cause youre squaring it, in fact this whole this is going to be negative
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right square on top and square on bottom its always gonna be 0 it cant be negative
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Okay so when X approaches 4 from the minus side
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From the minus side
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You're going to get plus infinity
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You all see that you have a number on top a number on the bottom
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That that's infinite
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And positive because it's square over square
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And there's other reasons why it's positive
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The limit as X approaches 4 from the left side
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of X squared over x -4
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Squared. Like I always said it's positive
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I get plus infinity again. since they agree
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The limit of this
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The limit is plus infinity. That's the kind of thing we are looking for
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We don't want to see you write positive infinity
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You have to do a little bit of explaining
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Now you can say. You could read this out and say this expression is either positive or zero
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And therefore the limit goes to post infinity. That would be good enough
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Just writing something probably wouldn't get you to full credit
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It's one of professors things. He also points out my thing to
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If you write stuff in mathematical nonsense that's bad
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Even if you get the right answer you write something that sort of
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That's totally not true you get the right answer
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Who rather you write something that is true I missed the answer a little bit
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Okay let's move onto the next page
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You see where it says name and then ID. That's ID
00:10:05.660 --> 00:10:06.800
Trying to know your ID
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Also I strongly suggest you where it says recitation
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Know what your recitation as an try know it before Thursday
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And don't feel like I don't know if some guy i never go
00:10:18.360 --> 00:10:22.360
If we can't find your exam at some point we will give you a zero
00:10:23.040 --> 00:10:25.760
So we suggest to be able two find your exam
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The best way to find it it's have the correct recitation number
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You heard the speech from me before
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It's the same speeches last time try remember what your recitation is
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It's not that hard to figure out it's up on blackboard
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Now we have
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Limit as X goes to infinity
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X squared -9
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Over five X-3
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No you look at the top and say
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There's an ask -3 and then the bottom has X -3
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But this is not as X goes two three this is an X goes to infinity
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Different kind of limit
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So what can we do with this
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You can look at the power for the highest term on top highest term on the bottom
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You can do a little algebra you can do a bunch of things the simplest thing
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Is to multiply out the bottom
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The highest term on top is X squared and the highest turn on the bottom is 5X squared that's a five
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So you can say this is just
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Liked the limit
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As X goes to infinity of X squared
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Over five X squared when you use that sign it means is behaving like
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Kind a like
00:12:09.360 --> 00:12:12.640
And now that the X squared cancel this would be 1/5
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That wasn't so bad
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This is the whole exam you be happy right
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How about limit as x goes to two
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X plus H squared minus X squared
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Over each that sort of looks derivativeish isn't it
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It's not quite a derivative as X goes to two
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It sure is close
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If you're not sure what's going on dive right in and do algebra. We love algebra
00:13:08.800 --> 00:13:11.440
So we multiply out the top X squared +2 XH
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Plus H squared minus X squared
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Over H
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Cancel a couple x squares
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Now we have the limit
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H approaches to 2XH
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Plus H squared over H
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You sure it's not derivative it kind of looks like a derivative keep going
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Taking H out of the top
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And we can cancel the H's
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Now when we let the limit as H goes to two we get
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2X plus H it's not a derivative but close
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By the way some of you are asking do I have to write the limit at every step
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You should like I said you shouldn't write mathematical nonsense
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Also you want to get as much partial credit as you can so make sure you can figure out what's going on
00:14:52.600 --> 00:14:53.960
2X +2 yes I apologize
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Alright next one
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Alright how about a limit as X goes to infinity
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E to the x cosin x
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This is a fun one what are we do with this
00:15:51.620 --> 00:15:52.260
The ideas
00:16:08.100 --> 00:16:10.100
Anyone have any idea what to do
00:16:17.020 --> 00:16:19.660
alright you are going to want to squeeze
00:16:19.660 --> 00:16:25.280
You're going to want to know the squeeze theorem how do we know to use the squeeze them you have to use your rules
00:16:25.420 --> 00:16:28.780
So when you see limits and you have signs in cosines
00:16:28.880 --> 00:16:34.480
Among other things use the squeeze theorem sin sine and cosine are between one and -1
00:16:35.880 --> 00:16:39.560
So what will be do here well dormant as X goes to infinity
00:16:41.780 --> 00:16:43.060
Of E to the X cosign X
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So it is less than or equal to
00:16:50.400 --> 00:16:52.800
The limit as X goes to minus infinity
00:16:55.040 --> 00:16:56.240
E to the X times one
00:16:56.240 --> 00:17:00.620
Because cosine is less than or equal to one this is less than or equal to
00:17:03.300 --> 00:17:06.500
Limit E to the X, X goes to negative infinity of -1
00:17:12.600 --> 00:17:16.440
Now would have happens to E to the x , know what it looks like
00:17:19.160 --> 00:17:22.440
That's what to the X looks like positive infinity
00:17:22.760 --> 00:17:26.280
E looks infinity when you go to the negative infinity
00:17:27.280 --> 00:17:28.240
It goes to zero
00:17:29.680 --> 00:17:31.200
So this term on the left
00:17:31.740 --> 00:17:34.860
Is going to look like zero times -1 which is zero
00:17:39.540 --> 00:17:42.500
Here on the right well it's just positive one
00:17:45.980 --> 00:17:48.780
It's just zero again that means less limit
00:17:51.880 --> 00:17:54.200
Is somewhere between zero and zero
00:17:57.260 --> 00:17:59.500
So it's between 0 and 0 so it's zero
00:18:04.040 --> 00:18:06.760
That is the squeeze theorem so far so good
00:18:11.880 --> 00:18:13.640
No we hating life right now
00:18:17.260 --> 00:18:20.300
Well think of what the cosine graph looks like
00:18:26.400 --> 00:18:30.880
Well since some of you took trigonometry recently well that didn't
00:18:31.440 --> 00:18:33.280
Sad before the fortune of me
00:18:34.760 --> 00:18:35.260
Cosine
00:18:38.200 --> 00:18:43.320
Looks like that and so does sign by the way . That is one and that is negative one
00:18:43.400 --> 00:18:47.960
So no matter what cosine and sign do they'll always between one and -1
00:18:48.020 --> 00:18:54.660
So whatever is going on here E to the X can't be bigger than one because cosine can't be bigger than one
00:18:54.660 --> 00:18:59.700
And it can't be less then e to the x times -1 because cosine cant be less than -1
00:19:00.360 --> 00:19:04.360
so whatever e to the x term is doing it is multiplied by cosine
00:19:04.360 --> 00:19:08.780
its never multiplied by anything greater than or multiplied by anything less than 1
00:19:08.780 --> 00:19:11.480
its always trapped between them thats why its squeezed
00:19:14.280 --> 00:19:19.320
so if you would graph tjis you would find that the graph is somewhere between
00:19:20.920 --> 00:19:21.420
0 and 0
00:19:43.780 --> 00:19:47.140
so before we do other stuff lets do some more limits
00:19:50.300 --> 00:19:51.100
so how about
00:19:54.720 --> 00:19:58.160
fall 08 exam did any of you look at the fall 08 exam yet
00:20:01.420 --> 00:20:04.780
how bout that firsy one limit as x goes to 0 sinex/10
00:20:07.800 --> 00:20:09.320
dont you just love trig
00:20:53.740 --> 00:20:56.620
getting 0/ arent we, gotta be a way around it
00:21:44.060 --> 00:21:45.820
alright thats long enough
00:21:45.880 --> 00:21:48.200
the problem is you know sine of 0 is 0
00:21:48.440 --> 00:21:50.040
you know tangent of 0 is 0
00:21:50.040 --> 00:21:52.760
if you dont you have four more days to remember the unit circle
00:21:54.100 --> 00:21:55.780
or what we suggest tattoo
00:21:56.020 --> 00:21:59.060
absolutely fine, nice tattoo of a unit circle
00:21:59.060 --> 00:22:01.180
make sure you can see it dont put it back here somewhere
00:22:02.120 --> 00:22:03.160
right out front
00:22:04.320 --> 00:22:07.120
have to explain that to your friends later
00:22:07.120 --> 00:22:10.380
why you have a tatto of a unit circle, because i love trig so much
00:22:10.800 --> 00:22:13.520
that i wanted to have it on my body forever
00:22:20.900 --> 00:22:22.420
so what could we do well
00:22:22.420 --> 00:22:25.560
we all remember the tangent sine over cosine of course
00:22:28.680 --> 00:22:29.800
this is sinx over
00:22:31.680 --> 00:22:32.400
sinx/cosx
00:22:36.820 --> 00:22:37.540
that means
00:22:37.540 --> 00:22:40.580
we can take the numerator and the denominator
00:22:44.040 --> 00:22:45.160
invert multiply
00:22:47.940 --> 00:22:48.440
that
00:22:53.700 --> 00:22:54.740
cancel the sine
00:22:57.820 --> 00:22:59.500
and now it would be cos of 0
00:23:00.880 --> 00:23:01.520
which is 1
00:23:03.300 --> 00:23:04.180
be right back
00:24:00.340 --> 00:24:01.700
im back so once again
00:24:01.700 --> 00:24:03.400
lets try some more of these
00:24:03.400 --> 00:24:05.300
these look painful dont they
00:24:06.020 --> 00:24:07.780
the second one is not to bad
00:24:08.200 --> 00:24:10.680
the limit as x goes to infinity of that
00:24:14.360 --> 00:24:16.840
we have the limit as x goes to infinity
00:24:18.460 --> 00:24:19.740
5x squared minus 4x
00:24:22.040 --> 00:24:22.540
-1
00:24:23.660 --> 00:24:25.100
over x squared minus 1
00:24:25.100 --> 00:24:28.780
we look at the power of the top and the bottom and its the same
00:24:30.440 --> 00:24:34.280
so this is going to behave as the limit as x goes to infinity
00:24:36.200 --> 00:24:38.040
of 5x squared over x squared
00:24:40.880 --> 00:24:42.240
which would just be 5
00:24:46.620 --> 00:24:48.380
these arent s bad are they?
00:24:50.120 --> 00:24:51.720
how about that third one
00:24:52.460 --> 00:24:55.340
what do you do with infinity minus infinity
00:24:56.500 --> 00:24:59.140
well you could be theres no denominator
00:25:02.400 --> 00:25:04.560
doesnt matter so lets do this one
00:25:12.760 --> 00:25:17.080
so you guys say what am i gonna do, you multiply it by the conjugate
00:25:17.680 --> 00:25:21.360
and you multiply it by the conjugate over the conjugate
00:25:34.700 --> 00:25:36.220
after all this is over 1
00:25:37.620 --> 00:25:40.020
it doesnt have to have a denominator
00:25:40.600 --> 00:25:42.360
you can multiply it by 1 but
00:25:42.360 --> 00:25:45.060
you can pretend theres a denominator there the denominator is 1
00:25:45.360 --> 00:25:46.320
so there you go
00:25:46.320 --> 00:25:50.080
so when you see that expression you should see conjugate
00:25:52.120 --> 00:25:54.440
the expression like that, radical
00:25:54.440 --> 00:25:56.240
some other stuff if you dont know wat to do
00:25:56.480 --> 00:25:59.920
cant squeeze it so maybe theres a conjugate in there
00:26:00.320 --> 00:26:01.920
so lets see what happens
00:26:05.420 --> 00:26:09.180
square root of 4x squared to square root of four x squared
00:26:10.620 --> 00:26:12.300
plus x is 4x suqared plus x
00:26:14.500 --> 00:26:15.860
middle terms cancel
00:26:15.860 --> 00:26:19.320
when you do the filing, thats the purpose of using the conjugate
00:26:19.760 --> 00:26:21.920
you get plus 2x times the radical
00:26:22.020 --> 00:26:23.780
minus 2x times the radical
00:26:26.260 --> 00:26:28.580
and the 2x times 2x minus 4x squared
00:26:33.120 --> 00:26:35.280
and the bottom we have this thing
00:26:37.020 --> 00:26:39.420
that we just introduced that we hate
00:26:44.820 --> 00:26:46.020
4z squared cancel
00:26:49.940 --> 00:26:51.940
the limit as x goes to infinity
00:26:54.920 --> 00:26:55.480
of x over
00:26:56.460 --> 00:26:58.300
squared of 4x sqaured plus x
00:27:01.020 --> 00:27:01.980
plus x now what
00:27:07.080 --> 00:27:09.640
oh 2x im sorry, good thing i brought you
00:27:18.320 --> 00:27:19.360
now what do we do
00:27:20.960 --> 00:27:24.240
i have that x on top can we take an x out of the bottom
00:27:25.760 --> 00:27:26.480
yea you can
00:27:28.420 --> 00:27:31.220
so what would that look like, well lets see
00:27:32.880 --> 00:27:34.640
nothing happens to the top
00:27:35.020 --> 00:27:37.580
you take an x out of the radical because
00:27:37.660 --> 00:27:39.740
the square root of x squared is x
00:27:39.740 --> 00:27:42.160
in fact the absolute value of x but we dont care
00:27:46.760 --> 00:27:49.880
what about the plus x well that would become 1/x
00:27:55.760 --> 00:27:57.280
how do we know that well
00:27:57.280 --> 00:28:01.180
pretend thats a 4x squared and put that back inside you get ax square
00:28:01.180 --> 00:28:03.320
plus x square root of x, x square root of x
00:28:04.760 --> 00:28:05.260
is x
00:28:05.520 --> 00:28:08.160
so you can take an x out of the whole thinf
00:28:08.160 --> 00:28:11.940
theres other ways around this but thats probably your best way
00:28:14.100 --> 00:28:15.380
remeber that trick
00:28:15.920 --> 00:28:18.080
cancel the x's now you have limit
00:28:20.800 --> 00:28:22.240
as x goes to infinity 1
00:28:26.700 --> 00:28:29.580
i couldve taken out the 4 but i didnt have too
00:28:35.000 --> 00:28:37.400
under the radical you have x squared
00:28:37.400 --> 00:28:41.700
is you take it outside the radical you get x because the square root is x
00:28:41.700 --> 00:28:44.340
so youre taking x squared out of this term
00:28:44.800 --> 00:28:48.960
that becomes the x and then you take the x squared out of this tem
00:28:51.380 --> 00:28:53.380
x is the same as x squared over x
00:28:57.360 --> 00:29:01.280
i see a lot of you guys are lost with that so we will do it again
00:29:01.420 --> 00:29:04.860
important you understand this step cause who knows
00:29:05.560 --> 00:29:06.600
someday in life
00:29:06.740 --> 00:29:09.700
this may be the only thing that matters to you
00:29:14.820 --> 00:29:18.260
you can think of that as the square root of 4x squared
00:29:19.160 --> 00:29:20.520
plus x squared over x
00:29:23.900 --> 00:29:27.740
which you can think of as the square root of x sqaured times
00:29:30.380 --> 00:29:30.940
4 plu 1/x
00:29:33.840 --> 00:29:37.120
now you can take the x squared out and that becomes
00:29:37.740 --> 00:29:38.620
x square root
00:29:42.100 --> 00:29:42.740
4 plus 1/x
00:29:55.620 --> 00:29:57.780
alright you understand that now
00:29:57.780 --> 00:30:01.080
were you just like ah ill write it down and look at it later
00:30:01.080 --> 00:30:04.540
lot of you guys do that, watch the video over and over again
00:30:07.460 --> 00:30:09.220
now if x is infinity thats 0
00:30:10.320 --> 00:30:12.400
the 1/x is 0, square root of 4 is 2
00:30:13.240 --> 00:30:13.960
2 plus 2 is 4
00:30:16.180 --> 00:30:19.300
thats 1/4, or how they say in brooklyn a quarter
00:30:44.120 --> 00:30:46.680
we are videoing this whole thing, matt
00:30:46.680 --> 00:30:49.460
back there putting this whole thing on video
00:30:49.460 --> 00:30:51.740
remember you can get to the videos from the web page
00:30:53.680 --> 00:30:55.200
if you go to our webpage
00:30:56.180 --> 00:30:57.540
remember this thing
00:30:58.700 --> 00:31:00.540
okay all the videos are here
00:31:00.760 --> 00:31:05.160
you have lecture 1, you have some of professor sutherlands videos
00:31:07.180 --> 00:31:09.340
all sorts of fun stuff okay thats
00:31:09.980 --> 00:31:11.180
on calculus a page
00:31:12.760 --> 00:31:16.520
lets do a few more of theses and then we do some other stuff
00:31:17.480 --> 00:31:19.880
i dont want to spend to long on limits
00:31:21.880 --> 00:31:25.000
i think we can skip d, e is just like the one we did
00:31:30.400 --> 00:31:34.400
eh we did enough limits for the moment lets do something else
00:31:37.280 --> 00:31:41.600
because limits, you can have enough of that i know some of you, yes
00:31:43.700 --> 00:31:45.460
there will be an answer key
00:31:45.460 --> 00:31:48.500
professor sutherland is in charge of putting that up at some point
00:31:48.640 --> 00:31:50.320
hopefully before friday
00:32:06.720 --> 00:32:09.280
where was i, i was doing the fall 09 exam
00:32:25.500 --> 00:32:27.180
back to the previous exam
00:32:30.880 --> 00:32:31.520
fall of 09
00:32:34.800 --> 00:32:35.680
lets look at 3
00:32:59.020 --> 00:33:03.900
remember monday and wednesday in class ill be reviewing problems aswell
00:33:04.580 --> 00:33:08.180
and they will be up on videos and those videos will be up
00:33:08.300 --> 00:33:11.420
so youll have lots of chances to watch me up here
00:33:25.720 --> 00:33:26.840
let f of x equal 5x
00:33:27.920 --> 00:33:29.280
cubed minus 8x plus 2
00:33:29.280 --> 00:33:31.060
fins the slope of the secant line
00:33:31.620 --> 00:33:36.420
you guys all know how to do that, take a second figure it out, not that hard
00:34:57.340 --> 00:35:00.620
find the slope of the secant line, the secant line
00:35:01.660 --> 00:35:03.420
from x equals 0 to x equals 1
00:35:05.300 --> 00:35:06.900
would be f of 1 minus f of 0
00:35:08.640 --> 00:35:09.200
over 1-0
00:35:10.900 --> 00:35:14.500
thats the slope of any line, change in y over change in x
00:35:16.420 --> 00:35:18.740
you just have to find f of 1 f of 0, 1-0
00:35:19.180 --> 00:35:20.940
f of 1 we get by plugging in 1
00:35:24.020 --> 00:35:24.900
you get 5-8+2
00:35:26.580 --> 00:35:27.460
is negative 1
00:35:29.880 --> 00:35:31.560
then you plug in 0 and get 2
00:35:36.860 --> 00:35:37.980
so that equals -3
00:35:39.900 --> 00:35:41.340
thats not so hard okay
00:35:43.340 --> 00:35:44.620
now find f prime of 1
00:35:44.620 --> 00:35:49.880
so for those of you who already taken calculus you can find this the fast way if you wanrt
00:35:50.140 --> 00:35:53.020
im also going to teach that tomorrow in clas
00:35:54.160 --> 00:35:59.280
in theory professor sutherland will be teaching it so youll get to see it, yes
00:36:03.260 --> 00:36:05.180
no you will not get points off
00:36:05.400 --> 00:36:08.200
i checked that with professor sutherland
00:36:09.420 --> 00:36:12.700
i fought for you guys, actually he said no problem
00:36:12.860 --> 00:36:14.220
wasnt much of a fight
00:36:17.780 --> 00:36:21.380
but just in case lets make sure you can do it the hard way
00:36:21.380 --> 00:36:23.940
because many of you did not take calculus in high school
00:36:25.760 --> 00:36:27.600
so we want to find f prime of 1
00:36:27.600 --> 00:36:29.480
we going to need to do the limit
00:36:31.200 --> 00:36:32.320
as h approaches 0
00:36:34.640 --> 00:36:35.520
of f of x plus h
00:36:37.180 --> 00:36:37.900
minus f of x
00:36:38.880 --> 00:36:41.120
all over h and we want to do this at 1
00:36:42.400 --> 00:36:44.640
so we really want to find the limit
00:36:46.180 --> 00:36:47.300
as h approaches 0
00:36:49.060 --> 00:36:49.780
f of 1 plus h
00:36:52.540 --> 00:36:53.900
minus f of 1 all over h
00:37:01.680 --> 00:37:02.880
what is f of 1 plus h
00:37:11.820 --> 00:37:16.060
would be the limit as h approaches 0 of 5(1+H) cubed minus 8(1+h)
00:37:17.560 --> 00:37:18.060
plus 2
00:37:22.660 --> 00:37:23.380
minus f of 1
00:37:23.440 --> 00:37:25.600
which we already found which is -1
00:37:27.580 --> 00:37:28.080
over h
00:37:33.600 --> 00:37:34.960
okay whats 1+h cubed
00:37:34.960 --> 00:37:37.740
well you can certainly multiply out 1+h cubed
00:37:37.960 --> 00:37:40.200
or if i showed in one of my lectures
00:37:42.580 --> 00:37:43.380
this is 1+3h
00:37:44.480 --> 00:37:45.280
+3h squared
00:37:47.080 --> 00:37:48.440
plus h cubed because
00:37:50.740 --> 00:37:51.240
x+hh
00:37:53.500 --> 00:37:54.540
cubed is x cubed
00:37:56.160 --> 00:37:57.280
plus 3 x squared h
00:37:59.220 --> 00:38:00.340
plus 3x h squared
00:38:01.500 --> 00:38:02.300
plus h cubed
00:38:03.600 --> 00:38:05.680
you might want to memorize that
00:38:07.280 --> 00:38:08.720
not as hard as it looks
00:38:09.840 --> 00:38:12.960
as i showed before the coefficients go 1,3,3,1
00:38:15.260 --> 00:38:16.540
x cubed, x squared x
00:38:17.480 --> 00:38:18.040
no x term
00:38:19.180 --> 00:38:21.500
no h term, h, squared cuber pattern
00:38:21.500 --> 00:38:23.920
that comes by the binomial expansion
00:38:23.920 --> 00:38:26.740
what were also going to do tomorrow in class
00:38:30.400 --> 00:38:31.680
now that we did that
00:38:32.900 --> 00:38:33.860
minus, minus 1
00:38:38.020 --> 00:38:40.260
all over h did i make a mistake agin
00:38:47.240 --> 00:38:50.280
i lost something in here, oh i forgot the 8 part
00:38:54.320 --> 00:38:54.820
minus 8
00:38:57.860 --> 00:38:58.900
minus 8h, plus 2
00:38:58.900 --> 00:39:00.000
plus 1, there we go
00:39:06.800 --> 00:39:08.000
distribute that 5
00:39:13.140 --> 00:39:14.260
you get 5 plus 15 h
00:39:17.760 --> 00:39:18.880
plus 15h squared
00:39:21.140 --> 00:39:22.020
plus 5h cubed
00:39:23.960 --> 00:39:25.480
minus 8 plus 2 is minus 6
00:39:27.700 --> 00:39:28.660
minus 8h plus 1
00:39:31.140 --> 00:39:31.780
all over h
00:39:35.020 --> 00:39:36.300
lets see 5+1=6-6=0
00:39:38.780 --> 00:39:39.500
15h-8h=7h
00:39:41.780 --> 00:39:43.300
this becomes the limit
00:39:44.820 --> 00:39:45.460
h goes to 0
00:39:48.460 --> 00:39:49.180
7h plus 15h
00:39:49.700 --> 00:39:50.260
squared
00:39:51.440 --> 00:39:51.940
plus 5
00:39:54.440 --> 00:39:54.940
over h
00:39:56.220 --> 00:39:57.420
just some algebra
00:39:58.040 --> 00:39:59.080
we love algebra
00:40:02.520 --> 00:40:03.320
pull out an h
00:40:04.940 --> 00:40:05.980
and then cancel
00:40:25.800 --> 00:40:28.920
now when you take the limit as h goes to 0 you get 7
00:40:39.020 --> 00:40:42.380
okay part c, write the equation of the tangent line
00:40:43.900 --> 00:40:45.500
the equation of any line
00:40:50.780 --> 00:40:51.280
y-y1
00:40:53.780 --> 00:40:54.280
equals
00:40:59.040 --> 00:41:02.320
y-y1=m(x-x1), we know have all the pieces we need
00:41:05.360 --> 00:41:07.760
what is y1, f of 1, which is negative 1
00:41:11.380 --> 00:41:12.260
the slope is 7
00:41:13.520 --> 00:41:14.720
and that is x minus
00:41:15.820 --> 00:41:16.320
also 1
00:41:16.320 --> 00:41:21.060
and you can leave it like that you dont have to simplify the equation of a line
00:41:21.060 --> 00:41:25.080
we always say dont do extra algebra and get yourself in trouble
00:41:30.400 --> 00:41:31.120
okay part d
00:41:36.760 --> 00:41:41.000
talks about concavity which is not on the test so you ignore that
00:41:51.920 --> 00:41:54.400
oh i think thats enough for the moment
00:41:59.940 --> 00:42:00.740
continuity
00:42:01.580 --> 00:42:04.620
how are we doing on this we understand this one
00:42:15.920 --> 00:42:18.240
were finding f prime of 1, remember
00:42:18.240 --> 00:42:22.460
every time you do the derivative you are shrinking the denominator as close to 0 as possible
00:42:22.460 --> 00:42:27.000
this is always going to be the limit as h goes to 0, the one part is up here
00:42:33.180 --> 00:42:35.740
how did we get -1, what is f of negative 1
00:42:37.920 --> 00:42:40.080
plus in 1, i lost the equation now
00:42:44.360 --> 00:42:45.880
its 5x cubed minus 8x+2
00:42:45.880 --> 00:42:48.920
you plug in 1, you get negative 1, we found that here
00:42:53.200 --> 00:42:53.760
thats y1
00:42:56.080 --> 00:42:57.280
which is f of 1 okay
00:42:58.300 --> 00:42:59.420
other questions
00:43:04.140 --> 00:43:05.100
were good okay
00:43:05.420 --> 00:43:07.020
we can come back to stuff
00:43:07.020 --> 00:43:09.740
we can always rewind the ideo at this point
00:43:19.860 --> 00:43:22.580
so for what values of x is this continuous
00:43:33.140 --> 00:43:34.660
any ideas, we have f of x
00:43:42.800 --> 00:43:46.960
e to the x e to the x looks like this, thats continuos everywhere
00:43:51.820 --> 00:43:53.980
so we want to know what values of x
00:43:53.980 --> 00:43:57.340
is the function continuous you can also ask where is the function not continuos
00:43:58.280 --> 00:44:01.080
and then its all reals except those values
00:44:05.640 --> 00:44:09.240
were definetely going to have a problem at 0, arent we?
00:44:10.100 --> 00:44:13.780
cause then we have e to the 1 over x, so we cant have a 0 here
00:44:15.240 --> 00:44:17.560
that would be 0 and that would be bad
00:44:18.480 --> 00:44:19.920
so its not continuous
00:44:25.240 --> 00:44:26.040
at x equals 0
00:44:28.300 --> 00:44:29.900
any other problem spots
00:44:30.300 --> 00:44:32.460
any other places you can think of
00:44:32.460 --> 00:44:35.920
when a teacher ask that, that means yes theres another spot
00:44:40.600 --> 00:44:43.480
got any ideas or do you want to jus wait for me
00:44:50.120 --> 00:44:52.200
when 2 equals e1/x its equal to 0
00:44:59.340 --> 00:45:01.900
when the denominators whole thing is 0
00:45:01.900 --> 00:45:04.020
because again youll be dividing by 0
00:45:04.640 --> 00:45:06.400
so where is that, well when
00:45:09.100 --> 00:45:09.980
2 equals e1/x
00:45:12.460 --> 00:45:14.700
take the natural log of both sides
00:45:18.980 --> 00:45:21.380
the natural log is 2 which equals 1/x
00:45:22.340 --> 00:45:22.840
or
00:45:24.820 --> 00:45:26.820
x equals the natural log of 1/2
00:45:29.020 --> 00:45:31.900
so thats the other place its not continuous
00:45:36.920 --> 00:45:39.800
other than that its continuous at all reals
00:45:39.800 --> 00:45:44.100
so if youre gonna write its not continuous at x=2, natural log of 1/2
00:45:44.600 --> 00:45:47.160
you should also write all reals except
00:45:49.660 --> 00:45:51.020
x equals 0, or 1/log2
00:45:55.540 --> 00:45:58.820
its not continuous because you can not divide by 0
00:46:02.900 --> 00:46:05.460
other wise its continuous everywhere
00:46:07.700 --> 00:46:09.620
we good should i do that again
00:46:09.620 --> 00:46:12.180
people understand, we have two problem spots
00:46:12.560 --> 00:46:14.320
the first is here when x is 0
00:46:15.100 --> 00:46:16.140
we do either 1/0
00:46:16.300 --> 00:46:19.900
you cant do e to the 1/0, cause it to the 1/0 is undefined
00:46:20.480 --> 00:46:22.320
thats problem spot number 1
00:46:22.320 --> 00:46:24.980
problem spot number 2 is when the whole denominator is s0
00:46:26.380 --> 00:46:28.300
so you set 2=e to the x equals 0
00:46:28.480 --> 00:46:30.160
put the 2 on the other side
00:46:30.620 --> 00:46:33.020
take the log of bot sides, natural og
00:46:33.820 --> 00:46:34.780
and solve for x
00:46:36.780 --> 00:46:38.860
its always asier when i do it yes
00:46:49.020 --> 00:46:51.500
e to the 1/x, i didnt say e to the 1/x is 0
00:46:51.640 --> 00:46:54.200
you want to know where e to the x equals 2
00:46:54.200 --> 00:46:56.820
because then you have 2-2 in the denominator
00:46:57.680 --> 00:46:59.200
so e to the 1//x equals 2
00:46:59.620 --> 00:47:01.060
when x eqauls 1/x log 2
00:47:01.140 --> 00:47:02.180
natural log of 2
00:47:41.220 --> 00:47:42.820
okay lets scorll around
00:47:45.840 --> 00:47:48.800
this thing is awesome got to tell ya, 8 points
00:47:50.400 --> 00:47:55.760
write a limit that represents the slope of the graph y= absolute value of x to the x
00:47:56.580 --> 00:47:59.620
when x is not equal to 0, and 1 when x is equal to 0
00:48:01.400 --> 00:48:03.640
what do you do with that, any ideas
00:48:13.980 --> 00:48:15.580
its not as bad as it looks
00:48:18.480 --> 00:48:20.720
now remember slope is derivative
00:48:20.720 --> 00:48:24.860
so we would have to do the definition of the derivative here right
00:48:25.360 --> 00:48:28.480
so whats the definition of the derivative, its
00:48:30.380 --> 00:48:31.820
the limit as h goes to 0
00:48:33.380 --> 00:48:34.100
f of x plus h
00:48:36.120 --> 00:48:36.840
minus f of x
00:48:39.300 --> 00:48:39.940
all over h
00:48:40.400 --> 00:48:42.560
write that down in a lot of places
00:48:42.560 --> 00:48:45.480
you never know what youll pick up partial credit for
00:48:50.080 --> 00:48:52.640
remember life is about partial credit
00:49:03.980 --> 00:49:06.860
well what is f of x plus h, were gonna do limit
00:49:08.480 --> 00:49:09.120
h goes to 0
00:49:10.620 --> 00:49:12.300
absolute value of x plus h
00:49:13.040 --> 00:49:16.160
to the x plus h, that kind of looks like an x plus h
00:49:18.760 --> 00:49:20.920
minus absolute value of x to the x
00:49:22.520 --> 00:49:25.720
all over h and your done, thats the whole problem
00:49:27.840 --> 00:49:29.280
thats it nothing else
00:49:35.620 --> 00:49:39.780
thats 8 points, the whole test is nine points thats pretty good
00:50:02.260 --> 00:50:05.060
what happens with the part when y=1 and x=0
00:50:05.060 --> 00:50:08.560
you dont care, youre only doing the limit you dont actually care what happens at 0
00:50:10.280 --> 00:50:11.800
you just want the limit
00:50:18.160 --> 00:50:21.040
i mean you can replace this with 1 if you want
00:50:21.480 --> 00:50:24.200
but you dont actually have to evaluate it
00:50:24.200 --> 00:50:27.280
i dont know maybe thats what professor sutherland wants but
00:50:27.900 --> 00:50:30.140
we dont have to evaluate the limit
00:50:32.640 --> 00:50:33.520
and by the way
00:50:36.180 --> 00:50:39.540
yea the limit as h approaches, yea id leave it alone
00:50:40.800 --> 00:50:43.600
ill check with it and see if he changes that
00:50:44.120 --> 00:50:46.680
now we move onto graphing which we love
00:50:49.440 --> 00:50:52.960
at the right is a graph of the derivative of a function
00:50:53.680 --> 00:50:56.880
you can ignore 6a theres no concavity on the test
00:51:01.360 --> 00:51:05.200
you know what a concavity is right , i have a hole in my tooth
00:51:08.520 --> 00:51:09.480
kahn-cavity,
00:51:32.700 --> 00:51:34.780
lets just draw that graph again
00:51:41.460 --> 00:51:46.260
so thats approximately what the graph looks like, thats the derivative
00:51:46.460 --> 00:51:47.500
of the fucntion
00:51:49.600 --> 00:51:53.120
which of the following best represents the function
00:51:59.380 --> 00:52:01.940
thats f prime of x, what do we know, well
00:52:04.360 --> 00:52:07.560
you look at the derivative of the function we say
00:52:11.600 --> 00:52:14.640
the derivative is negative up until this spot
00:52:15.400 --> 00:52:20.120
and then its 0, and if its negative that means the function is going down
00:52:20.120 --> 00:52:24.200
and then it gets to this spot and stops going down because its 0
00:52:24.260 --> 00:52:28.180
and now its going up, its positive all in this zone over here
00:52:29.920 --> 00:52:34.160
remember negative means going down and positive means going up
00:52:35.360 --> 00:52:37.760
so you need the graph to be going down
00:52:37.760 --> 00:52:39.960
and get that spot and go up for a while
00:52:41.020 --> 00:52:44.220
and then it goes up until you get to here another 0
00:52:48.100 --> 00:52:53.380
and then you go down for a while until you get to here where it makes a bottom again
00:52:55.120 --> 00:52:57.680
and then its positive and goes up again
00:52:58.280 --> 00:53:00.440
so its doing something like that
00:53:00.620 --> 00:53:03.100
so the closest one is the second graph
00:53:04.480 --> 00:53:08.880
remember theres more than one picture that can be that derivative
00:53:08.880 --> 00:53:10.760
but this is the one that you want
00:53:12.640 --> 00:53:15.200
graph number b, graph whatever that is
00:53:17.100 --> 00:53:20.700
so lets do that again, look at the derivative function
00:53:20.860 --> 00:53:26.060
the only difference is i didnt go down farther, you can go down as far as you want
00:53:28.020 --> 00:53:28.980
doesnt matter
00:53:30.340 --> 00:53:31.220
these are 0's
00:53:31.560 --> 00:53:36.120
the derivatives are 0 means it has a maximum and minimum or a flat spot
00:53:37.800 --> 00:53:38.300
okay so
00:53:38.720 --> 00:53:43.440
before hand we have a negative derivative so the function is going down
00:53:43.440 --> 00:53:47.180
then they have a positive derivative and the function is going up, so going down
00:53:47.840 --> 00:53:49.280
get to there, going up
00:53:50.060 --> 00:53:51.420
at the origin where 0
00:53:51.840 --> 00:53:54.480
so once again we have to have a max or a min
00:53:55.940 --> 00:53:57.060
at 0, at the y axis
00:53:57.240 --> 00:54:01.320
they were negative so we go down for a while and get to that spot
00:54:01.320 --> 00:54:03.820
so we go down for a while till we get to there
00:54:04.800 --> 00:54:05.300
go up
00:54:07.740 --> 00:54:10.620
another clue this is a polynomial and it has
00:54:11.240 --> 00:54:12.520
2 maxima and minima
00:54:13.440 --> 00:54:17.040
so the original function will have 3 maxima and minima
00:54:17.040 --> 00:54:19.880
the derivative every time you take the derivative you lose one
00:54:20.640 --> 00:54:23.360
if you started with 10 maxima and minimas
00:54:23.360 --> 00:54:27.200
then the first derivative will have 9, the second will have 8
00:54:28.080 --> 00:54:30.560
third derivative will have 7 and so on
00:54:32.980 --> 00:54:34.180
minor exceptions
00:54:34.980 --> 00:54:36.340
should i repeat that
00:54:37.500 --> 00:54:40.940
you have a function of f of x, and you have max and mins
00:54:40.940 --> 00:54:43.300
the first derivative will have 1 fewer
00:54:43.440 --> 00:54:46.640
and the 2nd derivative will have 2 fewer and so on
00:54:48.760 --> 00:54:51.640
so since i told you that if this is f prime of x
00:54:51.740 --> 00:54:54.780
f double prime of x will only have one max or min
00:54:56.020 --> 00:54:57.860
so what would that look like
00:55:00.160 --> 00:55:03.760
for the second derivative we would have to have a 0 here
00:55:04.540 --> 00:55:05.340
and a 0 there
00:55:07.840 --> 00:55:10.720
youd have to have a 0 somewhere around there
00:55:11.280 --> 00:55:12.880
somewhere around there
00:55:14.500 --> 00:55:18.820
the function is going up until we get to that spot, so its positive
00:55:18.820 --> 00:55:22.040
and its going down till it gets to this spot so its negative
00:55:22.700 --> 00:55:24.940
its going up again so its positive
00:55:28.080 --> 00:55:29.360
negative positive
00:55:29.760 --> 00:55:32.320
so that would look like the third graph
00:55:35.700 --> 00:55:36.980
this one right here
00:55:43.920 --> 00:55:46.240
these are fun, should i repeat that
00:55:46.240 --> 00:55:49.460
you look at the derivative graph now you preten this is a regular graph because
00:55:49.460 --> 00:55:54.020
we want the second derivative so the derivative of this, where this graph have maximums
00:55:54.140 --> 00:55:57.260
and minimums the derivative of it will have 0's
00:55:58.080 --> 00:56:00.080
it has a max there and a in there
00:56:00.620 --> 00:56:01.340
and 0 there
00:56:03.260 --> 00:56:06.140
the graph is going up so it would be positivw
00:56:06.280 --> 00:56:08.600
its going down so it can be negative
00:56:08.600 --> 00:56:10.460
its going up so it would be positive
00:56:10.560 --> 00:56:11.920
so you have positive
00:56:13.660 --> 00:56:14.300
negative
00:56:15.120 --> 00:56:15.760
positive
00:56:24.000 --> 00:56:27.520
oh your right its the first one, its the left one sorry
00:56:27.940 --> 00:56:32.340
i was just looking for a parabula, its the left one not the right one
00:56:33.580 --> 00:56:35.580
because the location of the 0s
00:56:50.300 --> 00:56:52.620
alright lets do a little more stuff
00:57:04.480 --> 00:57:07.040
actually i dont think we need to do this
00:57:10.240 --> 00:57:12.160
so lets pick a different test
00:57:19.020 --> 00:57:21.100
lets do another continuity one
00:57:40.400 --> 00:57:42.640
take a second and see if you can do 2
00:57:47.580 --> 00:57:49.580
this si the oct 08 exam, fall 08
00:57:49.580 --> 00:57:53.440
i did that kind of fast, ill back up for a second so you can see what i did
00:57:54.840 --> 00:57:56.680
this is here under fall of 08
00:57:58.400 --> 00:58:01.040
i remember when i was doing the fall of 08
00:58:01.520 --> 00:58:06.960
that was a long time ago for some of you, you were stil in elementary school oh my god
00:58:07.880 --> 00:58:11.240
so well do two, try and figure out if its continuous
00:59:43.380 --> 00:59:46.500
alright thats long enough, we suffered enough
00:59:49.740 --> 00:59:51.660
is this function continuous
00:59:54.240 --> 00:59:56.640
and if not where is it not continuous
01:00:14.740 --> 01:00:17.060
alright we go 3 things going on here
01:00:20.220 --> 01:00:23.100
x is less than negative 1, we have 3x squared
01:00:23.100 --> 01:00:24.900
if x is between 1 and negative 1 we have
01:00:26.240 --> 01:00:27.120
3tan(pi/4)x
01:00:29.320 --> 01:00:31.480
and x is gretaer than 1 is 3x cubed
01:00:31.480 --> 01:00:34.680
so first of all its going to be continuous everywhere
01:00:34.680 --> 01:00:37.840
1 and negative 1, thats the place we have to check
01:00:38.120 --> 01:00:41.080
but polynomials are continuous everywhere
01:00:43.120 --> 01:00:44.320
and the 3tan pi/4 x
01:00:44.320 --> 01:00:46.920
isnt going to be a problem between 1 and negative 1
01:00:48.000 --> 01:00:50.160
because youre between tan pi/-4
01:00:50.160 --> 01:00:52.820
and tan positive pi/4 its continuos there
01:00:52.820 --> 01:00:55.900
tangent doesnt give you a problem till you get to pi/2
01:00:58.420 --> 01:00:59.940
so you could write that
01:01:01.400 --> 01:01:04.840
so part of your answer is going to be all reals except
01:01:04.840 --> 01:01:06.940
you have to figure out where its not continuous
01:01:08.320 --> 01:01:10.880
so how would we test at negative 1, well
01:01:10.880 --> 01:01:12.660
first of all you just test in your head
01:01:12.660 --> 01:01:15.180
you plug in negative 1 in the top and you get 3
01:01:16.240 --> 01:01:18.960
you take -3 in the middle part you get 3 tan
01:01:19.620 --> 01:01:20.580
negative pi/4
01:01:20.580 --> 01:01:23.920
now youre really glad you tattooed the unit circle on your wrist
01:01:25.300 --> 01:01:28.180
and you look and you remember tan of pi/4 is 1
01:01:28.200 --> 01:01:30.280
so the tan of -pi/4 is negative 1
01:01:30.880 --> 01:01:32.160
so thats negative 3
01:01:32.740 --> 01:01:35.060
3 is not the same thing as negative 3
01:01:35.060 --> 01:01:37.360
its not going to be continuous at negative 1
01:01:37.780 --> 01:01:40.340
so how do you show that, you do the limit
01:01:41.500 --> 01:01:44.300
x approaches negative 1 from the left side
01:01:45.380 --> 01:01:45.880
of f of x
01:01:48.340 --> 01:01:49.780
3 times -3 squared is 3
01:01:54.980 --> 01:01:58.420
and the limit as x approaches 1 from the left side of x
01:02:03.060 --> 01:02:05.220
is 3tanpi/4, which is negative 3
01:02:09.880 --> 01:02:11.080
so not continuous
01:02:15.100 --> 01:02:16.540
at x equals negative 1
01:02:21.840 --> 01:02:23.440
alright lets test it at 1
01:02:23.800 --> 01:02:25.320
well 1, this would be 3
01:02:27.260 --> 01:02:28.540
3tan pi/4 would be 3
01:02:28.540 --> 01:02:30.840
its going to be continuous at 1 but lets check
01:02:32.620 --> 01:02:33.580
we do the limit
01:02:34.360 --> 01:02:37.000
x approaches 1 rom the minus side of f of x
01:02:39.600 --> 01:02:41.120
3 tan pi/4, 3tan pi/4 is
01:02:41.800 --> 01:02:42.300
1
01:02:44.380 --> 01:02:45.180
see you get 3
01:02:51.620 --> 01:02:54.580
limit as x approaches 1 from the plus side of x
01:02:56.280 --> 01:02:56.840
is just 3
01:02:57.660 --> 01:03:00.780
your not done though, you also have to show that
01:03:02.540 --> 01:03:04.300
f of 1 is 3 so its continuous
01:03:07.100 --> 01:03:10.220
at x equals 1, so you would say this is continuos
01:03:10.720 --> 01:03:11.520
at all reals
01:03:13.660 --> 01:03:14.160
except
01:03:17.720 --> 01:03:19.640
x=-1, or something like that
01:03:30.320 --> 01:03:31.760
whatd you say f of 1 is 3
01:03:31.760 --> 01:03:33.860
continuity you have to show the limit from the left
01:03:33.860 --> 01:03:37.400
plus the limit from the right equals the value of the function
01:03:38.260 --> 01:03:40.820
do bot parts, show that the limit exist
01:03:40.820 --> 01:03:42.020
and the two sides are the same
01:03:42.620 --> 01:03:45.100
and that equals the function its self
01:03:50.360 --> 01:03:51.800
alright we can skip it
01:03:53.700 --> 01:03:55.140
number 3, dont do that
01:03:58.820 --> 01:03:59.860
lets do number 5
01:04:02.820 --> 01:04:04.340
5a heres a fun function
01:04:05.320 --> 01:04:07.800
list all values where it has a maximum
01:04:21.140 --> 01:04:25.140
lets see if i can do a reasonable approximation of this curve
01:04:30.280 --> 01:04:31.640
something like that
01:04:54.500 --> 01:04:56.980
thats the graph of the derivative of x
01:04:59.800 --> 01:05:01.480
the derivative of f sorry
01:05:01.480 --> 01:05:03.000
thats the derivative of f prime
01:05:04.260 --> 01:05:08.180
what is the maximum that maximum means the graph is going up
01:05:08.180 --> 01:05:09.840
and the graph is going down
01:05:10.380 --> 01:05:12.060
so if the graph is going up
01:05:12.060 --> 01:05:15.220
the derivative would be positive, the graph is going down
01:05:15.220 --> 01:05:16.900
the derivative would be negative and at the to[
01:05:17.700 --> 01:05:18.260
you get 0
01:05:19.200 --> 01:05:20.720
so we have a 0 right here
01:05:21.600 --> 01:05:22.960
positive above that
01:05:23.380 --> 01:05:26.340
negative below that, so we have a maximum at 0
01:05:29.260 --> 01:05:30.220
here we have a 0
01:05:30.240 --> 01:05:32.160
but were negative to the left
01:05:32.220 --> 01:05:35.260
positive to the right so this will be a minimum
01:05:35.640 --> 01:05:36.520
not a maximum
01:05:37.100 --> 01:05:39.180
so the only place is at x equals 0
01:05:44.040 --> 01:05:45.480
thats it just that one
01:05:47.720 --> 01:05:49.800
can i explain that one more time
01:05:51.120 --> 01:05:52.400
sure okay a maximum
01:05:52.400 --> 01:05:54.320
if oyu think about the maximum on a graph
01:06:06.440 --> 01:06:08.600
whats going on right there, well
01:06:08.780 --> 01:06:09.820
to the left of it
01:06:10.140 --> 01:06:14.940
the slope is positive, the curve is going up, so you have a positive slope
01:06:16.540 --> 01:06:17.260
right here
01:06:19.460 --> 01:06:20.340
the slope is 0
01:06:21.740 --> 01:06:23.660
and now we get negative slope
01:06:28.780 --> 01:06:31.580
so if we look at our graph of our derivative
01:06:31.580 --> 01:06:34.960
we know that here its positive, were above the x axis
01:06:36.240 --> 01:06:38.800
we get a zero then we are below the x axis
01:06:38.800 --> 01:06:40.460
so that would be a maximum
01:06:41.640 --> 01:06:42.760
as oppose to here
01:06:42.840 --> 01:06:46.200
where at first were below the axis a negative value
01:06:46.200 --> 01:06:49.320
and then 0 and above the x axis so this would be a minimum
01:06:49.840 --> 01:06:52.240
because a minimum you are going down
01:06:52.640 --> 01:06:54.640
and you have 0 and your going up
01:06:54.640 --> 01:06:56.640
to the negative and you have a 0 and a poitive
01:06:58.880 --> 01:06:59.840
this one, good
01:07:02.640 --> 01:07:04.960
one more time, what are you missing
01:07:12.700 --> 01:07:14.860
the originally graph has to go up
01:07:14.880 --> 01:07:16.720
and then down to be a maximum
01:07:17.120 --> 01:07:21.200
the maximum value you reached the highest value for some zone
01:07:21.200 --> 01:07:22.240
its called a local max
01:07:22.420 --> 01:07:25.140
were going up till we get 0 were going down
01:07:25.140 --> 01:07:28.420
that means we have a slope on the left thats positive
01:07:28.420 --> 01:07:30.240
the slope on the right gets negative
01:07:30.300 --> 01:07:32.460
and right in the middle has to be 0
01:07:32.460 --> 01:07:34.680
so if we look at the derivative graph
01:07:35.300 --> 01:07:38.180
positive meets above the x axis so positive
01:07:38.320 --> 01:07:40.560
then were 0 and then were negative
01:08:01.300 --> 01:08:05.540
if you want a maximum or a minimum you have to cross the x axis okay
01:08:05.540 --> 01:08:07.880
it has to have a 0, the derivative graph has to have a 0
01:08:08.180 --> 01:08:10.340
the original graph well lets see
01:08:12.420 --> 01:08:13.700
here only positive
01:08:15.380 --> 01:08:17.460
so this will be getting steeper
01:08:18.600 --> 01:08:21.080
so your graph would be sort of going up
01:08:21.920 --> 01:08:25.040
and then it will be getting shallower like that
01:08:27.000 --> 01:08:29.400
and then here it would be the maximum
01:08:29.400 --> 01:08:32.220
and then it would turn arund and come down, thats what this piece would look like
01:08:35.280 --> 01:08:38.480
but im not really going to do that, thats painful
01:08:42.140 --> 01:08:48.380
if we go back and ask you to graph the orginal derivative, i dont think we would do that, we might
01:08:50.600 --> 01:08:52.840
i dont know, i didnt write the exam
01:08:56.460 --> 01:08:59.340
exam is really nice, well written, printed
01:09:01.080 --> 01:09:05.320
we are giving you guys a review session we have to save some stuff
01:09:05.920 --> 01:09:07.600
heres something we can do
01:09:09.280 --> 01:09:09.840
number 8
01:09:16.960 --> 01:09:18.400
no wrong test, hang on
01:09:19.960 --> 01:09:20.600
lets go to
01:09:22.780 --> 01:09:23.580
spring of 09
01:09:23.580 --> 01:09:26.460
although February is not exactly the spring
01:09:29.700 --> 01:09:30.660
heres a fun one
01:09:35.240 --> 01:09:39.320
sketch the graph of a function that does all that kind of stuff
01:12:30.800 --> 01:12:33.600
we had enough should i start drawing this?
01:12:49.100 --> 01:12:50.940
alright we got f of 0 equals 0
01:12:51.380 --> 01:12:52.340
pick up a point
01:12:58.320 --> 01:12:59.280
f of 7 equals 11
01:13:01.900 --> 01:13:02.400
so 7, 11
01:13:04.340 --> 01:13:05.540
okay so far so good
01:13:05.800 --> 01:13:09.640
and you should sort of cross these off as you are doing the,
01:13:09.720 --> 01:13:11.880
look at the bottom we got f of 1 is 3
01:13:12.540 --> 01:13:13.340
and f of 2 is 3
01:13:18.560 --> 01:13:20.640
alright those are the easy ones
01:13:25.520 --> 01:13:27.040
now lets do some limits
01:13:27.240 --> 01:13:29.560
the limit as x apporaches 7 from the
01:13:30.560 --> 01:13:31.520
minus side is 3
01:13:33.960 --> 01:13:37.240
but when we get it from the plus side its negative 3
01:13:38.840 --> 01:13:40.200
so put a circle there
01:13:41.480 --> 01:13:42.280
and lets see
01:13:42.280 --> 01:13:45.820
thats something, when we get it from the negative side
01:13:47.320 --> 01:13:50.920
but from the plus side youre doing something like that
01:13:53.660 --> 01:13:56.140
now the limit as x goes to infinity is 0
01:13:58.500 --> 01:13:59.380
so maybe that
01:14:00.080 --> 01:14:02.320
theres lots of ways you can get to 0
01:14:02.480 --> 01:14:04.560
you can go straight up like this
01:14:05.820 --> 01:14:07.260
you can dance for a bit
01:14:07.260 --> 01:14:08.440
doesnt really matter
01:14:13.400 --> 01:14:15.480
now we get from 2 to the plus side
01:14:16.600 --> 01:14:17.880
we go up to infinity
01:14:21.480 --> 01:14:23.560
but notice at 2 were at infinity
01:14:25.060 --> 01:14:26.660
and 2 from the minus side
01:14:27.600 --> 01:14:29.360
were at negative infinity
01:14:38.860 --> 01:14:40.140
and as f apporahces
01:14:41.660 --> 01:14:42.940
1 from the plus side
01:14:43.860 --> 01:14:46.740
is that right, that piece doesnt look right
01:14:51.860 --> 01:14:55.220
make sure i get all that correct 2 from the plus side
01:14:55.220 --> 01:14:56.660
up to positive infinity
01:14:57.460 --> 01:14:58.820
2 from the minus side
01:15:00.120 --> 01:15:01.400
negative infinity
01:15:04.800 --> 01:15:06.160
something like that
01:15:06.460 --> 01:15:10.460
so on the left side of 2 we are going down to negative infinity
01:15:10.460 --> 01:15:12.720
and the right side of 2 we are going up to positive infinity
01:15:15.220 --> 01:15:16.980
im sorry the right side of 1
01:15:17.840 --> 01:15:19.680
on the left side of 1 were at 5
01:15:25.140 --> 01:15:27.460
and we have to go through the origin
01:15:29.500 --> 01:15:31.180
and to negative infinity
01:15:31.400 --> 01:15:33.800
lets check all our conditions again
01:15:35.320 --> 01:15:35.880
f of 0 is 0
01:15:38.780 --> 01:15:41.020
f of 7 is 11, when were at 7 were at 11
01:15:41.900 --> 01:15:44.380
when we approach 7 from the minus side
01:15:45.260 --> 01:15:45.900
we get to 3
01:15:46.760 --> 01:15:49.160
when we approach 7 from the plus side
01:15:49.960 --> 01:15:51.640
thats this way we get to -3
01:15:52.880 --> 01:15:54.320
as we got to infinfity
01:15:54.640 --> 01:15:56.240
we get to 0, so far so good
01:15:57.580 --> 01:16:00.940
now when we approach 2 form the plus side, over here
01:16:00.940 --> 01:16:02.860
were going up to positive infinity
01:16:02.860 --> 01:16:07.580
when we approach two from the left side were going to negative infinity so thats a vertical asymptote
01:16:07.580 --> 01:16:09.600
except we actuallly have a value at 2
01:16:09.600 --> 01:16:13.900
so its kind of weird, you dont actually have to draw the dotted line if you dont want to
01:16:14.280 --> 01:16:17.080
because its not actually correct alright
01:16:17.080 --> 01:16:21.200
when we go up from 1, from the right side we go from positive infinifty
01:16:22.280 --> 01:16:24.600
1 from the left side we have to go ot 5
01:16:25.540 --> 01:16:29.620
then we have to go down, as we go down we go to negative infinity
01:16:31.180 --> 01:16:31.820
that hurt
01:16:37.800 --> 01:16:39.800
no you dont connect the points
01:16:40.880 --> 01:16:42.240
howd we like that one
01:16:46.140 --> 01:16:49.100
oh i went through the origin because f of 0 is 0
01:16:50.660 --> 01:16:54.020
you do need to go through that at the origin correct
01:16:57.520 --> 01:17:00.160
lets see if therees another one like thi
01:17:04.300 --> 01:17:07.180
february 07, we might have to do one of those
01:17:11.840 --> 01:17:14.400
october 08, do we have to do one of those
01:17:17.320 --> 01:17:19.240
take a picture it last longer
01:17:21.800 --> 01:17:22.600
february 07
01:17:31.580 --> 01:17:33.580
no okay so that was one like tht
01:17:33.580 --> 01:17:35.840
see if theres other good stuff to go through
01:17:42.480 --> 01:17:43.760
how are we on the e;s
01:17:45.200 --> 01:17:46.240
heres a good one
01:18:03.180 --> 01:18:05.020
limit as x approaches 0 plus
01:18:08.640 --> 01:18:09.840
this is which test
01:18:19.400 --> 01:18:20.920
this is the february 09
01:18:22.160 --> 01:18:22.880
the spring
01:18:39.720 --> 01:18:42.280
what si the limit as x approaches 0 plus
01:18:49.480 --> 01:18:50.840
e to the negative 1/x
01:18:51.000 --> 01:18:53.800
if youre not sure what e to the -1 looks like
01:18:53.800 --> 01:18:56.540
for now put it in your calculator and take a picture of it
01:18:56.820 --> 01:18:58.820
sor of plant taht in your brain
01:18:59.640 --> 01:19:01.400
e to the x e to the negative x
01:19:01.400 --> 01:19:03.660
e to the 1/x, you should know what those graphs look like
01:19:05.660 --> 01:19:08.460
so what happens to the graph og e to the -1/x
01:19:08.580 --> 01:19:09.860
when x approaches 0
01:19:10.060 --> 01:19:11.580
from the positive side
01:19:12.120 --> 01:19:13.880
well 1/0 would be infinite
01:19:15.640 --> 01:19:19.320
1 over positive number youd get positive but this is 1/x
01:19:19.580 --> 01:19:21.100
so this would look like
01:19:22.020 --> 01:19:23.780
e to the negative infinity
01:19:25.700 --> 01:19:30.100
e to the negative something is something over e positive infinity
01:19:31.320 --> 01:19:32.280
that would be 0
01:19:33.160 --> 01:19:36.760
because e to the infinity is a really big number so its 0
01:19:39.840 --> 01:19:40.880
1/infinity is 0
01:19:46.960 --> 01:19:47.840
so far so good
01:19:50.060 --> 01:19:53.500
as 0 plus means are denominator is a positive number
01:19:53.500 --> 01:19:55.460
approaching 0 so were getting infinity
01:19:56.300 --> 01:19:57.900
but this is negative 1/x
01:19:57.900 --> 01:20:01.000
so the whole thing becoes negative so you get negative infinity
01:20:02.000 --> 01:20:04.080
okay because you get negative 1
01:20:04.980 --> 01:20:06.500
over a positive number
01:20:09.260 --> 01:20:11.580
e to the negative infinity is 1 over
01:20:11.580 --> 01:20:14.260
e to the positive infinity, e to the positive infinity is positive infinity
01:20:14.740 --> 01:20:16.260
right thats our e graph
01:20:17.080 --> 01:20:17.800
so this is 0
01:20:20.860 --> 01:20:22.060
how bout the limit
01:20:22.840 --> 01:20:24.520
when x approaches 0 minus
01:20:26.480 --> 01:20:27.280
e to the -1/x
01:20:30.300 --> 01:20:32.700
well now this becomes positive -1/x
01:20:32.700 --> 01:20:33.720
so this would look like
01:20:34.200 --> 01:20:35.960
e to the positive infinity
01:20:36.940 --> 01:20:38.460
which is just infinity
01:22:03.340 --> 01:22:04.940
who has questions for me
01:22:11.560 --> 01:22:12.600
well do one more
01:22:12.600 --> 01:22:16.560
well do one more question and then well put an end to this pain
01:22:24.020 --> 01:22:25.700
lets go to the spring of 09
01:22:26.740 --> 01:22:27.620
one more time
01:22:32.680 --> 01:22:35.960
explain why the function is continuous at 3 or not
01:22:39.300 --> 01:22:42.660
ill give you a clue its either continuous or its not
01:24:02.180 --> 01:24:05.140
explain whether the function is continuous
01:24:05.840 --> 01:24:06.340
or not
01:24:10.200 --> 01:24:11.560
well lets do it, f of x
01:24:16.360 --> 01:24:18.040
equals x squared minus 3x
01:24:20.820 --> 01:24:22.260
over x squared minus 9
01:24:24.340 --> 01:24:25.220
when x is not 3
01:24:27.180 --> 01:24:27.680
and 21
01:24:29.420 --> 01:24:30.060
when x is 3
01:24:36.560 --> 01:24:40.160
well lets see were going to have to do the limit arent we
01:24:44.800 --> 01:24:46.560
so the limit when x goes to 3
01:24:48.460 --> 01:24:49.660
x squared minus 3x
01:24:51.160 --> 01:24:52.600
over x squared minus 9
01:24:53.840 --> 01:24:54.340
0/0
01:24:55.900 --> 01:24:58.860
that means there must be a x-3 factor in these
01:25:00.160 --> 01:25:00.880
we can pull
01:25:02.820 --> 01:25:04.420
an x minus 3 out of the top
01:25:07.040 --> 01:25:08.000
and the bottom
01:25:13.380 --> 01:25:15.940
look at that, now the x minus 3's cancel
01:25:18.060 --> 01:25:19.100
and so the limit
01:25:21.680 --> 01:25:22.800
is x approaches 3
01:25:24.520 --> 01:25:25.560
is now of x plus 3
01:25:26.760 --> 01:25:27.560
which is 3/6
01:25:29.020 --> 01:25:29.580
or a half
01:25:36.680 --> 01:25:37.560
but f of 3 is 21
01:25:42.620 --> 01:25:43.120
so f of 3
01:25:44.240 --> 01:25:45.200
does not equal
01:25:46.540 --> 01:25:48.780
the limit of x approaches 3 of f of x
01:25:50.280 --> 01:25:51.000
therefore
01:25:52.700 --> 01:25:57.420
therefore not ontinuous at x equals 3, its continuous everywhere else
01:26:00.360 --> 01:26:02.920
do that again so the limit you get a half
01:26:05.140 --> 01:26:06.020
but f of 3 is 21
01:26:06.020 --> 01:26:09.400
which means you have a little whole a a little dot at 21
01:26:09.940 --> 01:26:10.980
there not equal
01:26:11.840 --> 01:26:13.920
itsnot continuous at x equals 3
01:26:16.300 --> 01:26:19.180
good alright i think i had enough for one day