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## Calculus I

by: Dr. Taya Dickens

36

0

8

# Calculus I MAT 201

Dr. Taya Dickens

GPA 3.54

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
8
WORDS
KARMA
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## Popular in Mathematics (M)

This 8 page Class Notes was uploaded by Dr. Taya Dickens on Wednesday October 28, 2015. The Class Notes belongs to MAT 201 at Utica College taught by Staff in Fall. Since its upload, it has received 36 views. For similar materials see /class/230523/mat-201-utica-college in Mathematics (M) at Utica College.

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Date Created: 10/28/15
Section 23 February 22 2005 5 Identify all suspicious points and determine all points of discontinuity for zgi 71 Since f is a polynomial there is nothing suspicious going on and hence f is continuous everywhere 6 Identify all suspicious points and determine all points of discontinuity for 31 2171 r When we divide we must be careful not to divide by zero So we should be suspicious when the denominator equals 0 That is when 2r 7 1 0 When we solve for I we get I n 2 So I is a suspicious point To determine if this is a point of discontinuity we plug this value into the function But we determined that 12 is not in the domain of That is f12 does not exist So I 12 is a point of discontinuityl 12 Identify all suspicious points and determine all points of discontinuity for 3t2 iftgi gt 5 if1lttg3 3t271 iftgt3 1 get suspicious when a function changes the rule for its7 de nition For the piecewise de ned function 9 this occurs when t 1 and t 3 At t 1 the rule changes from gt 3t 2 to gt 5 and at t 3 the rule changes from gt 5 to gt 3t2 7 1 So the suspicious points are t 1 and t 3 However these need not be points of discontinuityl Recall the de nition when in doubt always refer to the de nition of continuity We say a function f is continuous at z a if the following three conditions hold 1 fa exists 2 limgcna exists 3 limgcna fal We need to check these three conditions for each suspicious point t 1 and t 3 First we will investigate t i H Since 91 31 2 5 91 indeed existsi E0 This is the part that is a little hard To determine if the limit exists both the lefthand limit and the righthand limit must exist and equal the same value So we rst need to compute limtnli 9ti As we approach from the left we are using the expression 3t 2 since numbers to the left of 1 are smaller than 1 So lim 9t lim 3t2 312 5 tal tal We also need to compute limtnl 9ti As we approach from the 7ight we are using the expression 5 since numbers to the right of 1 are larger than So 139 t 139 55 are 1 Since these two limits are equal limtnl 9t exists and equals 5 9 This part is easy Since we got 5 for both the rst two parts we have limtn19t 91i So 9 passed all three criteria for being continuous at t 1 Hence 9 is contin uous at t 1 We must also investigate t 3i 1 g3 5 so it exists 2 We follow the same reasoning as the rst part Here we must look at the lefthand and righthand limitsi To the left of 3 we use the rule 9t 5 and to the right of 3 we use the rule 9t 3t2 7 1 So lim 9t lim 5 5 LHS LHS and 139 t139 3t2713327126 AIM gt 13 lt gt Since these two limits are different limtn39t does not existl There is no need to check the third criterioni If it fails to satisfy one it fails So 9 is not continuous at t 3 To summarize The function 9 has suspicious points at t 1 and t 3 and 9 is discontinuous at t 3 20 Find the value that should be assigned to f2 any to guarantee that f will be continuous at z 2 for the function 571 fltzgtz 2 Note that out of the three criteria for continuity at a point z a two of them require something from fai The other one requires something out of the limit as I approaches a For this problem a is 2 So let s nd limgcng By cavalierly not naively plugging in z 2 we see that we get 1 on 11 7 277372 0 L which is a nice reasonable number 732 divided by 0 So this limit does not exist So there can be no value that we can assign to to make this function continuousi In particular this function has a pole at z 2 That is the graph will have a vertical asymptote at z 2 22 Determine whether or not the given function is continuous on the prescribed interval where mi 12 if0 zlt2 7 31 1 if 2 S I lt 5 Here the prescribed interval is 0 5 and the only place we should be suspicious of is z 2 since the function changes its de nition there To the left of 2 12 and to the right of 2 31 11 To be continuous at z 2 we require the three conditions to hold 1 32 1 7 which exists 2 We check the lefthand and righthand limitsi W I 22 4 and Ellir 0511r 31 1 321 7 Since these two values are different limgcng does not exist and so the function f is not continuous at z i 32 Show that the given equation has at least one solution on the indicated interval cosz 12 71 on 07r Let us consider the continuous function cosz 7 12 1 on this interval At the endpoints we have f0cos0702117012 which is positive and f7r cos7r 7 702 1 717 n2 1 7W2 Which is negative So by the Intermediate Value Theorem there exists some 5 E 07 such that 0 Or in other words7 There is a c E 07 such that cosc 7 02 1 07 or cosc c2 7 1i Notes on Evaluating Limits February 16 2005 This is meant to be a tip sheet for evaluating limits in Calculus l as a supplement for the assignment on section i 1 Almost all limits that exist are nice and can be done by plugging in the limiting value as in problem 6 of section 22 1231710 7 3233710 11313312475177 3325377 7 99710 271577 7 8 7 E Almost everybody got this problem correct The only mistakes were simple computational ones Thatls good Now it would not be much of a math class if everything were so easy Occasionally we run into a situation where if we plug in the limiting value we do not get a nice answer The nastiest answer we can get is the dreaded 0 This is an indeterminant form for the limit which is a fancy way of saying that the limit can be ANYTHING and so we need to do more work Sorry but for now there is no way around thisl Just wait until we learn some calculus then these problems will become EASY and dare I say fun to i The problems 12 32 even were meant to show you that by manipulating an indeterminant form of a limit we can nd out what the limit is Here is problem 12 from Section 22 1 12 7 4x 4 113 12 7 z 7 2 Now if we cavalierly plug in the limiting value 2 into the expression we see that we get 00 Go ahead and say your favorite fourletter word when you see this llll use the word drat Okay so we need to roll up our sleaves and get to work There are a variety of techniques that we can use For this problem it turns out that the numerator and denominator both factorl So we have 2 7 4 4 7 2 7 2 hm w hm W 0272 127172 172172Il 7 2 lim I 142 z 1 Factor top and bottom Cancel common factors And now we can cavalierly plug in the limiting value and get 172 727279 139 7 xl7gtmZxl 21 3 Problem 20 works a little bit differently Here we have V12 4 7 2 hm 170 I which also gives us the indeterminant form 00 DRAT when we plug in the limiting value 0 A technique that comes up quite a bit in mathe matics to make the expression harder than the original expression to get something nice to happen In this case7 if we multiply the top and bottom by the expression V12 4 2 note the change in the sign from 7 to then something nice does happen x12472 x124721242 hm hm 170 z 170 I I242gt 7 1 I2474 7 z1242 2 z 1 53 mg The nice thing about this is the troublesome factors of z in the numerator and denominator now canceli This now gives us 12 I 139 139 zm2 113mw 0 x0242 Problem 30 reads like 12 cos21 hm 7 170 l 7 cosz Once again we get OOi draw ah forget it Now we have two common trig limits to play with And it turns out we can multiply by the top and bottom by l cosz and get something nice Here we go lim 12 cos21 lim 12 cos21 1 cosz 7017 cosz 170 17 cosz l cosz lim 12 cos21 l cosz 170 1 7 cos2z 9 Now what the heck did that accomplish Well ifl rearrange our old friend the identity sin2 cos2z 1 we get 1 7 cos2z sin2zi Other than saying that you should learn to recognize such identities when they occur there isn7t much more that 1 can say about them So we will substitute sin2z for 1 7 cos2z in the denominator of our expression After that we will manipulate the expressions so that the factors of sinz are under the factors of z aiming to use the limgcno sinzz 1 identity So lim 12 cos2z 1 cosz 1 7 cos2z 70 lim 12 cos2z 1 cosz 40 sin2 11in cos2z 1 cosz 11111 2 Most of the trigonometric problems in this section amount to trying to manipulate the expression so that we get expressions of the form sinzz so that we can use the identity All the examples above were meant to show us the the limits do exist even when it appears that they do not But note that in each case that the limiting value is NOT in the domain of the expression In the following problems we are dealing with a different beast That of piecewise de ned functions When the de ning rule for the function changes we are suspicious of the activity of the function Problem 42 asks us to determine if a limit of a piecewise de ned function exists That is nd 3 7 2x 12 7 5 ifz 2 lgfw Where f ifzgt2 To do this we need to evaluate the lefthand limit liming and the righthand limit limggngJr For the lefthand limit we are taking values less than 2 so we use the rst expression 3 7 2x to evaluate the limit So lim lim 37 2x 37 22 71 172 172 For the righthand limit we are using values greater than 2 so we use the second expression 12 7 5 So 139 1 275227571 132flt1gt i n Since both of these number exists the limit exists and is equal to this value That is limQ 71 It is imperative that you check both the left and right hand limits to do this problemi Problems 56 and 57 deal with the function 2z 1 if z lt 3 4 39 7 3 12 7 1 if z gt 3 Problem 56 asks us to compute 13 1 Now 2 is nowhere near the place where the function f changes its7 de ning rule This happens at z 3 Since 2 is way less than 3 we only have to consider the de ning rule 2z l to compute the limit near 2 So for this problem we have lim lim 2z 1 22 1 6 EH2 EH2 Problem 57 asks us to compute the limit at the place where f does change its7 de ning ruler To properly do this we need to compute the lefthand limit and the righthand limiti For the lefthand limit we are using values less than 3 so we use 2z 1 And so ling lim72zl 23 l 8 For the righthand limit we are using values that are greater than 3 so we use 12 7 1 And so 139 1 27132718 132 W 0532 I Since these limits are equal the total limit exists and equals this value That is lim f 8 EH3 Okay my ngers are tired and 17m sure you are tired of reading my hack neyed prosei I hope that this was somewhat helpfuli

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