A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Anikate

  • one year ago

need help with limits, posting an equation. all help is appreciated. medals...

  • This Question is Closed
  1. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1433989546928:dw|

  2. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1433989546928:dw|\[(2x^2 -x -3)/ (2x^2 -5x +3)\]

  3. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    limit is x -> 1 and theres a + beneath the limit. how do i do this?

  4. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    \[\lim_{x\rightarrow ^+1} \frac{ (2x^2−x−3)}{(2x^2−5x+3)}\] This?

  5. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    So, are you familiar with what the + under the limit means?

  6. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    @Anikate you still there? :)

  7. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @jtvatsim yea sorry had to use the bathroom

  8. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I do not know what that means. sorry fpr late respoce

  9. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    You know that isn't allowed... jk :)

  10. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lolz

  11. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    OK, so the +1 just means that the limit is approaching 1 from the right side of the graph. It is a "one-sided limit" if you've heard of that term.

  12. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is there such thing as a two sided?

  13. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @jtvatsim

  14. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Yes, most limits are two-sided. This is what is indicated by having no + or - sign under the limit.

  15. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok, so what is the significance of the +?

  16. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    The + restricts the limit to coming only from the right side of the graph, a picture will help.

  17. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    |dw:1433990529989:dw|

  18. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm..... lost at that picture

  19. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    In this picture we have, \[\lim_{x\rightarrow ^+3} graph = 2\] \[\lim_{x\rightarrow ^-3} graph = 1\] \[\lim_{x\rightarrow ^3} graph = DNE\]

  20. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    As you approach x = 3 from the right the y values tend to 2.

  21. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    As you approach x = 3 from the left the y values tend to 1.

  22. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    If you approach x = 3 from both sides, you do not hit the same number. There is no two-sided limit in this picture.

  23. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what do u mean by grpah=2 or graph=1?

  24. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    I mean whatever the equation of that graph is. I'm using the word "graph" to stand for whatever the algebraic equation should be (I have no idea... :) ).

  25. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok

  26. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    In your case, you have been given the equation of some graph. And they are asking you to find out what the limit is from the right side.

  27. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and what about the "tend to 2"

  28. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    That's how I understand the concept of a limit. The y values never actually become 2, they just approach it "arbitrarily close." Those words are weird to me, so I think about it as the y values "approach 2" or "tend to become 2".

  29. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok so whats next? do we grpah my equation to find the asnwer?

  30. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    That is a very valid approach. We don't have to do that. If you recall, many limits can be found by simply plugging in the number that x is approaching. We might want to try that first (it's easier).

  31. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok, so how do do that?

  32. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Well, we are looking for \[\lim_{x\rightarrow ^+1} \frac{(2x^2−x−3)}{(2x^2−5x+3)}\] Let's try substituting x = 1 into the equation.

  33. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  34. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    \[\lim_{x\rightarrow ^+1} \frac{(2x^2−x−3)}{(2x^2−5x+3)} = \frac{2(1)^2-1-3}{2(1)^2-5+3}\]

  35. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i got 0

  36. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    But where did you get the zero?

  37. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    \[\frac{-2}{0}\]

  38. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yup... wait nvm, its UND

  39. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    UNDEFINED, cuz the 0 is on the bottom

  40. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Right, this is either positive or negative infinity.

  41. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    idk which one

  42. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Now is where that +1 thing comes in handy. We know that all the x's we plugged in are just a little bit bigger than 1. They are to the right of 1 after all.

  43. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    So, it might be best to test a number just a little bigger than 1 to see what kind of answer we get. Let's try x = 1.01.

  44. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    all the x's we plugged in WERE 1

  45. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    why 1.01? why not 2?

  46. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Well, yes I know, that's cuz we cheated what the limit actually tells us to do.

  47. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    If we go too far we could potentially get false results.

  48. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    And, yes, how do we define "too far" is the follow up to that.. :)

  49. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how come? doesnt the + mean anythign greater than 1?

  50. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Yes, but the limit is asking us to approach 1 as close as possible. x = 2 will approximate the limit, but it will probably be a bad approximation.

  51. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Welcome to the philosophy of Calculus. The precise science of imprecision and approximation where paradoxes happen left and right. :)

  52. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok, so what if it was - instead of +, would we use .99?

  53. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Yep, that's the idea.

  54. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    I mean it would be even better to use x = 1.0000000000001 in our current situation, but nobody wants to do that.

  55. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    x = 1.01 should get us close enough.

  56. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, sounds good. whats next?

  57. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    In any case, when we plug in x = 1.01, we get...

  58. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    201

  59. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yup thats what i got

  60. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Assuming that this behavior is continuous, I think that we have good reason to believe that the true limit is positive infinity.

  61. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how did we tell it was POSTIVE and how Infinity?

  62. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    I determined that it had to be either positive or negative infinity based on our first result (a number divided by 0). Then, since we got a positive answer just to the right of 1 (+201) it is reasonable to believe that the true answer is positive infinity.

  63. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so any number divided by 0 is inifinity?

  64. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Any NONZERO number divided by 0 is positive or negative infinity. 0/0 is indeterminate and requires additional tests.

  65. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok so the answer is positive infinity?

  66. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    cn

  67. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Yes. This can also be confirmed via a graph. (Try Desmos). :)

  68. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you please help me with one more? and i hav

  69. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    a graphing clalc next to me haha

  70. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Yay!

  71. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1433992014016:dw|

  72. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Is that a superscript -, or an actual negative 3?

  73. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    not even sure what im supposed to find an actual negative 3

  74. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Same question as before, find the limit as x goes to -3.

  75. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    its an actual negative 3

  76. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is 0/0

  77. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    So this one is very unfriendly... :)

  78. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    haha ill try to u

  79. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    rst

  80. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    understand it

  81. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    In these situations we have two options: 1) factor or 2) use L'Hospital's Rule. If you haven't heard of 2 yet, then ignore that.

  82. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    havent heard of 2, but is it easier than 1?

  83. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Yes, but it usually isn't introduced until Calc II. Most Calc I teachers don't like to give it away early, but we can do it both ways if you want.

  84. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im in precalc, but if its easier pleace tell me :D

  85. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    If you are in Precalc, it may not be easier yet, it depends on if you know what derivatives are?

  86. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sounds familiar, what does it look like real quick

  87. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is hospitals like a formula?

  88. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Yes, but it requires you to know derivatives well. For example, if you have x^2, then the derivative is 2x. Is that something you've done?

  89. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    nope

  90. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    will it work if i dont know?

  91. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Well, I would have to tell you the specific derivatives for this equation, but you would probably not be able to use it on different questions: In short, derivatives change one algebra equation into another, they work like this (a short intro): Any number -> becomes 0 x -> becomes 1 x^2 -> becomes 2x x^3 -> becomes 3x^2 x^4 -> becomes 4x^3 Multiplies of expressions -> become multiples of derivatives example 3x -> becomes 3(1) = 3 4x^2 -> becomes 4(2x) = 8x

  92. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    If you can memorize these facts, you will be able to use L'Hospital's rule.

  93. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    got it, keep going

  94. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    why 3x become 3(1), where did the 1 come form?

  95. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Because x always becomes 1 after applying the derivative.

  96. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    o ok, gotcha

  97. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    So, for instance, the equation x^3 + 27 becomes 3x^2 + 0 after applying the derivative.

  98. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    gotcha

  99. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    does the denominator become -4?

  100. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    How does this help? Well L'Hospital's Rule says this: "If you take a limit and the answer is 0/0, you may apply the derivative to the top and bottom equation and the answer will be equivalent".

  101. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Close, -x - 3 becomes -1 - 0 since any number becomes 0 after the deriviative.

  102. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh yea

  103. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    The derivative sort of disintegrates algebra equations into simpler and simpler equations.

  104. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    By L'Hospital's Rule, we can say this:

  105. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    -3x^2

  106. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    \[\lim_{x\rightarrow -3} \frac{x^3 + 27}{-x-3} = \lim_{x\rightarrow -3} \frac{3x^2 + 0}{-1 - 0}=\lim_{x\rightarrow -3} -3x^2\]

  107. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Yes, you are right.

  108. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Now, if you plug in x = -3, you will get the answer.

  109. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I dont understand what my answers should look like?

  110. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Answers to limit questions are numbers or infinities.

  111. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    why do some look like infinites and some jus regular numbers?

  112. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how do i decide if it will be a infinite or a number ?

  113. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Well, those are all the possible numbers that exist. Sometimes y values shoot up very fast into infinity land (like the previous graph). Other times y values level off and reach an actual number.

  114. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    If you get a limit that says (number/0) the answer will be infinity (or Does Not Exist depending on your teacher).

  115. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    If you get a limit that equals an actual number than the answer is that number.

  116. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what if i got -1 for one of them?

  117. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    after plugging in the x value into the equation

  118. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Then just treat it like dividing by -1, I think you are talking about what we just did above. Nothing special happens. We know what to do when we see a -1 dividing. We have no idea what to do if we see a 0 dividing.

  119. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no after plugging in the x value and everythign my answer was -1, what do i do?

  120. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    If after plugging in the x value your answer is -1, then that is the answer to the limit question. You would write: lim_x-> whatever (some equation) = -1.

  121. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok gotcha, can you help me with a piecewise function?

  122. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Sure, and by the way our answer to this previous question is -27. Just checking.

  123. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yup i got -27

  124. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    just a min i will post a pic

  125. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    OK cool, your teachers will not expect you to use that L'Hospital method, but it just happened. :)

  126. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

  127. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol it will be fun surprising my math teacher with that lol math hacks

  128. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i didnt do the piece wise right, can u plz explain to me how to graph and get the answers?

  129. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    OK, let's see...

  130. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    So, would you know how to graph -x^2 + 1 if it was just by itself?

  131. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I just plug in random x values and get the y values to plot

  132. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    OK, that works. The only thing you will do differently is only pick random x values that are less than 0.

  133. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    That is what the piecewise function is telling you "Plot -x^2 + 1" for x values less than 0.

  134. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea i didx that

  135. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Good. We can imagine that at x = 0 (right on the not allowed zone) we would get a value of 1, so something like this picture ...|dw:1433993796086:dw|

  136. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Alright, now we do the same thing with the |2x-1| for x >= 0 graph.

  137. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    why would we plot at 0? do we really need it? i know the inequality says x<0 not lesser than or equal to

  138. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    We don't "really" need it, it's just a nice visual barrier to have an 'open dot' at 0. As you say, there isn't really anything plotted there, it is just a reminder of where the plot ends.

  139. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok, it cant just end on the x-axis at -1?

  140. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and leave a open circle there?

  141. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Only if you are considering whole numbers. I get the feeling that we should include all decimal numbers as well.

  142. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    That's the standard practice anyways.

  143. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    why decimals ?

  144. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Well, certainly x = -0.0000001 is less than 0 and should be included as part of the graph

  145. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea

  146. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh so, we just start at 0 to be friendly?

  147. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Right. This gives us a feeling for the shape of the graph. Since, of course, you can't plug in all the decimals that are out there, so we end up connecting the dots just to get a sense of the shape.

  148. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    got it, how do we do the next peice wise functions?

  149. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Well, let's use the same strategy since it worked for the last one. Now we are allowed to plug in x = 0 and greater.

  150. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yup quick question

  151. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the absolute values

  152. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    does that automatically turn the equation into 2x+1

  153. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Very good question.

  154. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Try a few numbers and see what you think... :)

  155. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    or does the answer just turn positive if its negative?

  156. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    What you just said is the best way to start. Take numbers, plug in, and make them positive if they are negative.

  157. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so does the equation turn postive from the beginning itself, or do we wait for the answer then turn it positive?

  158. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Wait for the answer. The absolute value bars are like parentheses and don't act until after the inner answer.

  159. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok

  160. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    You've had very good questions so far! You have the making of a great mathematician!

  161. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    :D thanks!!

  162. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok so theres an issue here

  163. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    when i plot x=0 and x=1 it appears to be the same y value? looks odd on the graph

  164. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    That feels weird right? But the math doesn't lie. You have it correct so far. :)

  165. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so should i make it a curved line or a straight line through the points?

  166. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Good question. You could confirm by graphing points like x = 1/2, x = 1/4, and x = 3/4, but I won't torture you. :)

  167. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    The inner equation is that of a line, 2x - 1. The absolute value graph will act the same except it will be the shape of a "V" now. Here's a picture.

  168. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh yea i plotted it, i see it now

  169. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    |dw:1433994694244:dw| Perfect!

  170. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    now what are these conintues or dis continues things?

  171. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and jump discont etc,

  172. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Continuous functions are basically those that you could draw without lifting your pencil. In other words, they have no "holes" and no "gaps" or "jumps".

  173. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Here are two examples.

  174. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    |dw:1433994827549:dw|

  175. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    A discontinuous graph is "broken" in some sense.

  176. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    mhm

  177. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    When we first plotted -x^2 + 1, we had a hole at x = 0, y = 1 since we weren't really allowed to plot there. However, after we plotted |2x - 1| we filled that exact hole in with x = 0, y = 1 which was allowed to plot. There is no hole at x = 0, obviously no jump, so the graph is continuous at x = 0.

  178. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok

  179. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Does that make sense? How about does the limit exist?

  180. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what do u mean by does the lmit exist?

  181. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea idk how to do limits or anything like those on the piecewise functions

  182. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    OK, so here's a quick test to tell if a limit exists.

  183. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

  184. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Take your right finger and place it on the right side of the pencil graph. Take your left finger and place it on the left side of the pencil graph.

  185. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Now, trace the graph with your fingers (moving them towards each other) going to x = 0.

  186. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    pencil graph?

  187. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    The graph you drew.

  188. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok

  189. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    done

  190. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    If they touch when you reach x = 0, then the limit exists at x = 0.

  191. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    any other way of checking other than that

  192. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Yes, as long as there are no jump discontinuities at the point, the limit exists.

  193. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    This is exactly what the tracing trick tests for.

  194. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok, so how do i say that a limit exists?

  195. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Your teacher probably will say something like: "The limit exists when the left-hand limit equals the right-hand limit" Again, this is the same thing as the tracing trick.

  196. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yea he does, but just in case, is there any other way of saying it?

  197. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can i just say yes a limit exists, cuz it only

  198. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    asks if a limit exists

  199. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Sure. It's a good practice to give your reason for saying that. You could simply say, the left-hand limit as x approaches 0 is 1, and the right-hand limit as x approaches 0 is 1. Since they are equal, the limit as x approaches 0 exists and is equal to 1.

  200. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol jsut a sec trying to translate all that in my head

  201. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    lol, I guess that is a bit much... :)

  202. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so x=1?

  203. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    The limit is 1. (The limit talks about y-values, if that helps).

  204. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    So as x approaches 0 (x = 0), y goes to 1, (y = 1).

  205. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how do we tell if the limit is talking the x value or the y value?

  206. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    In general, the limit always talks about the y value. I suppose there are cases where it could talk about the x value, but this is not standard practice.

  207. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh o k

  208. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    If you like symbols, you could write \[\lim_{x\rightarrow 0}f(x) = 1\]

  209. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    f(x) is the piecewise function (that produces y values). Our x is moving toward 0, so we expect that y will end up being 1.

  210. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh gotcha

  211. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait wh

  212. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what wou

  213. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    be the answer to part a?

  214. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ugh sorry if my sentences break apart in messages, my laptop has an oddly placed mousepad hahah

  215. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    no worries, I had to find the graph back.

  216. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    OK, so the left-hand limit would be just tracing the graph with your left finger until you get to x = 0 and report the y value you get (it will be 1). You would write: \[\lim_{x\rightarrow 1^-} f(x) = 1\] Note the - sign to indicate from the left (if you ask me L would make more sense, but...)

  217. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    That should be \[\lim_{x\rightarrow 0^-}\] sorry

  218. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oooh ok gotcha, and b.) is yes c.) is continuous

  219. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    why cant it be lim f(x)=1?

  220. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    without any x information underneath you mean?

  221. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no with a x info, but why is it lim, why cant we add the f(x)=1?

  222. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    So, you are saying why make this complicated by saying lim f(x) = 1? Why not just say f(x) = 1 when x = 0 instead?

  223. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    noo

  224. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you said to write lim(underneath "x -> 0")

  225. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    why cant I write lim f(x)=1 (underneath "x -> 0")?

  226. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    OH! No that's right, you are doing it fine! I was being lazy on that reply. Teaches me... :)

  227. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh ok, lol just makin sure, cuz u corrected urself and I was wondering why haha

  228. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so c.) continuous? right?

  229. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Yeah, I correct that I had written lim(x->1) It should have been going to 0.

  230. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Yes, your answers to b and c are correct.

  231. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok, now how do I do D?

  232. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Well, is it discontinuous?

  233. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    nope

  234. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    So, there you have it. Just say, "Not discontinuous."

  235. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what are all the type of discontinuities?

  236. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    You have "hole" (a missing value), "jump" (a gap), and "infinite" (an asymptote) where the graph shoots up or down to infinity.

  237. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thats the only 3 right? are there any different types of continuities?

  238. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    There are different forms of continuity, but that is usually not discussed until advanced calculus at university.

  239. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh thank goodness lol

  240. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    lol, the first good news all night! :)

  241. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can we go over one more?

  242. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait actually 2? second one is easy hopefully

  243. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    do u live east coast USA?

  244. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Go ahead and close this question. I may have to run, but if you post a new question, others can see it. I'll stick around as long as possible though. :)

  245. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Nope, west coast.

  246. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    OH! what timei s it for you?

  247. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    9:15 or so.

  248. Anikate
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    should I open a new question, are u gonna join it?

  249. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    Yeah, I'll join it. :)

  250. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 6

    just @ me.

  251. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.