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Function problem involving IRC

  1. Jan 21, 2013 #1
    1. The problem statement, all variables and given/known data


    See attachment - Question3
    2. Relevant equations



    3. The attempt at a solution

    Not sure. I believe it's an easy plug and chug, but I could be wrong. I'm having trouble where to begin
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jan 21, 2013 #2

    Dick

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    Sure plug and chug part b). Then part c) says to take a guess. Part d) says to verify it using the definition of derivative. Where are you stuck?
     
  4. Jan 21, 2013 #3
    Part B: I did a similar one earlier, but alright so I'm using the g(x) equation right?

    Double check:

    So to get the value for 1.9 for the table I plug that number into this equation

    5(1.9)^2+3(1.9)-4 - 5(2)^2 + 3(2) - 4 / (1.9-2)
     
  5. Jan 21, 2013 #4

    Dick

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    Ok. You aren't really using enough parentheses to indicate what you mean, but I'll trust you know. What do you get?
     
  6. Jan 21, 2013 #5
    I got for question C - a number around 23.

    The values I got for the table is in order from left to right: 22.5, 22.95, 23, 23, 23.05, 23.5

    I need help for question D: I'm not familiar with the chain rule
     
  7. Jan 21, 2013 #6

    Dick

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    Ok, so you must have a pretty good idea for c). Why do you think you need a chain rule for d)?
     
  8. Jan 21, 2013 #7
    Not sure, a friend told me - he may be wrong.


    It says to calculate the IRC at x=2, do I just plug in 2 for 5(x)^2+3X-4 ?
     
  9. Jan 21, 2013 #8

    Dick

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    The IRC is the limit of g(x) as x->2, isn't it? That's what part b) was on about. g(x) is a difference quotient for f(x). What do limits of difference quotients have to do with derivatives? Check back to the definition of derivative.
     
  10. Jan 21, 2013 #9
    a derivative is a function changing due to a different input or also how one quantity is changing due to another quantity.
     
  11. Jan 21, 2013 #10

    Dick

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    That's vague. The definition of f'(a) is limit x->a of (f(x)-f(a))/(x-a). Don't you have that definition someplace?
     
  12. Jan 21, 2013 #11
    Yes, it's on the THQ question.

    The reason why I'm sort of lost on these questions is 1.) I just started calc and 2.) We only covered 2 sections so far barely going over any problems. Matter of a fact our first problem section is tom w/ a classroom quiz over this material that he hasn't even covered fully.

    Sorry for my lack of knowledge regarding this material. This is my last question and the packet is due tomorrow.

    As for your question - yes i have it - not in classroom notes because he didn't go over it. But there is a question in the packet that involves the difference quotient and I solved it since it I knew it from college alg.

    I'm still unsure for this last part. My 3rd class for calc is tom so we literally started a week ago. Also, we didn't brush up on any precal either (never took it).
     
  13. Jan 21, 2013 #12

    Dick

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    All ok. Part b) is asking you to compute values of (f(x)-f(2))/(x-2) for values of x that get close to 2. The closer x gets to 2 the closer you should be to the IRC. You should have guessed that the IRC is 23. To find it analytically you need to compare that with f'(2). Do you know how to find the derivative of f(x)=5x^2+3x-4 and put in x=2?
     
  14. Jan 21, 2013 #13
    hold on one sec - is it 10x +3? then 10(2) + 3 = 23
     
    Last edited: Jan 21, 2013
  15. Jan 21, 2013 #14

    Dick

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    Yes, but no chain rule. It's just differentiating really. Can you differentiate 5x^2, 3x or 4? Just checking what you know.
     
  16. Jan 21, 2013 #15
    I got 10X +3

    10(2) +3 = 23 IRC

    I wasn't even taught this, but figured it out. Is this a precal thing?
     
  17. Jan 21, 2013 #16

    Dick

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    No, it's a full calculus thing. The precalculus thing would have been to derive lim x->2 (f(x)-f(2))/(x-2) from first principles. Any idea how to do that? Not having taken precalc is not going to help. But try and tough it out.
     
  18. Jan 21, 2013 #17
    No idea how to do that. Did I get the right answer though?

    And yea I'ma try. I really need to. This is my last semester in order to graduate, and I've heard horror stories about people failing and it being there last class ;/ I never failed a class though so hopefully I'll do alright, but then again I made a 43 in Organic chemistry 2 but that was considered a C- that being my only C in general and lowest grade.
     
  19. Jan 21, 2013 #18

    Dick

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    It's the right answer. The derivative of 5x^2+3x-4 is 10x+3. It's nice to know how to really derive things, but if it's the last semester just try and bluff your way through. Good luck! It's easier than organic chemistry.
     
  20. Jan 21, 2013 #19
    Thanks, I appreciate the help! And I'll do my best to do so!
     
  21. Jan 22, 2013 #20

    Dick

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    By the way, the precalc way is to find the limit (f(x)-f(2))/(x-2) is not so bad. f(2)=22. So you get 5x^2+3x-26=(x-2)(5x+3). Putting that into the quotient and cancelling the (x-2) gives (5x+3). Finally putting x=2 give 23. Just like the calculus way.
     
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