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Quadratic vs linear function

  1. May 28, 2016 #1
    Mod note: Changed thread title from "exponential vs linear function" to "quadratic vs linear function.
    basic question, won't give the full question as i don't need it solved for me, but rather need a conceptual overview of HOW to solve so i can do it myself


    given two functions
    A f(n)= c * n
    B f(n)= c * n^2

    i need to figure out when equation b is greater than equation a for the same values of n (when it is more efficent)

    i tried some simple things that got me nowhere and i need some help

    i found the point of intersection of the two functions but that doesn't help as the exponential function still grows faster

    i tried differentiating and finding when the rate of change is equal (for n), that value was 40500
    i'm not sure how that helps me
     
    Last edited by a moderator: May 28, 2016
  2. jcsd
  3. May 28, 2016 #2

    Math_QED

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    I don't see an exponential function?
     
  4. May 28, 2016 #3
    i meant function B, where the degree of the function =2

    sorry if i said that wrong
     
  5. May 28, 2016 #4
    i assumed this would be a precalc question, if i'm incorrect in that regard, please move this to the appropriate subforum
     
  6. May 28, 2016 #5

    Math_QED

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    Well, define:

    f(n) = cn
    g(n) =cn²

    You can use: f>g <=> f-g>0
    I think you should be able to continue now on your own.
     
    Last edited: May 28, 2016
  7. May 28, 2016 #6

    Mark44

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    As already noted by another member, neither of these function is an exponential function. The second function is a quadratic function. BTW, you shouldn't give both functions the same name: f.
    The graphs of the two functions intersect at (1, c). The slope of the first function is always c. The slope of the second function is 2cn, so that should give you an idea of how the graphs are different.
    I don't see how 40500 plays a role here at all. You don't need concepts from calculus to answer your question. Just a graph of the two functions would give you an idea of where cn2 > cn.
     
  8. May 28, 2016 #7
    ok i done goofed
    here's a different and more accurate set of functions

    f(x)=0.002n^2
    g(x)=700n

    i need to find out when g(x) is larger than f(x) for identical values of x


    i tried to be intentionally vague to avoid having the question outright answered but i didn't state things properly. my bad.
     
  9. May 28, 2016 #8

    SammyS

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    I suppose you mean:
    f(x)=0.002x2
    g(x)=700x
     
  10. May 28, 2016 #9

    Math_QED

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    1) g(x) is a function in x. This implies that what you have written is a constant function. Probably, you meant g(n)
    2) In the original question, you did not mention that the c in f(n) and g(n) were different.

    Now, let's get started.

    We have 2 continuous functions f and g.

    You want to know the n values of f(n) and g(n) for which: f(n) > g(n)

    Thus: f(n) > g(n) <=> f(n) -g(n) > 0
    Try to use this.
     
  11. May 28, 2016 #10
    It's when g(n) is bigger than f(n), so when the linear is bigger than the quadratic, the linear has a greater c value.

    I can use the above method for trial and error but I was wondering if there was a method to generate an exact value of x at which g(x) is greater than f(x).
    Hence my misguided attempt to find the derivative. Please ignore the values of the derivative that I have in my first post.

    If I was to grind it out via trial and error there would be no method and thus no reason for my thread.
     
  12. May 28, 2016 #11
    Sorry if my previous sounded ungrateful, I just want to know how to compute this without using trial and error in order to learn something.
     
  13. May 28, 2016 #12

    mfb

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    You can solve f(x)=g(x) for x. For values smaller than that, the linear function is larger, for values larger than that the quadratic function is larger (assuming both have a positive coefficient).

    Alternatively, simplify f(x)<g(x) to get the region where f(x) is smaller than g(x) - nearly the same calculation, but needs more care in some aspects.
     
  14. May 28, 2016 #13
    :smile::cool:Great reply, thanks :)
     
  15. May 28, 2016 #14

    SammyS

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    Actually, trial and error is a method. In this case, it might be good as a method for showing how some other methods work. However, ...

    In post #5, @Math_QED gave the following suggestion.
    That can lead to the answer, even though you have corrected the problem as follows:
    So, if g(x) > f(x), then 0 > f(x) - g(x) .

    In other words, f(x) - g(x) < 0 .

    There are several ways to solve this.

    A way I like is to recognize that the function f(x) - g(x) is continuous. So as x increases, f(x) - g(x) Cannot change sign without passing through zero.

    So solve for f(x) - g(x) = 0 . Then check at "test points".

    By the way: You are at the point of having 100 posts on PF.
    1. Congratulations.
    2. You should learn to use the reply feature so readers know which post you are replying to.
    3. You should learn to use the superscript & subscript features. X2 X2 in the darker blue banner above the message composing box.
    4. Also in that banner the ∑ (Sigma) character will give you access to many useful symbols.
     
  16. May 28, 2016 #15
    is there a button to quote/reply to a post or do i have to use code?

    i'm familiar with using sub/superscript
     
  17. May 28, 2016 #16

    SammyS

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    It's in the lower right hand corner of the following image.

    upload_2016-5-28_21-20-7.png
     
  18. May 31, 2016 #17

    Svein

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    Assuming n≠0 and a(n) = c⋅n; b(n) = c⋅n2, then b(n) > a(n) ⇔ b(n)/a(n) >1. Since b(n)/a(n) = n (continue yourself).
     
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