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Homework Help: A Basic Function Problem with a Limit

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data

    For all real numbers x and y, let f be a function such that f(x+y)=f(x)+f(y)+2xy and such that the limit as h approaches zero is f(h)/h=7. [tex]\stackrel{Lim}{h\rightarrow0}\frac{f(h)}{h} = 7[/tex]

    a) Find f(0) and justify your answer.

    b) Use the definition of the derivative to find f ' (x).

    c) Find f(x).


    2. Relevant equations

    Limit as h->0 is (f(x+h)-f(x))/h

    3. The attempt at a solution

    To solve part a, I looked at the original equation. To find f(0), I just put in 0 for x and 0 for y.

    Let x=y=0.

    And thus, I got f(0)=0, so now part a should be cleared. Right?

    Then I get to part b and I'm a little stuck. Find f ' (x) ? Should I be using the equation I put in the relevant equations section to help me out with this?
    Last edited: Sep 24, 2009
  2. jcsd
  3. Sep 24, 2009 #2
    The definition of the derivative is the equation that you gave, so yes.
  4. Sep 24, 2009 #3
    OKay, so here goes my shot at part b (do you think part a is right?):

    f ' (x) is ....

    (f(x+7) - f(x)) / 7

    Hm ... I'm getting a little lost here. I just substituted h for 7.

    Also, is there a way to post these fractions using the latex reference?
  5. Sep 24, 2009 #4
    you want to start with:
    [tex]\stackrel{Lim}{h\rightarrow0}\frac{f(x + h) - f(x)}{h}[/tex]

    What can you say about f(x + h)?

    to do the fractions, just use \frac{}{} where the numerator goes in the first bracket pair and the denominator goes in the second.
  6. Sep 24, 2009 #5
    Thanks for letting me know about the fractions part.

    For f(x+h), I was thinking that would just be equal to f(x) since h is approaching zero. However ...

    [tex]\stackrel{Lim}{h\rightarrow0}\frac{f(h)}{h} = 7[/tex]

    Which doesn't make sense anyway since that would be undefined as h approaches zero. Hm ....
    Last edited: Sep 24, 2009
  7. Sep 24, 2009 #6
    You have the formula for f(x + [strike]y[/strike]h) in your original post ;)
  8. Sep 24, 2009 #7
    Also, [tex]\stackrel{Lim}{h\rightarrow0}\frac{f(h)}{h} = 7[/tex] according to your original post. You edited it but left the old version. Which one is it?
  9. Sep 24, 2009 #8
    Sorry, the limit as h goes to zero is definitely SEVEN. :smile:

    So how about this. I was thinking that since I could do this ....

    f(x+y) = f(x) + f(y) + 2xy
    f(x+y) - f(x) = f(y) + 2xy

    Then couldn't I say that since h is approaching zero, then it figures that:

    x=y=0=h as well. I can then substitute the h's in for x or y?

    Am I on the right track here?
  10. Sep 24, 2009 #9
    well f(x+h) - f(x) = f(h) + 2xh is right but that's inside a limit and over h as h goes to zero.

    now, as I'm sure you know, the sum of limits is the limit of sums. So do it like that. There is no y in this equation and you can't set x to a value because you're taking the limit for all x so it needs to remain a variable.
  11. Sep 24, 2009 #10
    I'm not sure what to do what you said, because we never discussed limits in my class yet.
  12. Sep 24, 2009 #11
    Ok. I don't think there's a way to do this without limits as they're in the problem as stated so the rule you want is:
    [tex]\lim_{h\to a}(f(h) + g(h)) = \lim_{h\to a} f(h) + \lim_{h\to a} g(a) [/tex]

    See if that helps.
  13. Sep 24, 2009 #12

    I think this problem requires knowledge of limits, although I never learned that last year (I think my teacher expected us to know from last year).

    So for part b:

    f ' (x) = [tex]\lim_{h\to 0} (f(x+h)/h) + \lim_{h\to 0} f(x)[/tex]

    Hopefully this is on the right track.
  14. Sep 24, 2009 #13
    That's the idea but do you remember how to expand f(x + h)?

    Just keep it in one limit until you have expanded f(x + h) and the f(x) cancels out like it did before. And remember that when a sum is on top of a fraction, the denominator goes with both sides of the sum if you split it up!

    [tex]\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}[/tex]
  15. Sep 24, 2009 #14
    Revised version.

    f ' (x) = [tex]\lim_{h\to 0} ((f(x+h)/h) + f(x/h))[/tex]
  16. Sep 24, 2009 #15
    And, no I do not know how to expand f(x + h). I know this is easy once I understand the concept but I'm really new to it, sorry.
  17. Sep 24, 2009 #16
    That's ok. Look at your original post and post #8 and #9 in this thread. You've already done it
  18. Sep 24, 2009 #17

    So do I substitute f(h) + 2xh for f(x+h) - f(x) now?

    I'm still a bit confused, sorry.
  19. Sep 24, 2009 #18
    No problem.

    Try it and see what happens. Post your results!
  20. Sep 24, 2009 #19
    [tex]\stackrel{Lim}{h\rightarrow0}\frac{f(y) + 2xy}{h}[/tex]

    Do I put an equal sign after or before the limit part?
  21. Sep 24, 2009 #20
    Ok, where is the y coming from?
  22. Sep 24, 2009 #21
    The y comes from ( f(y) + 2xy ) being equal to ( f(x+h) - f(x) ).
  23. Sep 24, 2009 #22
    Ok, but you're taking the limit with respect to h so it starts of as f(x + h) - f(x). Why are you switching to y?
  24. Sep 24, 2009 #23
    Not sure, really, since a classmate of mine told me that I might have to do something like that. Yeah, clueless.

    Should I set h equal to something to help me find the derivative of f ' (x) ?
  25. Sep 24, 2009 #24
    No, you can't set anything equal to anything.

    you have [tex]\lim_{f \to 0}\frac{f(h) + 2xh}{h}[/tex] given the rule I gave you earlier for the sum of a limit, can you figure it out from there? you already know what [tex]\lim_{f \to 0}\frac{f(h)}{h}[/tex] is. How does the value of 2x change when h changes? (assuming that x is fixed)
  26. Sep 24, 2009 #25
    [tex]\lim_{h \to 0}\frac{f(h)}{h} + \lim_{h \to 0}\frac{2xh}{h}[/tex]

    The second part would just cancel out since it's a straight up 2xh/h problem wher ethe h's cancel. 2x remains.

    f ' (x) = 7 + 2x

    Just wondering, but how did you replace f(x) with 2xh?
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