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Calculus: Limits

  1. Mar 23, 2008 #1
    [SOLVED] Calculus: Limits

    1. 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 lim (h -> 0) (f(h)/h) = 7

    a. Find f(0). Justify your answer.
    b. Use the definition of the derivative to find f'(x).
    c. Find f(x).

    Dang... I'm really stumped because I am unsure what it is asking :(. Thank you.
  2. jcsd
  3. Mar 23, 2008 #2
    The first part is downright simple. Just put x=0, y=0 in the equation and solve for f(0). For the second and third part..

    What does the first principle method of derivative tell you?

    f'(x) = \lim_{h\rightarrow 0} \frac{f(x + h) - f(x)}{h}

    Using the fact that [itex]f(x + y) = f(x) + f(y) + 2xy[/itex], you have:

    f'(x) = \lim_{h\rightarrow 0} \frac{f(h) + 2xh}{h}

    Now.. solve this question of limits keeping the first result in mind. Also, you have been given that [tex]\lim_{h\rightarrow 0} \frac{f(h)}{h} = 7[/tex]. You should be able to do this now easily.

    Once you have done that, integrate it and then see if you can use the given data to determine the constant of integration. [i'm not so sure if u can determine it.. but do try it]
  4. Mar 23, 2008 #3
    wow! Thank you very much =)
  5. Dec 11, 2008 #4
    Re: [SOLVED] Calculus: Limits

    I got 7 + 2x after I continued from where you left off, but Im not sure if thats f'(x). Also I put 0 in for x and y, and got f(o) = f(0) + f(0) + 2(0)(0) = 0, which im pretty sure doesnt mean anything. So if you could please break it down further, it would be much appreciated. Thank you.
  6. Dec 11, 2008 #5


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    Re: [SOLVED] Calculus: Limits

    f(0)=f(0)+f(0) does mean something. What does it tell you about f(0)? If you've now got f'(x)=7+2x (and it is) and you want to find f(x) that's just an antiderivative problem, isn't it?
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