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I want to understand how to use the general equation [f(x) - f(x+h)]/h

  1. Oct 25, 2009 #1
    I am in a 12th grade Physics class and in the beginning of the year, we learned the calculus based equation: [f(x+h) - f(x)]/h. I understand that I am trying to simplify/solve for the answer when H goes to 0 (h -> 0), and I understand that I must substitute f(x) into the given fields f(x) and f(x+h). However, I can never manage to completely simplify the problem.

    Here is a sample problem we were given in Physics class earlier this year:

    f(x) = x2+1/x

    Solve for: [f(x+h) - f(x)]/h


    Can someone please help me understand how to do these sort of problems? Is there any kind of technique that can help me solve them? I need to know ASAP, as I am struggling in class.
     
  2. jcsd
  3. Oct 25, 2009 #2

    arildno

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    Well, f(x+h) then replaces "x" in the general formula woth "x+h" at all instances:
    [tex]f(x+h)=(x+h)^{2}+\frac{1}{x+h}[/tex]
    The difference is then:
    [tex]f(x+h)-f(x)=x^{2}+2xh+h^{2}+\frac{1}{x+h}-(x^{2}+\frac{1}{x})=2xh+h^{2}+\frac{x}{x(x+h)}-\frac{(x+h)}{x(x+h}}=h*(2x+h-\frac{1}{x^{2}+xh})[/tex]
    Whereby dividing by "h" yields, for every non-zero "h":
    [tex]\frac{f(x+h)-f(x)}{h}=2x+h-\frac{1}{x^{2}+xh}[/tex]

    As h tends to 0, we will get:
    [tex]f'(x)=2x-\frac{1}{x^{2}}[/tex]
     
  4. Oct 25, 2009 #3
    arildno,

    Thank you for helping me solve the problem. While I understand how you arrived at the answer, as you probably know, every physics problem is different. Is there any technique I can use, or any tip you can provide me when solving future problems and more complex problems using the same formula? Any advice you can give would be much appreciated.
     
  5. Oct 25, 2009 #4

    arildno

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    1. "every physics problem is different. "
    Not so.

    For example, now you should know how to calculate the difference expression for ANY function f.

    Try a few more here on PF, and we will help you if you get stuck in the middle.

    Now is the time to hone your skills on THESE types of problems. :smile:
     
  6. Oct 25, 2009 #5
    So every problem isn't different? Oh. We were taught in Physics class that every problem is.

    Thanks so much for your help. I will take your advice and check out other problems on the forum. :smile:
     
  7. Oct 25, 2009 #6

    arildno

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    It depends on what you mean by "different".

    The same general technique can be used to solve a lot of similar problems.
     
  8. Oct 25, 2009 #7

    this is a general method in itself, the idea is to algebraically manipulate the equation until you can get something that isn't a division by zero

    think of "excluding" the case where h is "literally" zero, since you want to consider the behaviour of the function as h APPROACHES zero. So, if you see something that doesn't work because "it might be zero", then ignore that it might be zero and do it anyway
     
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