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Another introductory calc problem

  1. May 14, 2014 #1
    Given a function f(x), I'm wondering if the slope of the secant between points a, f(a) and b, f(b) is equal to the average of the derivative at a and b.

    Mathematically, I want to know whether:

    [itex]\frac{f(a) - f(b)}{a - b}[/itex] = [itex]\frac{f'(a) + f'(b)}{2}[/itex]

    is true. I can see that it is for polynomial functions, but I can't seem to prove this relationship for a general function, f(x).

    This isn't homework, I'm just curious. Thanks for the help!
     
  2. jcsd
  3. May 14, 2014 #2
    Have you learned about Taylor series expansions yet?

    Chet
     
  4. May 14, 2014 #3

    Simon Bridge

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    You could try to come up with a counter-example.
    Also think about what "average" means.

    If you have a proof for all polynomials, then can you apply the proof to any arbitrary sum of polynomials?
    Can you write an arbitrary function as a sum of polynomials?
    [Chester beat me to it]

    Notes:
    http://homepage.math.uiowa.edu/~idarcy/COURSES/25/2_7.pdf
     
  5. May 14, 2014 #4
    Nope. I was literally just introduced to calculus four weeks ago.

    EDIT: It's late here, I'm going to have a look at this thread tomorrow (and the comments already posted).
     
  6. May 14, 2014 #5

    Simon Bridge

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    In a nutshell ... any function f(x) can be written as a sum like this:
    $$f(x)=a_0+a_1x+a_2x^2+\cdots $$

    If you already have a proof that applies to polynomials - you can apply it to that.
     
  7. May 14, 2014 #6
    Your relation is not exact. It is a good approximation, but the difference between the two sides of the relationship increases with increasing distance between the two points a and b.

    Chet
     
  8. May 14, 2014 #7
    [tex]f(a)=f(b)-(b-a)f'(b)+\frac{(b-a)2}{2}f''(b) + ...[/tex]
    [tex]f(b)=f(a)+(b-a)f'(a)+\frac{(b-a)2}{2}f''(a) + ...[/tex]
    Subtracting:
    [tex]2(f(b)-f(a))=(b-a)(f'(a)+f'(b))-\frac{(b-a)2}{2}(f''(b)-f''(a))+...[/tex]
    Or,
    [tex]\frac{f(b)-f(a)}{b-a}=\frac{f'(a)+f'(b)}{2}-\frac{(b-a)}{4}(f''(b)-f''(a))+...[/tex]
    A better approximation is:
    [tex]\frac{f(b)-f(a)}{b-a}≈f'\left(\frac{a+b}{2}\right)[/tex]
    This relation is exact if f is quadratic.

    Chet
     
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