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Differentiation using a constant question

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data
    If f'x = 0 [tex]\forall[/tex] x [tex]\in[/tex] (a,b), show that f is constant there.

    I've gotten a final result, but I'm not entirely sure if I actually showed what I intended to show...


    2. Relevant equations



    3. The attempt at a solution

    Fix a point c [tex]\in[/tex] (a,b) and choose a point x [tex]\in[/tex] (a,b) different from c

    From definition of differentiable:

    f'(x) = lim (x[tex]{\rightarrow}[/tex]c) (f(x)-f(c))/ x - c

    From problem, f'(x) = 0 = lim (x[tex]{\rightarrow}[/tex]c) (f(x)-f(c))/ x - c

    From sums/products of limits,

    lim (x[tex]{\rightarrow}[/tex]c) (f(x)-f(c))/ x - c = lim (x[tex]{\rightarrow}[/tex]c) f(x)/(x-c) - f(c)/(x-c)

    implies 1/(x-c)lim (x[tex]{\rightarrow}[/tex]c) f(x) - 1/(x-c) lim (x[tex]{\rightarrow}[/tex]c) f(c) = 0

    implies lim (x[tex]{\rightarrow}[/tex]c) f(x) = lim (x[tex]{\rightarrow}[/tex]c) f(c)

    implies lim (x[tex]{\rightarrow}[/tex]c) f(x) = f(c)

    This implies that the function will have the same value as f(c) when evaluated anywhere inside (a,b) and is therefore constant.

    Sorry if my typing is messy, but I am not too proficient with LaTex! :-(
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 7, 2009 #2
    Mean value theorem, son.
     
  4. Nov 8, 2009 #3
    The mean value thm states that the function needs to be continuous on the closed interval [a,b].... however, the problem does not state anything about the continuity of the function on the closed interval.... it only says that the derivative inside the open interval (a,b) is 0.... doesn't this mean that i cannot assume anything about the nature of the function at the endpoints and therefore cannot use the mean value thm?
     
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