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Constant Rule Proof?

  1. Jun 28, 2014 #1
    1. The problem statement, all variables and given/known data
    For the Constant function f(x)=c prove that f'(x)= 0

    2. Relevant equations
    Alright so the relevant equation would be the Difference quotient or definition of a derivative.


    3. The attempt at a solution
    This is what I said

    I restated the equation as this
    f(x)=c >>>> f(c)= c(x^0)

    Then I imputed into definition of derivative

    (c(x+h)^0 - c(x^0))/ h

    Then I solved it to be

    (c-c)/h = 0/h
    and then set h to 0 and then you get 0/0 which is non determinant form. I am not sure if this "proves" the statements to be true or if since it is 0/0 that doesn't show anything.
     
  2. jcsd
  3. Jun 28, 2014 #2

    LCKurtz

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    You don't need an artificial ##x^0## factor. It is just ##c-c##.
    Then I solved it to be

    (c-c)/h = 0/h = 0
    and then set h to 0 and then you get 0/0 which is non determinant form. I am not sure if this "proves" the statements to be true or if since it is 0/0 that doesn't show anything.[/QUOTE]

    That comes out zero before you let ##h\to 0## so the limit is zero. It is not indeterminante 0/0. So you're done.
     
  4. Jun 28, 2014 #3
    So by including the artificial x^0 I actually made the proof incorrect? I am curious what is the most proper way to represent this proof? As I am going to be a math major, I don't really know how they would prefer this to be represented.
     
  5. Jun 28, 2014 #4

    LCKurtz

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    I would write it like this. Given ##f(x)=c## then$$
    f'(0) = \lim_{h\to 0}\frac {f(0+h)-f(0)}{h} = \lim_{h\to 0}\frac {c-c}{h}=
    \lim_{h\to 0}\frac {0}{h}=\lim_{h\to 0} 0=0$$to be explicit and correct.
     
  6. Jun 29, 2014 #5
    did you mean f'(c) ?
     
  7. Jun 29, 2014 #6

    LCKurtz

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    Woops, I was thinking you wanted f'(0). Just put ##x## instead of ##0## in that difference quotient in the first two expressions. It's the same arithmetic.
     
  8. Jun 29, 2014 #7

    verty

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    Also you need to understand or be able to see that the limit is 0. I mean this expression: ##lim_{h \to 0} {0 \over h}## is 0 for every h, it is a horizontal line heading toward the origin, so of course the limit is 0.
     
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