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Question about a derivative theorem proof

  1. Jun 23, 2011 #1
    Good afternoon,

    I would like to ask a doubt concerning the proof of a theorem which I found in a Calculus book. The doubt is about a particular detail in the proof.
    First, I have to state the definition of relative minimum, which is given at the same source:
    "A function f has a relative minimum in c if there is an open interval containing c, where f is defined, such that f(c) ≤ f(x) for all x in this interval."
    Now, the theorem and its proof:
    If f(x) exists for all values of x in the open interval (a,b) and f has a relative extremum in c, where a < c < b, if f'(c) exists, then f'(c) = 0.
    Proof: (For the case in which there is a relative minimum in c.)
    If f'(c) exists, then:
    [tex]f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}[/tex]
    Since f has a relative minimum in c, by the definition of relative minimum, [tex]f(c) \leq f(x)[/tex]. So:
    [tex]f(x)-f(c) \geq 0[/tex]
    If x approaches c from the right, x - c > 0, therefore:
    [tex]\frac{f(x)-f(c)}{x-c} \geq 0[/tex] (*)
    If this limit exists, then:
    [tex]\lim_{x\to c^+} \frac{f(x)-f(c)}{x-c} \geq 0[/tex]
    Analogously, if x approaches c from the left, x - c < 0, therefore:
    [tex]\frac{f(x)-f(c)}{x-c} \leq 0[/tex] (*)
    such that, if this limit exists:
    [tex]\lim_{x\to c^-} \frac{f(x)-f(c)}{x-c} \leq 0[/tex]
    Since f'(c) exists, the limits of the above inequalities must equal, and both have to equal f'(c). Then, we have:
    [tex]f'(c) \geq 0[/tex]
    [tex]f'(c) \leq 0[/tex]
    Since the two inequalities above are true, we conclude that:
    [tex]f'(c) = 0[/tex]
    as we wished to demonstrate.
    My doubt is this: the definition says that, if there is a relative minimum in c, then f(c) ≤ f(x). I imagine that the "equal" in "less or equal than" part comes from the fact that this definition holds "for all x in this interval", including c. But the text above the parts that I marked with (*) states that "x approaches c from the left" or "x approaches c from the right". Then, since this x is not equal to c, it only approaches c, shouldn't it be possible to write the asterisked parts with "less than" and "greater than", without the "or equal" part? That is:
    [tex]\frac{f(x)-f(c)}{x-c} < 0[/tex] and
    [tex]\frac{f(x)-f(c)}{x-c} > 0[/tex]
    instead of:
    [tex]\frac{f(x)-f(c)}{x-c} \leq 0[/tex] and
    [tex]\frac{f(x)-f(c)}{x-c} \geq 0[/tex]
    It seems to be reasonable, since x approaches c, and c is a relative minimum. But it can't be done, because it would make the proof senseless.
    What is wrong with this reasoning?

    Thank you in advance.
  2. jcsd
  3. Jun 23, 2011 #2
    Hi pc2-brazil! :smile:

    What if f is a constant function, i.e. what if f(x)=3 for all x? Then


    So we need the "or equal" part here.
  4. Jun 23, 2011 #3
    OK, thank you, I get it now. I assumed that, for a function to have a relative minimum (c,f(c)) in an interval, it should have only one value c which is smaller than every other value in the interval.
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