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Function is continous question

  1. Nov 24, 2004 #1
    Hi,

    Can anyone help me with teh following question?

    What can be said about the function f continous on [a,b], if for some c in (a,b), f(c) is both a local max and a local min?

    Thanks in advnace.
     
  2. jcsd
  3. Nov 24, 2004 #2

    kreil

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    If f(c) is the biggest and smallest value in that area, it sounds to me like the function is just a constant i.e. f(x)=1

    Seems too simple, probably more to be said...
     
  4. Nov 24, 2004 #3

    Astronuc

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    Certainly a constant funtion, f(x)=const, on the interval would meet the criteria.

    But

    A local minimum is the smallest value in some local neighborhood.
    The largest value of a set, function, etc., within some local neighborhood.

    So could c be a stationary point with f'(c)= 0? Consider the function f(x)=x^3, where x=0 is a local maximum for x<=0, and a local min for x>=0.

    Example: http://mathworld.wolfram.com/StationaryPoint.html
     
  5. Nov 24, 2004 #4

    quasar987

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    See this thread. Are you and trap taking the same class?
     
  6. Nov 25, 2004 #5

    learningphysics

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    Astronuc : An inflection point such as you mentioned in x^3 is not a local max or min. It is a stationary point though. But the second derivative is 0 (and not negative -as required for local max or positive-as for local mins).

    Another way to verify this is that there needs to be an "open" interval containing the value a, where f(x)<=f(a) for a to be a local max. x<=0 and x>=0 are "closed" intervals.

    I think the answer to the original question is that the function is constant in the open interval (a,b).
     
  7. Nov 25, 2004 #6

    quasar987

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    Why not constant on the closed interval [a,b] ??
     
  8. Nov 25, 2004 #7

    JasonRox

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    It said local max and local minimum.

    It didn't say global max/min.

    A local max is a foint c where f(c)>=f(c+h), where +-h is the interval.

    The question said that the function is continuous on interval [a,b] and c is in (a,b). It did not say that the interval of the local max/min is [a,b]. Because of this, we know that the function does not have to be constant. What this tells us is that there is some interval such that f(c)=f(c+h), where +-h defines the interval of the local max/min.

    Take this sloppy function as an example.

    \
    .\_________
    ad c b

    Ignore the dot.

    Looking at this graph we see that it is continuous on [a,b] and c is in (a,b). Clearly c is a local max/min in the interval [d,b].

    Basically this tells us that there is some horizontal line in the function with length unknown.
     
  9. Nov 25, 2004 #8

    Astronuc

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    learningphysics, thanks for the reminder.

    If f(x) = constant, f''(x)=0, so it couldn't satisfy f"(x)>0 for a min and f"(x)<0 for max.

    In JasonRox example, f'(c) is discontinuous, f"(c) = zero, both functions being linear,

    and f'(c-h) not = f'(c+h) for h approaching 0, same thing as discontinuous.

    Why does there have to be a horizontal line, which presumably includes c. Could two lines with different slopes (but both slopes either postive or negative) satisfy the requirement?
     
  10. Nov 25, 2004 #9

    JasonRox

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    The interval in which c is a local max/min is the interval of the horizontal line.

    What's not to understand?

    Let's say the function is continuous on the interval [a,b] and c being in (a,b). If c is the local max/min of [a,b], then the funtion is a constant. Because they did not say that on what interval c is a local max/min, we do not know and can not assume that c is the local max/min on [a,b]. In fact, it could be the local max/min of [a,b], but that's not the only possibility.

    All we know is that there is some horizontal line in the interval [a,b] that contains c with length unknown because the length can be from [a,b]. We do not know.

    There must be an horizontal line if c is a local max/min. The definition of the local max and min IMPLIES this. So, why does it have an horizontal line what includes c? Because the definition said so.
     
  11. Nov 25, 2004 #10

    JasonRox

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    If f(c)>=f(c+h), then c is a maximum.

    Look at the graph.

    /\

    Let's say the point at the top is where f(c) is located. Clearly f(c)>=f(c+h).

    The min is f(c)<=f(c+h).

    The question says that...

    f(c)>=f(c+h) and f(c)<=f(c+h), which implies that f(c+h)>=f(c)>=f(c+h). I guess you can use the Squeeze Theorem to show that f(c+h)=f(c), where +-h determines the interval.

    I'd advise you to speak of the local max/min without using derivatives.
     
  12. Nov 27, 2004 #11
    Thanks for the help guys. I don't know who trap is, since there's close to 400 ppl in the course.
     
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