Function is continous question

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In summary, the function is continuous on [a,b], but the interval of the local max/min is [a,b+h], where h is the interval of the local max/min. Because c is in the interval of the local max/min, the function is a constant.
  • #1
KataKoniK
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Hi,

Can anyone help me with teh following question?

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

Thanks in advnace.
 
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  • #2
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...
 
  • #3
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
 
  • #4
See this thread. Are you and trap taking the same class?
 
  • #5
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).
 
  • #6
Why not constant on the closed interval [a,b] ??
 
  • #7
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.
 
  • #8
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?
 
  • #9
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.
 
  • #10
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.
 
  • #11
Thanks for the help guys. I don't know who trap is, since there's close to 400 ppl in the course.
 

1. What is a continuous function?

A continuous function is a function that has no sudden jumps or breaks in its graph. This means that the value of the function changes smoothly and continuously as the input changes.

2. How can I determine if a function is continuous?

A function is continuous if it meets the three criteria of continuity: it is defined at every point in its domain, the limit of the function as x approaches a given value is equal to the value of the function at that point, and the function has no sudden jumps or breaks in its graph. If all three criteria are met, the function is continuous.

3. What is the difference between a continuous and a differentiable function?

A continuous function is one that has no sudden jumps or breaks in its graph, while a differentiable function is one that has a derivative at every point in its domain. This means that a differentiable function is also continuous, but a continuous function may not necessarily be differentiable.

4. Can a function be continuous at a single point?

Yes, a function can be continuous at a single point. This means that the function meets the criteria of continuity at that specific point, but may not be continuous at any other points in its domain.

5. Why is continuity important in mathematics?

Continuity is important in mathematics because it allows us to make predictions and analyze functions with ease. It also allows us to use tools such as the Intermediate Value Theorem and the Mean Value Theorem to find important information about a function, such as its extrema and critical points.

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