# Function is continous question

1. Nov 24, 2004

### KataKoniK

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?

2. Nov 24, 2004

### kreil

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. Nov 24, 2004

### Staff: Mentor

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. Nov 24, 2004

### quasar987

See this thread. Are you and trap taking the same class?

5. Nov 25, 2004

### learningphysics

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. Nov 25, 2004

### quasar987

Why not constant on the closed interval [a,b] ??

7. Nov 25, 2004

### JasonRox

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.

\
.\_________

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. Nov 25, 2004

### Staff: Mentor

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. Nov 25, 2004

### JasonRox

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. Nov 25, 2004

### JasonRox

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. Nov 27, 2004

### KataKoniK

Thanks for the help guys. I don't know who trap is, since there's close to 400 ppl in the course.