# Local max/min

1. Mar 28, 2006

### endeavor

Sketch the graph of a function f that is continuous on [1,5] and has the given properties:
f has no local maximum or minimum, but 2 and 4 are critical numbers.

my professor drew a constant function (horizontal line) from x=1 to x=5. How come there is no local maximum or minimum for a horizontal line? My professor said that all the values of a horizontal line are the absolute maxima and minima...

According to my book, "A function f has a local maximum at c if f(c) >= f(x) when x is near c. [This means that f(c) >= f(x) for all x in some open interval containing c.]" Let's say the open interval is (1,5), and since f(x)=f(c) (because we have a horizontal line), shouldn't f have a local maximum????

2. Mar 29, 2006

### Tom Mattson

Staff Emeritus
The function does have a local max--and a local min--at 2 and 4. But what a boring example! Surely we can come up with a more interesting one. Note that critical numbers are points at which the derivative is zero or at which it fails to exist. One way to achieve this is to graph a function with vertical asymptotes at x=2 and x=4.

3. Mar 29, 2006

### endeavor

so, in drawing a horizontal line, there are local maxima and minima at every location on the horizontal line??

4. Mar 29, 2006

### Tom Mattson

Staff Emeritus
According to the definition, yes.

5. Mar 29, 2006

### Nimz

One of the problem with using vertical asymptotes is that you still have to consider the endpoints of the interval. If the derivative is negative at 5, for instance, there would be a local minimum at x = 5, with the open interval being (5-$\epsilon$, 5]. I know it doesn't look open, but it is, since [1, 5] is the interval we started with. I'm pretty sure a horizontal line is as interesting of an example as is possible. Although it fails when you apply the strict definition of local maxima and local minima, it's about as close as you can get.:yuck:

6. Mar 29, 2006

### J77

Well, the text for a constant, continuous function $$f(x)$$ on $$[a, b]$$ is that $$f'(x)=0,\;\forall x\in[a, b]$$.

The classification for local extrema are strict inequalities in the second derivative, ie. $$f''(x)>0$$ and $$f''(x)<0$$.

For the constant function, we have $$f''(x)=0$$. Therefore, no conclusion can be drawn to whether it's a maximum or minimum.

I think this is what your teacher was getting at.

Last edited: Mar 29, 2006
7. Mar 29, 2006

### J77

No.

If you look at an interval containing the critical point, you have to have the opposite strict inequalities on the first derivative either side of it.

8. Mar 29, 2006

### HallsofIvy

Staff Emeritus
But the original poster said:
Since that is the definition in his book, that is the definition that should be used. In fact, I have never seen any definition of "local max" or "local min" that said anything about the first derivative.

9. Mar 29, 2006

### J77

OK, my mistake.

And you're right, I used a theorem

10. Mar 29, 2006

### endeavor

Does this mean that my professor is wrong in drawing a horizontal line??

What are the possible ways to draw this graph?

11. Mar 30, 2006

### HallsofIvy

Staff Emeritus
f(x) increases from x= 1 to 2, levels off (becomes horizontal) at x= 2 then continues to increase again until it levels off at x= 4, then increases after that.

Think of two y= x3 graphs patched together.

12. Mar 30, 2006

### endeavor

Is it wrong to draw a horizontal line for a solution to this problem??