# Math Help - maximum and minimum values

1. ## maximum and minimum values

Sketch the graph of f by hand. (Do this on paper. Your instructor may ask you to turn in this graph.) Use your sketch to find the absolute and local maximum and minimum values of f. (If there is not one, enter NONE.)

on [-4,0)
f(0)=16 (hole)
f(-1)=15
f(-2)=12
f(-3)=7
f(-4)=0

on [0,4]
f(0)=-1
f(1)=1
f(2)=3
f(3)=5
f(4)=7

What I was thinking:
absolute maximum value_________
f(0)=16 is the highest point on the graph, but isn't in the top functions interval, so I put f(-1)=15 and it was marked wrong. ??

local minimum value=____________
The graphs don't intersect and since the top function is only from [-4,0), this part of the function is only concave down. I put 'none' and it was also marked wrong.

2. The first answer should be “none”. $f$ has no absolute or local maximum value at all (think about it).

However, although it has an absolute minimum value at $x=0,$ I agree that $f$ has no local minimum value.

3. So the first one was due to it not being in the domain. As for the second part, I still don't see how there could be any local extrema either.

4. Any other ideas?

5. It has a local minimum at x=-4.

Let S be the domain of f.
Then there exists I an open interval about x=-4, such that for all $x\in I\cap S$
$f(x)\geq f(-4)$

e.g $I=]-5,-3[$

6. hmm...it marked that wrong as well.

7. I don't know why it marks that one as wrong because there is a local minimum at x=-4.
Mabey it is asking you to put x=0 in too, because a global minimum is also a local minimum.

8. Originally Posted by hjortur
It has a local minimum at x=-4.
Yes, so it has. I had the erroneous impression that $f$ didn’t have a local minimum (or maximum) – it didn’t occur to me to check the end points.

In this case, $f$ should also have a local maximum at $x=4.$

9. Originally Posted by proscientia
Yes, so it has. I had the erroneous impression that $f$ didn’t have a local minimum (or maximum) – it didn’t occur to me to check the end points.

In this case, $f$ should also have a local maximum at $x=4.$
No, no local maxes.

10. Originally Posted by hazecraze
No, no local maxes.
It looks like there are different defenitions around. According to the definition at wikipedia
there are no local extrema at endpoints.

The definition I learned was the one I gave in a post above, and is the same one given in
calculus (Spivak 3ed), and there you can have local extrema at endpoints.

So it just depends on what defenition you are working with.

There still is a local minimum at the global min. because global extrema are also local.

11. Here's a response I got on the class discussion board:
It does not need to be continuous to have local extrema. The local minimum is actually the point where the graph is discontinuous, since it is lower than the values immediately to the right and left of it.
The discontinuity occurs at f(0), so that would make the f(0) from [0,4] the local minimum? That value being -1?

12. Mabey you should read better what I have been saying.
I have said 2 times before that there is a local minimum at x=0.