# Okay, a stupid question about curve sketching

1. Jun 27, 2006

### Hollysmoke

Is it possible to have 2 global minimums? I'm just having trouble determining whether this quartic has minimums or not =/

Last edited: Jun 27, 2006
2. Jun 27, 2006

### NateTG

No. There is only one global minimum, however, a function can be minimal in more than one place.

For example, the function:
f(x)=0
is minimal everywhere.

3. Jun 27, 2006

### Hollysmoke

For the function y=x^4-2x^2-2, does this look right, then? I know the IPs are right but I'm not sure about the minimums.

http://img174.imageshack.us/img174/3466/graphpickup5mj.png [Broken]

Last edited by a moderator: May 2, 2017
4. Jun 27, 2006

### Hurkyl

Staff Emeritus
(0, -2) isn't a local minimum.

5. Jun 27, 2006

### Hollysmoke

err...it should be maximum, right?

6. Jun 27, 2006

### Hurkyl

Staff Emeritus
Right! (I wasn't sure if you were marking it as a minimum or not, but I wanted to be sure you noticed)

7. Jun 27, 2006

### Hollysmoke

That was a typo on my part (thank you for noticing it!)
So there are no minimums in this case?

Becaue when I try to calculate it, the 3 critical numbers I get are 2,-2, and 0. But if I sub in 2 or -2, I get 6, which doesn't seem right...

8. Jun 27, 2006

### prace

No, there are minimums, just no absolute minimums. There are actually 2 local minimums, and one local maximum between them.

Last edited: Jun 27, 2006
9. Jun 27, 2006

### Beam me down

But isn't the definition of the minimum (not at a domain endpoint) that:

$$f(x \pm \epsilon) > f(x)$$ for sufficiently small $$\epsilon$$

But $$f(x \pm \epsilon) = f(x)$$ if $$f(x)=0$$ for all x and so would not have any minimum.

Last edited: Jun 27, 2006
10. Jun 27, 2006

### shmoe

Your plot should make it obvious that there is in fact a global (absolute) minimum so you should either distrust your plot or your work.

Check your critical points again! (in the plot we trust)

Nope, it's a less than or equal to, $$\leq$$, for a minimum. Or $$\geq$$ if you're looking in a mirror.