Question

Find an ex. of a bdd disc. fn. \[f:[0,1]\rightarrow R\] that has neither an absolute min. nor an absolute max.

Answer 1

@zzr0ck3r

Answer 2

So my issue is, even a horizontal line achieves an absolute max and min, would it have to be a vertical asymptote?

Answer 3

bdd ?

Answer 4

bounded sorry

Answer 5

see the vertical asymptote doesn't allow for bounded

Answer 6

sorry my new computer is hating this site, I got to install chrome and then ill be back.

Answer 7

it's ok, the site is muffed up currently

Answer 8

So you need a bounded discontinuous function that doesn't have an absolute min or an absolute max?

Answer 9

yea, but none exists

Answer 10

Since your function needs to be discontinuous, you might be able to be creative and pull it off.

Answer 11

I mean, it needs to have a domain of [0,1] the only thing I can think of is something that is disc at 0 and 1 but I don't have an ex of that

Answer 12

like needs to hit infinity

Answer 13

unlessss Tangent? a modification on it?

Answer 14

Something like:\[f(x)=x \text{ if x is irrational,}0\text{ if x is rational }\]

Answer 15

but that achieves it's min and max

Answer 16

at the irrational before 1 right?

Answer 17

imn would be 0

Answer 18

ah, it achieves its min. It doesnt achieve its max.

Answer 19

it has to be both

Answer 20

Then change the value at x=0 to something else. like 1/2. This is a discontinuous function. It can be defined however you want.

Answer 21

ah, its still not right. I see what you are saying.

Answer 22

so to get rid of the min that would work because the zeroes would be near zero and the ones would be near one... wow, i can't believe I didn''t think of that

Answer 23

but 0 and 1 are rational.... I still think it may work

Answer 24

I guess change the value at the rationals? make it, f(x)= x if x is irrational, f(x)=1/2 if x is rational

Answer 25

yea, I think that is what I'll do

Answer 26

thanks!