Question

calculus question @zepdrix

Answer 1

The extreme value theorem states that if f is contuous on a closed interval [a,b], then f has both a maximum and minimum value on the interval. But I'm not sure if this function is continuous over the entire interval, because I always get confused over the whole open dot, and closed dot above it thing

Answer 2

continuous*

Answer 3

I think it is..... hmm not so sure why don't u search it up on google?

Answer 4

I think I'll wait for someone else on here to answer if you don't mind @ilyvowls2004

Answer 5

ok that's cool ... lol

Answer 6

The function is certainly not `differentiable` at x=a.
And also not at x=c, we have this sharp corner, problem with slope there.

But continuous?
Mmm I'm trying to remember how we define continuity XD
This is embarrassing, I should know this. lolol

Answer 7

Yeah, I knew the differentiable part, but I know a function can be continuous but not differentiable, so I'm not sure about this one

Answer 8

I mean I guess we would describe this value at x=a as a `removable discontinuity`, yes?

Answer 9

yeah...does that make the function continuous or not continuous over [a,b]? I always get confused over this

Answer 10

So I guess our function is not continuous at x=a.
We are continuous on (a,b] but not on the interval [a,b].

Answer 11

Okay, so then the extreme value theorem does not apply?

Answer 12

Mmmmmmmmmmmmmmmmmmmmmmm
ya that sounds right...

Answer 13

But there is still a absolute max and min, right?

Answer 14

Well, maybe not max in this case? Is there an absolute minimum at x = c?

Answer 15

I think use the EVT on the continuous part, + the end point so the EVT applies, find max and min. Then compare it to the lonely point to get global max and min.

Answer 16

I mean the natural end point, not the one in the function.

Answer 17

what's the natural end point?

Answer 18

@Redcan so is the absolute minimum at c and the absolute maximum at a?

Answer 19

The function isn't continuous at x = a, so the EVT doesn't work. The condition is that the function must be continuous on the closed interval, so it has to be continuous at x = a too

the EVT won't apply, but you can just look at the graph to determine the absolute extrema. I'm not sure what your teacher is going for when s/he says "Explain how your answer is consistent with the EVT". To me, it seems like your teacher is implying that the theorem is going to be applicable here, but it's not

Answer 20

This page shows why continuity is important

http://oregonstate.edu/instruct/mth251/cq/Stage4/Lesson/EVT.html

look at example 3

Answer 21

Thank you. But am I right in what I said were the absolute extrema? (I got them from just looking at the graph)

Answer 22

yes the absolute max is at x = a and the absolute min is at x = c