Question

\(g:[0,1]\rightarrow \mathbb{R}\) is continuous, \(g(0)>0\) and \(g(1)=0\). Show that there is an \(a\in (0,1]\) s.t. \(g(a) = 0\) and \(g(x)>0\) for all \(0\le x < a\) .

Answer 1

A student brought this to the tutoring center where I work, and the problem is in a section that comes right before the intermediate value theorem, it is in the section for extreme value theorem. So they basically have the definition of continuity and the E.V.T . I would love some help to find a not to intense solution. It is from a advanced calculus class.

Answer 2

Well, IVT would say what exactly? That \(b\in (0,1]\) where \(0 \leq g(b)<0\)?

Answer 3

that there exists a x in (0,1) s.t. f(x) = 0.

Answer 4

my real point about the IVT is that we cant use it.

Answer 5

I thought IVT was something like \[
c \in (a,b) \implies f(c) \in(f(a), f(b))
\]or \((f(b), f(a))\) depending on which interval is smaller.

Answer 6

not which interval is smaller, I mean depending on how you'd order the endpoints.

Answer 7

This tablet is killing me, brb

Answer 8

if \(f:[a,b]\rightarrow \mathbb{R}\) is continuous. Let \(x\) be in \((f(a),f(b))\), then there exists \(x_0\) in \((a,b)\) s.t. \(f(x_0)=c\)

Answer 9

that c should be an x

Answer 10

But anyway, I cant use it...

Answer 11

why is that?

Answer 12

Because it is in a section in the book that comes before IVT. So they don't have that tool.

Answer 13

Oh, they just have EVT?

Answer 14

I can use EVT and continuity, and anything about sequences.

Answer 15

Correct.

I was thinking if I could get the function down to an interval where we only have one zero, then we could take the abs value of the function and find the min. (assuming f(x) is not always positive for x not equal to 1).

Answer 16

We know there is a minimum on \(g\), that minimum can't be at \(0\) because \(g(0) > g(1)\).

Answer 17

I mean \(g(0)\) is not the minimum.

Answer 18

Now it is possible that \(g(1)\) is the minimum, which means \(a=1\).

Answer 19

If it isn't, then we would say there is some \(a\neq 1\) where \(g(a)\leq 0\)

Answer 20

yep

Answer 21

or actually let's use \(b\) as not to confuse with the \(a\) we claim to exist.

Answer 22

I think in the case where \(g(1)\) is not the minimum, then we will change the interval to \([0,b]\).

Answer 23

lets just assume that its not always positive outside of g(1)

Answer 24

Sure.

Answer 25

This is a bit complicated without intermediate value theorem isn't it.

Answer 26

Yeah, apparently the student brought it to his prof and the prof could not solve it.

Answer 27

I'm thinking we would take the limit like... \[
\lim_{x\to b^+}g(x)
\]

Answer 28

We know this limit exists and equals \(g(b)\) because it is continuous.

Answer 29

In terms of epsilon delta, what does that tell us?

Answer 30

Is it something like:\[
x-b<\delta \implies |g(x) - g(b)|<\epsilon
\]

Answer 31

Hmm, I think I should have put \(x-b>\delta\).

Answer 32

I want it to be a limit going toward the positive direction.

Answer 33

Uhg. I actually have two issues with this problem.

Answer 34

The first is that it could hit 0 and infinite amount of times.

Answer 35

yeah, I agree. So its like there is some 'first" time on a conceivably uncountable list.

Answer 36

The second is trying to show that if the minimum is negative, we would want to throw it out while maintaining a closed interval.

Answer 37

Like consider:

Answer 38

I think we can work with this case a bit...