Closed Interval Puns

I have to prove the following. Just wanted to know if my approach makes sense.

If f (x) is continuous in a closed interval and if M and m are, respectively, the least upperbound (l.u.b.) and greatest lower bound (g.l.b.) of f (x), there exists at least one value of x in the interval forwhich f(x) = M or f(x) = m.

Approach: Just show this for the l.u.b

Fix some point x_0 in [a,b] (no boundary point) such that f(x_0)<=M. Then M-f(x_0)>=0. Since f is continuous in x_0 we can find some delta neighborhood (x_0-delta,x_0+delta) where M-f is >=0. Suppose that M-f(x)>0 for all x in (x_0-delta,x_0+delta). Then by definition of supremum for all eps>0 we can find some x_1 in (x_0-delta,x_0+delta) such that f(x_1)+eps>M. Therefore 0<M-f(x_1)<eps. Contradiction.

Help would be appreciated, thanks. I know nothing about how to solve this.

Been trying to do this for a while and am stuck because I can't use the result that a function is continuous if lim(x - > c) = f(c) because lim(x -> c) is undefined at endpoints of the function. How else can I go about this?? Thanks.

Hi all. I've been trying to place a Grubhub order for 2 days now and haven't been able too because there's never any drivers. I'm guessing this is linked to the drivers but my.buddy is online and it's still not showing any stores open at all. Can some e help me understand this mess? I have a $28 gift card and I'd rather just get it out of the way because GH sucks a fat one and I hate using it. I always tip big for those poor souls driving.

I was looking at some past AP Calc FRQs and saw that in the scoring rubric, an absolute maxima was justified using the change in sign of derivatives, not by evaluating its critical points and endpoints. Is this an actual method to evaluate the absolute extrema of a function?

The intersection is [0, 1], because that is the smallest closed interval containing (0,1). So I need to somehow show that [0, 1] = ⋂ A where A is all closed intervals containing (0,1). To do this, we were shown to prove equality by proving that each side is a subset of the other. I'm having trouble wrapping my head around doing that because it seems to me that I need to prove what the intersection is before I can begin proving that it is a subset of something else, which brings me back to the start of the problem. I was told upon asking to apply the definition of intersection (given in lecture as A ⋂ B = { x ∈ X | x ∈ A and x ∈ B } where X is an ambient set), but again that seems to require knowing or having proved what B is. Any help getting started would be appreciated!

Also, does being continuous or differentiable on the endpoints matter?

Edit: Obviously if the interval is continuous, then every points except possibly the end points are differentiable. But why does the interval have to be continuous?

Hi guys! I’m working on a circuit and a problem says the following: “Solve for x on the closed interval [2,4]. 20/(3+e^tanx)=5.3” That’s all the information I was given. I tried to plug 2 and 4 in which didn’t align with any of the answers on my circuit. Does anyone have experience with something like this?

I know there's the outdoor stadium, are the gates for that open? Or is it off limits to students/public.

https://en.m.wikipedia.org/wiki/Mean_value_theorem

In this they mentioned that MVT states that if a function, say f(x), is contiuous throughout [a,b] and differentiable throughout (a,b) then there exists a value for x, say c, such that f’(c) = f(b)-f(a)/b - a.

In this, why is there a closed interval for continuity but open for differentiability?

Thanks in advace

Basically I have an electronic paintball gun that has a circuit inside it. I want it to shoot at full auto (say, 15 balls per second) just by holding down the trigger (which closes the gun's circuit). Thanks in advance.

I'd be grateful if anyone could explain the question asked in the title.

Thanks.

John.

How do I prove a closed interval [a,b] in R is compact using the method of contradiction i.e assuming that some open cover of [a,b] has no finite sub-cover ?. I know it's pretty easy to prove this using Heine-Borel.

Hello everyone.

I have the following function:

> f(x) = \frac{\sin(x) \cdot e^{x}}{x}

And I'm supposed to show that it's integrable in the [0,1] interval, then verify the inequality of:

> 0 \leq \int_0^1 f(x) dx \leq e-1

I don't really know how to do this. My plan was to try out if the function was Riemann integrable by comparing the upper sum and the lower sum.

But my upper sum:

> U(P_n,f) = \frac{1}{n} = \frac{1}{n}[f(x_1) + f(x_2) + ... + f(x_n)]

Then ends up being:

> \frac{1}{n} [ \frac{\sin(x_1) \cdot e^{x_1}}{x_1} + \frac{\sin(x_2) \cdot e^{x_2}}{x_2} + ... + \frac{\sin(x_n) \cdot e^{x_n}}{x_n}

And I have no clue how to go from there.

Wouldn't it just end up looking like this? I'm very stuck here :(

And also not sure how to verify the inequality that is asked for in the second part.

If anyone could help me out, I'd really appreciate it!

Please feel free to correct me if I'm wrong anywhere. Are we allowed to use mathematical induction to prove P(k) for k→∞ (k∈N)? Thank you!

Lemma: Let I = [a , b] and J = [c , d] be closed intervals (of R). Then I ⊆ J if and only if c ≤ a and b ≤ d.

Proof (of lemma): (⇒) Suppose that I = [a , b] and J = [c , d] are two closed intervals such that I ⊆ J. Then, by definition, x∈I ⇒ x∈J. Since a,b∈I, we have that a,b∈J = [c , d] = {x∈R : c ≤ x ≤ d}. It follows that c ≤ a ≤ b ≤ d. Thus, c ≤ a and b ≤ d. (⇐) Conversely, suppose that I = [a , b] and J = [c , d] are two closed intervals such that c ≤ a and b ≤ d. Let x∈I. (We want to show that x∈J.) By definition, a ≤ x ≤ b. This implies that c ≤ a ≤ x ≤ b ≤ d since c ≤ a and b ≤ d (by assumption). Because we have that c ≤ x ≤ d, x∈J is satisfied. Hence, I ⊆ J. ■

Proof (of the nested intervals theorem [by induction]): Let I₁ = [a₁ , b₁] and I₂ = [a₂ , b₂] be such that I₁ ⊇ I₂. Then (by the lemma), a₁ ≤ a₂ and b₂ ≤ b₁. Consequently, a₁ ≤ a₂ ≤ x ≤ b₂ ≤ b₁ whenever x∈I₂. (We know that any closed interval is nonempty -- that is, it is either a singleton or a nontrivial closed interval of R.) Hence, x∈I₁ and x∈I₂. This implies that x ∈ I₁ ∩ I₂ ≠ ∅. Now, let I₁ ⊇ I₂ ⊇ I₃ ⊇ ... ⊇ Iₖ (where k∈N) and let Iₖ₊₁ = [aₖ₊₁ , bₖ₊₁]. Then Iₖ ⊇ Iₖ₊₁ (by assumption of the theorem). By the lemma, this implies that aₖ ≤ aₖ₊₁ and bₖ₊₁ ≤ bₖ. Thus, we have that aₖ ≤ aₖ₊₁ ≤ x ≤ bₖ₊₁ ≤ bₖ whenever x∈Iₖ₊₁. Since x∈Iₖ whenever x∈Iₖ₊₁ and since I₁ ⊇ I₂ ⊇ I₃ ⊇ ... ⊇ Iₖ, this means that x ∈ I₁ ∩ I₂ ∩ I₃ ∩ ... ∩ Iₖ and x∈Iₖ₊₁ ⇒ x ∈ I₁ ∩ I₂ ∩ I₃ ∩ ... ∩ Iₖ ∩ Iₖ₊₁ (for k∈N). Therefore, x ∈ I₁ ∩ I₂ ∩ I₃ ∩ ... ≠ ∅. ■

Edit:

Updated Proof (of theorem): Let Iₙ = [aₙ , bₙ] ⊇ Iₙ₊₁ = [aₙ₊₁ , bₙ₊₁] be a nested sequence of closed intervals (n∈N). (We want to show that I₁ ∩ I₂ ∩ I₃ ∩ ... ≠ ∅, that is, ∃x∈R such that x ∈ I₁ ∩ I₂ ∩ I₃ ∩ ... . Equivalently, it suffices to show that ∃x∈R such that x∈Iₙ [for all n∈N].) By the (previously proven) lemma, we have that aₙ ≤ aₙ₊₁ and bₙ₊₁≤bₙ for all Iₙ⊇Iₙ₊₁. Furthermore, since x∈Iₙ ⇒ aₙ ≤ x ≤ bₙ, we have that aₙ ≤ bₙ (by transitivity). Let {aₙ} = {a₁,a₂,a₃,...} and {bₙ} = {b₁,b₂,b₃,...} (for any n∈N). Since aₙ ≤ bₙ and since a closed interval is either a singleton or a nontrivial subset of R, {aₙ} is bounded above and nonempty. By the axiom of completeness, ∃sup{aₙ} = λ∈R. Hence, aₙ ≤ λ for any aₙ∈{aₙ}. Let bₙ∈{bₙ}. We have aₙ ≤ bₙ, and so bₙ is an upper bound of {aₙ}. Since λ = sup{aₙ}, it means that λ ≤ bₙ. I

From what I have read about the definition of differentiability, in my mind there should never be any mention of differentiability in a closed interval as it simply does not make sense because the left hand derivative must exist and be equal to its right hand derivative and since our argument is concerned only with the end points of an interval, there we cannot compare/equate right hand derivative of a function at a lower point of an interval with its left hand derivative as that would be outside the interval and similarly in the other case...

I need confirmation of whether my understanding is correct and if not, what is wrong in what I said. I was unable to find any true and straight forward confirmation of this online elsewhere...

As always I greatly appreciate your answers!

So when you have integral of f(x) from 0 to 1 does it calculate the area under [0,1] or (0,1)?

RolIe’s Theorem

Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b).

If f(a) = f(b), then there is at least one number c in (a, b) such that f’(c) = 0.