Calculus memories... :eek: (pre-decided forfeit)

[Cheeky Aahya]

Hello all : ))

Thought this might be unique...

This is a person above solve calculus problem, or get punished thread.
(pre-decided punishments)
School memories? lol...

How to play--

1. Person A says they are in and posts a punishment they'll do if they lost.
2. Person B posts a question.
3. If person A is unable to solve the question, they do the forfeit
4. Next person in is person B.

Rules--

1. One question per post.
2. The person posting the question must know how to solve the questions themselves.


i'm in, i'll do 300 spanks if i'm unable to solve the problem

[Hukky]

I’ll try to stump you though it’s been so long I’m going to have to start with an easy one.

Please utilize the definition of continuity or discontinuity to tell me how it is possible for f(5)=5. But for the function itself other than this one point to never go above f(x) = 4.

Does the function still have a limit at x=5? Why?

In respect to this question, and the next person who hopefully isn’t too mean, I would put a forfeit of 20 edges up watching the porn of the next persons choosing.

[herpderp42]

I like how you formulated the question did basically include a hint.
This question should be able to be solved with basic knowledge about differentiation, even if you didn't have seen it so far:

Suppose f and g and two functions (conti. diff. from from R to R, so just perfect in any way you want them to be ). We define a third function h(x) as the integral from point 0 to point f(x) of the function g(x) [so h(x)=integral_0^(f(x)) g(s) ds ].
What is h'(x)?

I am betting a week of denial for me if you can stump me.

Oh and please message me via PM if you have added a question for me. i like the idea for the thread but I doubt it will get much attention.