The discussion revolves around finding intervals of length 1 that contain roots for the equations \( xe^{x}=1 \) and \( x^{3}-6x^{2}+2.826=0 \). Participants explore the application of the Intermediate Value Theorem in continuous functions to identify these intervals.
There is ongoing exploration of potential intervals, with some participants suggesting intervals like [-1,1], [0,1], and [-0.5,0.5]. Questions about the validity of these intervals based on function evaluations are raised, and some guidance is provided regarding checking conditions for the Intermediate Value Theorem.
Participants note that the intervals must have a length of 1 and question whether certain function evaluations satisfy the conditions for the Intermediate Value Theorem. There is a focus on ensuring that the chosen intervals contain roots based on the behavior of the functions involved.
Pere Callahan said:Do you know any method to find roots of an equation? If not you can always guess. Observe that the given functions are continuous so the intermediate value theorem applies. How dies this help you?
Erm, yes, and in particular there exist x between a and b, such that f(x)=y.kathrynag said:So, we have f(a)<y<f(b) or f(a)>y>f(b)
Pere Callahan said:Erm, yes, and in particular there exist x between a and b, such that f(x)=y.
What is your f here? It is the function of which you want to find the root, so
f(x)=x ex-1
Would it help you if you could find numbers a,b with
f(a)<0 and f(b)>0 ?
Yes, except that is not an interval of length 1! What about x= 0?kathrynag said:Ok, so -1 works for a and 1 works for b. So could we have the interval [-1,1]?
kathrynag said:Ok, so could I have [0,1] or [-0.5,0.5]?
Pere Callahan said:You should be able to verify this proposed solution yourself!
Do the intervals have lengths one?
Is f(0)<0<f(1) ?
Is f(-.5)<0<f(.5) ?
If you can answer these last question with yes, then the Intermediate Value Theorem asserts that there exists x in [0,1] or [-.5,.5] respectively with f(x)=0. This x would be the root, of course.
kathrynag said:Yes, it works for [-0.5,0.5], but [0,1] only works if we are letting 0<0<f(1)
Pere Callahan said:No it does not work for [-.5,.5]! Remember f(x)=xex-1.
What is f(-.5), f(0), f(.5), f(1)...?
kathrynag said:f(-.5)=-.30326
f(0)=-1
f(.5)=-.1756
f(1)=1.71
So, [-.5,.5] does not work but [0.1] works because f(0)<1 and f(1)>1
Pere Callahan said:It works because f(0)<0 and f(1)>0; this means that there is a number x between zero and one with the property f(x)=0. This again means that xex-1=0 which is equivalent to xex=1.
This is correct.kathrynag said:So, now I wnat to do the same for x^3-6x^2+2.826=0
[-1,0]?