Interval of length 1 root

[kathrynag]

1. The problem statement, all variables and given/known data
Find an interval of length 1 that contains a root of the equation xe^{x}=1



2. Relevant equations



3. The attempt at a solution
I'm not quite sure how to find these intervals...
1. The problem statement, all variables and given/known data
Find an interval of length 1 that contains a root of the equation x^{3}-6x^{2}+2.826=0


2. Relevant equations



3. The attempt at a solution

[Pere Callahan]

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?

[kathrynag]

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?

So, we have f(a)<y<f(b) or f(a)>y>f(b)
So f(a)=ae^a=1
f(b)=be^b=1

[Pere Callahan]

So, we have f(a)<y<f(b) or f(a)>y>f(b)

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 ?

[kathrynag]

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 ?

Ok, so -1 works for a and 1 works for b. So could we have the interval [-1,1]?

[HallsofIvy]

Ok, so -1 works for a and 1 works for b. So could we have the interval [-1,1]?
Yes, except that is not an interval of length 1! What about x= 0?

[kathrynag]

Ok, so could I have [0,1] or [-0.5,0.5]?

[Pere Callahan]

Ok, so could I have [0,1] or [-0.5,0.5]?

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]

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.

Yes, it works for [-0.5,0.5], but [0,1] only works if we are letting 0<0<f(1)

[Pere Callahan]

Yes, it works for [-0.5,0.5], but [0,1] only works if we are letting 0<0<f(1)

No it does not work for [-.5,.5]! Remember f(x)=xex-1.

What is f(-.5), f(0), f(.5), f(1)...?

[kathrynag]

No it does not work for [-.5,.5]! Remember f(x)=xex-1.

What is f(-.5), f(0), f(.5), f(1)...?

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]

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

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.

[kathrynag]

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.

Oh yeah. That's actually what I meant to type, haha.
So, now I wnat to do the same for x^3-6x^2+2.826=0
[-1,0]?

[Pere Callahan]

So, now I wnat to do the same for x^3-6x^2+2.826=0
[-1,0]?
This is correct.