# Interval of length 1 root

1. Nov 18, 2008

### 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

2. Nov 18, 2008

### 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?

3. Nov 18, 2008

### kathrynag

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

4. Nov 18, 2008

### Pere Callahan

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

f(a)<0 and f(b)>0 ?

5. Nov 18, 2008

### kathrynag

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

6. Nov 18, 2008

### HallsofIvy

Staff Emeritus
Yes, except that is not an interval of length 1! What about x= 0?

7. Nov 18, 2008

### kathrynag

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

8. Nov 18, 2008

### Pere Callahan

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.

9. Nov 18, 2008

### kathrynag

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

10. Nov 18, 2008

### Pere Callahan

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

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

11. Nov 18, 2008

### kathrynag

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

12. Nov 18, 2008

### Pere Callahan

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.

13. Nov 18, 2008

### kathrynag

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]?

14. Nov 18, 2008

### Pere Callahan

This is correct.