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]?
An "Interval of length 1 root" refers to a mathematical concept in which the difference between the upper and lower bounds of an interval is equal to 1. This interval can be represented as [x,x+1], where x is any real number.
The length of an "Interval of length 1 root" can be calculated by subtracting the lower bound from the upper bound. For example, if the interval is [2,3], the length would be 3-2 = 1.
"Interval of length 1 root" has various applications in calculus, probability, and number theory. It helps in understanding the concept of limits, probability distributions, and prime numbers.
Yes, an "Interval of length 1 root" can be negative. For example, the interval [-2,-1] is an "Interval of length 1 root" as the difference between -2 and -1 is 1.
An "Interval of length 1 root" is represented on a number line by marking the lower bound and the upper bound, and then drawing a line segment connecting them. The length of the line segment will be equal to 1 unit.