How do I know a function has "a zero for x" between 1 & 2 (Newton's method?)?

Question

I have this question:

Show that the function f(x) = x^3 - x^2 -x -1 has a zero for x between 1 and 2.

I looked up newton's method on the internet and found this explanation

"Consider the problem of finding the positive number x with cos(x) = x3. We can rephrase that as finding the zero of f(x) = cos(x) − x3. We have f '(x) = −sin(x) − 3x2. Since cos(x) ≤ 1 for all x and x3 > 1 for x>1, we know that our zero lies between 0 and 1. We try a starting value of x0 = 0.5."

I'm not sure how in the example it just makes the statement since cos(x) <= 1 for all x. I am not sure how this leads to knowing the zero is between 0 and 1... how does this apply to my question... could someone please explain the method

Answer 1

f(1) = 1-1-1-1= -4
f(2) = 8 -4 -2-1 = 1
Since f(x) changed from - to + between x=1 and x=2, it must have crossed the y-axis and so there is a root (zero) in that interval.

Answer 2

If you don't need to find the exact root of the equation f(x)=0, all you have to use is Bolzano's theorem, which states:

If a function f is continuous in [a,b] and f(a)*f(b)<0, then there is at least one x0 in (a,b) for which: f(x0)=0.

On this notion lies the method you described above. Newton's method is a way of making [a,b] as small as possible. If we know for a fact that the equation f(x)=0 has a root in (a,b) all you need to do is "cut" (a,b) in half: you get (a,a+b/2) and (a+b/2,b). Apply Bolzano's theorem on these two spaces. If f(a)*f(a+b/2)<0 then there is a root within (a,a+b/2), so all you have to do is keep on "cutting" (a,a+b/2) in half and so on.

Answer 3

Answer: x = 1.839286 is a root.

See source for another explanation and a mathlet.