Answer:
We can find a root in the average point (a+b+c)/3
Step-By-Step Explanation:
We can use the Rolle Theorem. Since f is two times deribable on both [a,b] and [b,c], then there exists points x in (a,b) and y in (b,c) such that f'(x) = f'(y) = 0. Now, again by using Rolle Theorem, since f' is derivable in [x,y] (because it is a closed interval in I), then there exists s in [x,y] such that f''(s) = 0. This proves a.
The cube (x-a)(x-b)(x-c) has 3 roots, a, b and c and it is two times derivable because it is a polynomial. Hence we can use (a) to ensure that there is a root on I. Nevertheless, we can try to find the root manually:
f'(x) = (x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)
f''(x) = (x-c)+(x-b)+(x-c)+(x-a)+(x-b)+(x-a) = 2(x-a)+2(x-b)+2(x-c) = 2( (x-a) + (x-b) + (x-c) ) = 6x - 2 (a+b+c).
We want x such that
6x - 2(a+b+c) = 0, or, equivalently,
3x - (a+b+c) = 0
Hence
x = (a+b+c)/3
Is a root of f'' in I.
