Question

a) Suppose f"(z) exists on an interval I and f(s) has a zero at three distinct points a < b< c on I. Show there is a pbint p on [a, c] where f"(p) = 0. b)mustrate part (a)on the cubic f(s) = (x- a)(x - b)(x - c)

Answer 1

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.