Proving Existence of a Point c for Taylor Expansion | Homework Statement

[talolard]

Homework Statement

Let f be differentiable on [a,b] and f'(a)=f'(b)=0. Prove that if f'' exists then there exists a point c in (a,b) such that
$$test$$
$$|f''(c)| \geq \frac{4}{(b-a)^2}|f(b)-f(a)|$$

Homework Equations

All of the equations are supposed to be in absolute value but I had trouble getting tha tto apear so please assume that everything is bsolute value. I didnt use any inequalities so it shouldn't make much difference.

The Attempt at a Solution

Assume f'' exists
Via Taylor we get that
1. $$f(x) (at a) = f(a) + f'(a)x + \frac{f''(c)}{2}(x-a)^2 = f(a)+\frac{f''(c)}{2}(x-a)^2$$
2. $$f(x) (at b) = f(b) + f'(b)x + \frac{f''(d)}{2}(x-b)^2 = f(a)+\frac{f''(d)}{2}(x-b)^2$$
Then Via 1 we get that
3. $$|f(b)|= |f(a)+\frac{f''(c)}{2}(b-a)^2| \iff |\frac{2(f(b) -f(a))}{(b-a)^2}| = | f''(c)|$$
via 2 we get that
4. $$|f(a)|= |f(b)+\frac{f''(d)}{2}(b-a)^2| \iff |\frac{2(f(b) -f(a))}{(b-a)^2}| = | f''(d)|$$
adding equation 3 and 4 together we get
$$|f''(c)|+|f''(d)|=|frac{4(f(b) -f(a))}{(b-a)^2}|$$
then $$\frac |{f''(c)+f''(d)}{2}|=|f''(e)| =| \frac{2(f(b) -f(a))}{(b-a)^2}|$$

At this point I am stuck. I'm thinking that I can show that c and d are actualy the same point.

 

[kurt.physics]

If this is true then we could say that |f''(e)|=|f''(c)|. But I dont know how to show that c and d are the same point. Any help would be much appreciated. Thanks