(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

[tex] test [/tex]

[tex] |f''(c)| \geq \frac{4}{(b-a)^2}|f(b)-f(a)| [/tex]

2. Relevant 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 shouldnt make much difference.

3. The attempt at a solution

Assume f'' exists

Via Taylor we get that

1. [tex] 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 [/tex]

2. [tex] 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[/tex]

Then Via 1 we get that

3. [tex] |f(b)|= |f(a)+\frac{f''(c)}{2}(b-a)^2| \iff |\frac{2(f(b) -f(a))}{(b-a)^2}| = | f''(c)|[/tex]

via 2 we get that

4. [tex] |f(a)|= |f(b)+\frac{f''(d)}{2}(b-a)^2| \iff |\frac{2(f(b) -f(a))}{(b-a)^2}| = | f''(d)| [/tex]

adding equation 3 and 4 together we get

[tex] |f''(c)|+|f''(d)|=|frac{4(f(b) -f(a))}{(b-a)^2}| [/tex]

then [tex] \frac |{f''(c)+f''(d)}{2}|=|f''(e)| =| \frac{2(f(b) -f(a))}{(b-a)^2}| [/tex]

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

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Taylor expansion

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