1. ## Taylor.

Let f be 2 times differentiable function in open interval (0,1) with following conditions:

1. lim(x-->0)f(x) = lim(x-->1)f(x) = 0

2.Exist M>0 so that |f ''(x)|<=M for all x in (0,1)

Prove that |f '(x
)|<=M/2

Thank you all!

2. Originally Posted by Also sprach Zarathustra
Let f be 2 times differentiable function in open interval (0,1) with following conditions:

1. lim(x-->0)f(x) = lim(x-->1)f(x) = 0

2.Exist M>0 so that |f ''(x)|<=M for all x in (0,1)

Prove that |f '(x
)|<=M/2

Thank you all!
It looks like you have that $f(0)=f(1)=0$ thus by Rolle's theorem there exists some $c\in (0,1)$ such that $f'(c)=0$. Form the first degree taylor polynomial around that point $f'(x)=f'(c)+\frac{f''(\xi)}{2!}(x-c)^2=\frac{f''(\xi)}{2}(x-c)^2$. Now, combining $\left|f''(\xi)\right|\leqslant M$ and $(x-c)^2\leqslant 1$ we see that $\left|f'(x)\right|=\left|\frac{f''(\xi)}{2}(x-c)^2\right|\leqslant\frac{M}{2}$

3. Originally Posted by Also sprach Zarathustra
Let f be 2 times differentiable function in open interval (0,1) with following conditions:

1. lim(x-->0)f(x) = lim(x-->1)f(x) = 0

2.Exist M>0 so that |f ''(x)|<=M for all x in (0,1)

Prove that |f '(x
)|<=M/2
Use the Taylor series with remainder $f(x+t) = f(x) + tf'(x) + \tfrac12t^2f''(\xi)$, where $\xi$ lies between x and x+t. This implies that $|tf'(x) +f(x) - f(x+t)| = |\tfrac12t^2f''(\xi)|\leqslant\tfrac12Mt^2$. There are now two separate cases to consider.

If f(x) and f'(x) have the same sign, take $t = 1-x$. Then $|tf'(x) +f(x)| = |(1-x)f'(x) +f(x)| \geqslant |(1-x)f'(x)|$. Also, $f(x+t) = f(1) = 0$. So the inequality in the previous paragraph tells us that $|(1-x)f'(x)| \leqslant \tfrac12M(1-x)^2$. Hence $|f'(x)|\leqslant \tfrac12M(1-x)\leqslant \tfrac12M$.

If f(x) and f'(x) have the opposite signs, take $t = -x$. Then $|tf'(x) +f(x)| = |-xf'(x) +f(x)| \geqslant |xf'(x)|$. Also, $f(x+t) = f(0) = 0$. So the inequality tells us that $|xf'(x)| \leqslant \tfrac12Mx^2$. Hence $|f'(x)|\leqslant \tfrac12Mx\leqslant \tfrac12M$.

Originally Posted by Drexel28
Form the first degree taylor polynomial around that point $f'(x)=f'(c)+\frac{f''(\xi)}{2!}(x-c)^2$.
The second term in that Taylor polynomial is $f''(\xi)(x-c)$, not $\frac{f''(\xi)}{2!}(x-c)^2$.