A negative third derivative implies a positive first derivative at a point. 
Let $f$  be three times differentiable on $\mathbb{R}$ with $f(0)=f(1)$, and for all $x\in[0,1]$, $f'''(x)<0$. Prove that $f'(\frac{1}{2})>0$

I actually have a proof of this question using the Taylor expansion
But looking at the problem intuitively (as much as it fails in analysis), it just doesn't really add up to me, I mean what I get from the third derivative is that the first derivative is convex (and the second is strictly increasing), and what I get from $f(0)=f(1)$ is that $f'(c)=0$ for some $c\in[0,1]$.
Why do these add up to the derivative being positive specifically at $^1/_2$?
(The solution I already have is:
Looking at the Taylor exapnsion of the series around $^1/_2$, 
$$T_{^1/_2}(x)=f(\frac{1}{2})+f'(\frac{1}{2})(x-\frac{1}{2})+\frac{f''(\frac{1}{2})(x-\frac{1}{2})^2}{2}+\frac{f'''(c)(x-\frac{1}{2})^3}{6}$$
For some $c$ in the interval of $x$ and $^1/_2$. 
Then setting $T_{^1/_2}(0)-T_{^1/_2}(1)=0$ we get 
$$f'(^1/_2)=-\frac{f'''(c_0)}{6\cdot8}-\frac{f'''(c_1)}{6\cdot8}>0$$
As required)
 A: I'm not sure this is the kind of intuition you're looking for, but another way to approach the problem is to let
$$g(x)=f\left({1+x\over2}\right)-f\left({1-x\over2}\right)$$
It's easy to see that $g(x)$ is an odd function, and $g(-1)=g(0)=g(1)=0$.  Moreover, 
$$g'(x)={1\over2}\left(f'\left({1+x\over2}\right)+f'\left({1-x\over2}\right)\right)\implies g'(0)=f'\left({1\over2}\right)$$
and
$$g'''(x)={1\over8}\left(f'''\left({1+x\over2}\right)+f'''\left({1-x\over2}\right)\right)\lt0$$
Note that the derivatives of an odd function toggle back and forth between being even and odd, so $g''(x)$ is an odd function -- hence if the function $g$ is concave up at $x$ it is concave down at $-x$ and vice versa.  The strict negativity of the third derivative means that $g$ is concave up for $x\lt0$ and concave down for $x\gt0$ (and, obviously, $g''(0)=0$).  From $g(-1)=g(0)=g(1)=0$ it follows that $g$ is strictly positive with a unique maximum on $0\lt x\lt1$ and strictly negative with a unique minimum on $-1\lt x\lt0$.  Hence $g'(0)\ge0$.  Finally, we rule out $g'(0)=0$ by appealing to the strict negativity of $g''(x)$ for $x\gt0$, in particular on the interval from $0$ to wherever the maximum of $g$ occurs.  The net conclusion is
$$f'\left({1\over2}\right)=g'(0)\gt0$$
as desired.
