Question: Suppose $f(x)$ is derivable on$[a,b]$, and $f(a)=f(b)=0$, $f\_+'(a)\gt0$ and $f\_-'(b)\gt 0$. Prove there exists a point $c$ such that $f(c)=0$ and $f'(c)\le0$.

---

It's easy to get the first one. Since $f\_+'(a)\gt0$ and $f\_-'(b)\gt 0$, there is a $\delta \gt 0$, $f(x)-f(a)\gt0$ for $0\lt x-a\lt\delta$, $f(b)-f(x)\gt0$ for $0\lt b-x\lt\delta$. Hence there is a $c$ such that $f(c)=0$. But how to estimate the sign of $f'(c)$.

Answer: 
Since $f'\_+(a)>0,$ then $$\lim\limits\_{x \to a+}\dfrac{f(x)-f(a)}{x-a}=\lim\limits\_{x \to a+}\dfrac{f(x)}{x-a}>0,$$ which implies that $\exists \delta\_1>0$ such that $\dfrac{f(x)}{x-a}>0$,namely $f(x)>0$ when $0<x-a<\delta\_1$. Likewise, since $f'\_-(b)<0$, then $\exists \delta\_2>0$ such that $f(x)<0$ when $-\delta\_2<x-b<0$. 

Thus, we may take $\delta=\dfrac{\min(\delta\_1,\delta\_2)}{2}$ such that $f(x)>0,\forall x \in (a,a+\delta)$ and $f(x)<0, \forall x \in (b-\delta,b)$ .

Now, we define a set $C$ as $$C=\{r|r \in (a,b);f(x)>0,\forall x\in (a,r)\},$$which apparently is nonempty and bounded above. Hence, $C$ has a supremum, whom we denote as $\sup C=c.$ Apparently, $a<c<b.$ Thus, we claim that $f(c)=0$ and $f'(c) \leq 0.$

If $f(c) > 0,$ then $f(x) > 0$ in some neighborhood of $c$ and so $\sup C > c,$ which contradicts. Similarly, if $f(c) < 0,$ then $\sup C < c,$ which also contradicts. Therefore, $f(c)=0.$

Notice that $$f'(c)=f'\_+(c)=\lim\limits\_{x\to{c+}} \dfrac{f(x)-f(c)}{x-c}.$$ If $f'(c) > 0$, then $f(x)>0$ in some right-neighborhood of $c$ and thus $\sup C> c,$ which contradicts. Therefore, $f'(c) \le 0.$

The proof is completed so far.
Suppose that $f(c)=0, f'(c)>0$, repeat the argument, there exists $c\_1\in (c,b)$ such that $f(c\_1)=0$, if $f'(c\_1)\leq 0$, done, otherwise there exists $c\_2\in (c\_1,b)$ such that $f(c\_2)=0$. Suppose constructed $c\_n$ such that $c\_n\in (c\_{n-1},b)$ and $f(c\_n)=0$, if $f'(c\_n)\leq 0$, done, otherwise we have a sequence $c\_n$ which increases such that $f(c\_n)=0$ and $f'(c\_n)>0$. This sequence is bounded by $b$, so it converges towards $d$.
Since $f$ is continuous, $f(d)=0$.

$f'(d)=lim\_{n\rightarrow+\infty}{{f(c\_n)-f(d)}\over{c\_n-d}}=0$ since $f(c\_n)=f(d)=0$.