Prove there exists $\xi \in(-\infty,a) $ such that $f'(\xi)=0$. Suppose $f(x)$ is differentiable on $(-\infty,a]$, $f(a)f'_{-}(a)\lt0$, and $\lim_{x\to-\infty}f(x)=0$. Prove there exists $\xi \in(-\infty,a) $ such that $f'(\xi)=0$.  
Frome $f(a)f'_{-}(a)\lt0$ we know $f(a)\gt0 ,f'_{-}(a)\lt0$ or $f(a)\lt0 ,f'_{-}(a)\gt0$. Let's assume  $f(a)\gt0 ,f'_{-}(a)\lt0$ , then $f(x)$ is decreasing to $f(a)\gt 0$.  But $\lim_{x\to-\infty}f(x)=0$ , hence $f(x)$ achieves maximum at first then decreases from $-\infty$ to $a$. Therefore the derivatives at maximum is zero. After these imagination, I still don't know how to make a strict proof.
 A: Hint: First show that there exists a $b<a$ with $f(b)=f(a)$, then use Rolle's theorem. To show that there must be a $b<a$ with $f(b)=f(a)$, assume the contrary and derive a contradiction either with $\lim_{x\to -\infty} f(x)=0$ or with $f_-'(a)<0$ (assuming $f(a)>0$; otherwise you can just consider $-f$ instead). 
A: You have well identified the two cases. Suppose that $f(a)\gt0$ and $f^\prime(a^-)\lt 0$, the second case being similar.
Denote $b = f^\prime(a^-) <0$. By definition of derivability, it exists $\delta > 0$ such that
$$\left\vert \dfrac{f(x)-f(a)}{x-a} - b \right\vert < \dfrac{-b}{2}$$ for $x \in (a-\delta,a)$, which is equivalent to
$$\dfrac{3b}{2}(x-a) +f(a)> f(x) > \dfrac{b}{2}(x-a)+f(a).$$
In particular $$f(a)<f(a)-\dfrac{b\delta}{4}<f(a-\delta/2)$$
As $\lim\limits_{x\to-\infty}f(x)=0$, it exists $X < a-\delta$ such that $f(X) < f(a)$. By Intermediate Value Theorem, it exists $y \in (X,a- \delta/2)$ such that $f(y) = f(a)$. By Rolle's Theorem, $f^\prime(\xi) = 0$ for $\xi \in (y, a)$. This concludes the proof.
A: Proof
According to the assumption $f(a)f'_-(a)<0$, we may know that:1) $f(a)>0$ and $f'_-(a)<0$; or 2) $f(a)<0$and $f'_-(a)>0.$ Here, we only deal with the case 1). As for the case 2), the proof ought to be similar.
If there exists a point $p \in (-\infty,a)$ such that $f'(p)>0$. Then we may apply G.Darboux's theorem, which asserts the intermediate value property of the derivative, and claim that: since $f'(p)>0$ and $f'(a)<0,$ then there necessarily exists a point $\xi \in (p,a)\subset (-\infty,a)$ such that $f'(\xi)=0$, which is what we want.
If not, then apparently we only need to discuss the case that $f'(x)<0$ for all $x \in (-\infty,a).$ If so, then $f(x)$ is decreasing over $(-\infty,a)$. Hence，$f(x)<\lim\limits_{x \to -\infty}f(x)=0$ for all $x \in (-\infty,a),$ which contradicts $f(a)>0$. That is because that, $\lim\limits_{x \to a-}f(x)=f(a)>0$ implies that, there exists a $\delta>0$ such that $f(x)>0$ when $a-\delta<x<a$.
So far, the proof is completed.
