prove that for any $\alpha\in \mathbb{R}$ there is point $c\in(a,b)$ st $f'(c)=\alpha$ If  $f:(a,b)\to R$ is differentiable and bounded suppose that the limits $\lim _{x\to a} f(x)$ and $\lim _{x\to b} f(x)$ do not exist prove that for any $\alpha\in \mathbb{R}$ there is point $c\in(a,b)$ st $f'(c)=\alpha$
i really have no idea can some give me any hint
 A: I suppose $a$ and $b$ are finite real munbers. (Actually they must be finite, otherwise the result is not ture, consider the example $f(x)=\sin x, x\in(-\infty,+\infty)$)
Suppose it is not ture, then there exists $\alpha_0\in \mathbb{R}$ such that: for any $x\in(a,b)$ , $f'(x)\ne \alpha_0$.
BY Darboux's theorem http://en.wikipedia.org/wiki/Darboux%27s_theorem_%28analysis%29, we know 
$$f'(x)>\alpha_0 \ \text{or}\ f'(x)>\alpha_0,\ \forall\ x\in(a,b).$$
Let $$F(x)=f(x)-\alpha_0x,\ x\in(a,b),$$
then we know $F'(x)>0$ or $F'(x)<0$ for $x\in(a,b)$,
and $F(x)$ is increasing or decreasing function in open inieral $(a,b)$.
Due to the boundedness of $f(x)$, we konw $F(x)$ is bounded, so
the two limits $$\lim _{x\to a} F(x),\ \lim _{x\to b} F(x)$$
both exist, so do limits $$\lim _{x\to a} f(x),\ \lim _{x\to b} f(x).$$
A: Here is another approach. Since $f$ is bounded but $\lim_{x\to a^{+}} f(x) $ does not exist, it follows that the numbers $$A=\liminf_{x\to a^{+}} f(x), B=\limsup_{x\to a^{+}}f(x) $$ exist and $A<B$. Also by definition of $\limsup, \liminf$ there exist sequences $\{x_n\}, \{y_n\} $ with values lying in $(a, b) $ such that $x_n\to a, y_n\to a$ as $n\to\infty $ and $f(x_n) \to A, f(y_n) \to B$.
Let's consider the ratio $$\frac{f(x_n) - f(y_m)} {x_n-y_m} $$ Clearly there exist infinitely many pairs $(m, n) $ such that the denominator is non-zero and the ratio is well defined for these values of $m, n$. And clearly as $m, n$ tend to $\infty$ the numerator tends to $A-B<0$ and denominator tends to $0$. Further we can ascertain that there will be infinitely many pairs $(m, n) $ such that the denominator is positive and another set of infinitely many pairs $(m, n) $ such that the denominator is positive. Therefore given any $M>0$ the ratio takes values larger than $M$ as well as values smaller than $-M$. By mean value theorem the same holds for derivative $f'$. Choose $M$ such that $\alpha\in(-M, M) $ and then the conclusion easily follows from Darboux Theorem which says that derivatives possess intermediate value property.
Note that the assumption about limit of $f$ not existing at $b$ is not needed. 
