Prove that if $\exists c \in (a,b)$ s.t. $\frac{f(b)-f(c )}{f(c ) - f(a)} < 0$, then $\exists s \in (a,b)$ s.t. $f'(s) = 0$ $f:[a,b] \to \mathbb{R}^{1}$ be a function continuous on $[a,b]$ and differentiable on $(a,b)$. Prove that if $\exists c \in (a,b)$ s.t. $\frac{f(b)-f(c )}{f(c ) - f(a)} < 0$, then $\exists s \in (a,b)$ s.t. $f'(s) = 0$
Before I ask my question, I've search here in MSE and found this. Even though there are many nice hints available, I am still having trouble understanding those. Also, since it is not polite to ask there, I have no other choice but asking the same question. So, all I understand from that post is that since f is continuos on [a,b] and $f(c ) > f(b)$ and $f(c ) > f(a),$ for some $c \in (a,b),$ then, $f(a) = f(b),$ and from the Rolle's theorem, we can conclude that $\exists s \in (a,b),  s.t. f'(s) = \frac{f(b)-f(a)}{b-a} = 0$. Please can anyone tell whether or not I am on the right track? If not, is there much simpler way to unedrstand this? Thanks. 
 A: I don't see how you infer $f(a)=f(b)$.
One way to reason through the problem is the intermediate value property of $f'$ (see Darboux's Theorem).
Case 1: $f(b)>f(c)$ and $f(c)<f(a)$. Then the MVT says that there is $u\in(c,b)$ and $v\in(a,c)$ such that $f'(u)>0$ and $f'(v)<0$. Now invoke Darboux's Theorem to see that there is $s\in(v,u)$ s.t. $f'(s)=0$.
Case 2: $f(b)<f(c)$ and $f(c)>f(a)$: similar.
A: Case 1: $f(c)>f(a)$ and $f(c)>f(b)$. Then there are two subcases.
Subcase 1a: $f(a)\in[f(b)\ f(c)]$. Then by Bolzano theorem we can conclude that there exists $d\in[c\ b]$ such that $f(d)=f(a)$. Now apply Rolle's theorem to $a$ and $d$.
Subcase 1b: $f(b)\in[f(a)\ f(c)]$. By similar argument there exists $d\in [a\ c]$ such that $f(d)=f(b)$. Apply Rolle's theorem again.
Case 2: similar argument.
A: I think the question is much easier than it seems. The equation $$\frac{f(b) - f(c)}{f(c) - f(a)} < 0$$ shows that $f(b) - f(c)$ and $f(c) - f(a)$ are of opposite signs. This clearly means $f(c) - f(b)$ and $f(c) - f(a)$ are of same sign and hence their product $\{f(c) - f(b)\}\{f(c) - f(a)\}$ is positive.
Thus it means that the polynomial $g(t) = (t - f(a))(t - f(b))$ is positive at $t = f(c)$. It follows that $t = f(c)$ does not lie between the roots $f(a), f(b)$ of quadratic polynomial $g(t)$. Thus in case $f(c)$ lies to the right of interval with end-points $f(a), f(b)$ then clearly the maximum value of $f$ is greater than $f(a), f(b)$ and hence is attained at some interior point $d \in (a, b)$. Similarly if $f(c)$ lies to the left of interval with end points $f(a), f(b)$ then the minimum value of $f$ is attained at an interior point $e \in (a, b)$.
In either case we have $f'(d) = 0$ or $f'(e) = 0$ (remember that derivative vanishes at maxima/minima).
A: I know that you didn't ask for a solution, but this was just too much fun not to provide!  This result, however, requires that $f\in C^1$ which was not given!  One, however, can use the answer of @KimJongUn to remove this condition.
Suppose there exists a $c\in(a,b)$ such that
$$
\frac{f(b)-f(c)}{f(c)-f(a)}<0.
$$
Now, consider,
$$
\frac{c-a}{b-c}.
$$
Note that this fraction is always positive.  Therefore,
$$
\frac{f(b)-f(c)}{b-c}\cdot\frac{c-a}{f(c)-f(a)}=\frac{c-a}{b-c}\cdot\frac{f(b)-f(c)}{f(c)-f(a)}<0.
$$
By the mean value theorem, there are points $d\in(a,c)$ and $e\in(c,b)$ such that
$$
f'(d)=\frac{f(c)-f(a)}{c-a}\qquad\text{and}\qquad f'(e)=\frac{f(b)-f(c)}{b-c}.
$$
Therefore, 
$$
\frac{f'(e)}{f'(d)}<0,
$$
but then the derivatives have different signs and by the intermediate value theorem (since $f$ is $C^1$), there must be a point between them where the derivative is zero.
A: Yes, you are on the right track. 
Here are the details.  Since $\exists c \in (a,b)$ s.t. $\frac{f(b)-f(c )}{f(c ) - f(a)} < 0$, we have two possible cases: 
Case 1: $f(c ) > f(b)$ and $f(c ) > f(a)$.  Let $d$ such that $\max\{f(a),f(b)\} < d < f(c)$. Since $f$ is continuos, there is $a_1\in (a,c)$ such that $f(a_1)=d$ and there is $b_1\in (c,b)$ such that $f(b_1)=d$. NOW you can apply Rolle's Theorem and deduce $\exists s \in (a_1,b_1)\subseteq (a,b),  s.t. f'(s) = 0$. 
Case 2: $f(c ) < f(b)$ and $f(c ) < f(a)$. It is completely analogous (or you can multiply $f$ by $-1$ to reduce this case to Case 1). 
