If $f'(c) \neq \frac{f(b)-f(a)}{b-a}$, then find number of such $c$. 
Let $f(x)=x^3+3x+2$ and $x=c$ is a point such that
$$f'(c) \neq \frac{f(b)-f(a)}{b-a}$$
for any two values of $a$ and $b$, where $a,b$ and $c \in \mathbb R$. Then find the number of such points $c$ for which this is true.

By analyzing the graph of $f(x)$, I think that such points exist on $f(x)$ (and possibly all doubly differentiable curves) only when $f''(x)=0$ at those points. But I can't seem to prove it. Using that, I got $c=0$ and thus the number of such points is $1$. Is my claim true ? If so, then how can one prove it ?
 A: Your statement is ambiguous; I shall interpret it as saying that $\exists c$ such that $f'(c) \neq \frac {f(b)-f(a)} {b-a}$ for $\forall a, b$ with $a \neq b$. Well, in this case there is no such $c$, by the following argument: note that $f'(x)=3c^2+3$ and $\frac {f(b)-f(a)} {b-a} = a^2 + ab +b^2 +3$. Equating the two expressions, you claim that there is a $c$ such that $c^2 \neq \frac {a^2 + ab +b^2} 3$ $\forall a,b$. But if $c \neq 0$, this cannot be true since one can always find some $a$ and $b$ making this equality true, for instance $a=0, b=c\sqrt 3$. If $c=0$, this is equivalent to $a^2 + ab + b^2 \neq 0 \space \forall a, b$ with $a \neq b$ and this is clearly true (at least one of the two numbers must be $\neq 0$ because otherwise they would be equal so assume $b \neq 0$; then $({\frac a b})^2 + {\frac a b} +1 \neq 0$ which is obviously true since this equation does not have real roots).
In conclusion, there exist a single such $c$, namely $0$.
A: Let us prove that this can only happen when $f''(c) = 0$:
Since $f$ is continuous and $\frac{f(a) - f(b)}{a - b} \ne f'(c)$ we have either $\frac{f(a) - f(b)}{a - b} \ge f'(c)$ for all $a,b$ OR $\frac{f(a) - f(b)}{a - b} \le f'(c)$ for all $a,b$ (because $f$ is continuous and hence it takes all intermediate values). So WLOG let us assume that $\frac{f(a) - f(b)}{a - b} \ge f'(c)$ for all $a,b$. Now, for any $d$ we have $f'(d) = \lim_{e \to d}\frac{f(e)-f(d)}{e-d} \ge f'(c)$. So $c$ is local minimum of $f'(x)$ and hence $f''(x) = 0$. Now, for a cubic $f$ we have only one possible solution which we have to check.
A: If $f$ is twice differentiable and $f''(c)\ne 0$ then we can find two distinct $a,b$ with $f'(c)=\frac{f(b)-f(a)}{b-a}$ - why?
Assume wlog. that $f''(c)>0$. Then for sufficiently small positive $h$ we have $\frac{f'(c+h)-f'(c)}{h}>0$ and $\frac{f'(c-h)-f'(c)}{h}<0$, hence $f'(c+h)>f'(c)>f'(c-h)$. Again, for sufficiently small $\eta$, we have $\frac{f(c+h+\eta)-f(c+h)}\eta$ and $\frac{f(c-h+\eta)-f(c-h)}\eta$ close enough to $f'(c\pm h)$ that the inequalities $\frac{f(c+h+\eta)-f(c+h)}\eta>f'(c)>\frac{f(c-h+\eta)-f(c-h)}\eta$ hold. Apply the IVT to $\frac{f(x+\eta)-f(x)}\eta$ on the interval $[c-h,c+h]$ to find $a\in(c-h,c+h)$ and $b=a+\eta$ with $\frac{f(b)-f(a)}{b-a}=f'(c)$.
We conclude that only points with $f''(c)=0$ are candidates. For each such candidate, ohwever, one should perform the check "manually" as weird things can happen globally.
A: Such a $c$ does exist and $c=0$
Here is a graph of the function (red) and its tangent at $x=0$ (blue):

