Does $f~'$ have to be continuous for application of Mean Value Theorem? If $f$ is a continuous function in $[a,b]$, then for any $x,y \in [a,b];~y>x$, we say that $\exists~c \in [x,y]$ such that $f~'(c)= \dfrac {f(y)-f(x)}{y-x}$.
Does $f~'$ have to be necessarily continuous to apply the mean value theorem for derivatives?
If not, why is $f~''$ needed to be continuous in this problem here
Thank you very much for your help.
 A: No, the continuous assumption is not necessary. As pointed out by others in the comments, the essential reason is the derivative satisfies the intermediate value property. If $f'$ takes value $a,b$ and $a<b$ at an interval, then $f'$ would have to take any value $c$ between $(a,b)$. 
A heuristic way to think about it might be as follows: by elementary arguments you would have
$$
f(c)-f(d)=\int^{c}_{d}f'(t)dt\le (c-d)\max_{t\in [c,d]} f'(t)
$$
and by analogous arguments $$
(c-d)\min_{t\in [c,d]}f'(t)\le f(c)-f(d)\le (c-d)\max_{t\in [c,d]} f'(t), c>d
$$ 
Therefore by intermediate value property there must be some element $e\in [c,d]$ such that $$(c-d)f'(e)=f(c)-f(d)\leftrightarrow f'(e)=\frac{f(c)-f(d)}{c-d}$$ and this is the mean value theorem. This is only heuristic because we assumed maximum and minimum exist for $f'$ on $[c,d]$, which is not necessary, but I hope the geometry is clear. The usual (rigorous) proof goes by constructing an ancillary function $h(t)=(f(c)-f(d))t-(c-d)f(t)$ and apply Rolle's theorem. 
Some related discussions can be found at here (but might be too technical and not really related to the question you are concerning at this point):
Discontinuous derivative.
How discontinuous can a derivative be?
A: No,  it is necessary that $f$ be continuous on $[a,b]$ and that $f^{\prime}$  merely exists everywhere on $(a,b)$. The Mean Value Theorem is an easy consequence of Rolle's Theorem, so I discuss the proof of Rolle's Theorem. So assume that $f(a) = f(b)$, and it suffices to prove that $f^{\prime}(c) = 0$ for some $c \in (a,b).$
  The essential point is that by the continuity of $f$, with no assumption on $f^{\prime}$, but using the fact that we are dealing with a closed interval, there is some $c \in [a,b]$ such that $f(x) \leq f(c)$ for all $x \in [a,b].$ If $c = a$ or $c = b,$ then $f$ is constant on $[a,b]$ and $f^{\prime} = 0$ everywhere on $(a,b)$. Suppose then that $c \in (a,b)$. Now if $x < c$ then $f(x) - f(c) \leq 0$ and $\frac{f(x)-f(c)}{x-c} \geq 0$ and hence $\lim_{x \to c^{-}}\frac{f(x)-f(c)}{x-c} \geq 0 .$ Similarly, if $x > c$ then $f(x) - f(c) \leq 0$ and $\frac{f(x)-f(c)}{x-c} \leq 0$ and hence $\lim_{x \to c^{+}}\frac{f(x)-f(c)}{x-c} \leq 0 .$ But since $f$ is differentiable at $c$, these two limits must be equal. Since one is $\geq 0$ and the other is $\leq 0$, they must both be $0,$ and $f^{\prime}(c) = 0.$
A: The function's derivative does not have to be continuous. The classic example is the function given by $$f(x) = \left\{ \begin{array}{cc}x^2 \sin(1/x) & x\neq 0\\ 0 & x = 0 \end{array}\right.$$
This function can be shown to be differentiable on $[0,1]$ but it's derivative is not continuous at $x=0$. However, the hypothesis of the mean value theorem applies for any function differentiable on a closed interval.
The reason we required $f''$ to be continuous in the question you are referencing is so that we may use the fact that a continuous function on a compact set has a bounded image.
A: $f'$ does need to be continuous to apply the theorem for derivatives.  For example, consider
$[a,b] = [0,1]$ and 
$$f(x) =\left\{ \begin{array}{cl}
\frac{x}{4} & 0 \leq x < \frac{1}{3} \\
2x - \frac{7}{12} & \frac{1}{3} \leq x < \frac{2}{3} \\
\frac{3x+1}{4} & \frac{2}{3} \leq x \leq 1
\end{array}\right.
$$
$f$ is continuous, but $f'$ is not. 
Then the MVT for derivatives (if applicable) says that somewhere in the interval, $f'(x) = 1$. But for this function, $f'(x) \in \{\frac{1}{4}, 2, \frac{3}{4} \}$.
However, we would not say that this function is everywhere differentiable.  If the function is everywhere differentiable, then the MVT does hold.
A: i don't think so. what about $f(x) = x^2\sin(1/x) , x \neq 0$ and $f(0) = 0$ on the interval $[-1/\pi, 1/\pi]?$
we have $f^\prime(x) = -\cos(1/x) + 2x\sin(1/x)$ which is not continuous at $x = 0.$
