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.
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):
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 $f^{\prime}(c) = 0.$
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.
$f'$ 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.
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 $x = 0.$
