Intuitively and graphically what does the inequality of the mean value mean? Let $f: [a, b] \to \mathbb{R}$

*

*$f$ is continuous on $[a, b]$

*$f$ is differentiable in $(a, b)$

*$f'$(the derivative) is continuous on $[a, b]$
If all this is true, the mean value theorem can be expressed as follows:
$$|f(b)−f(a)|\le (b−a)M$$
(Where $M$ is the bound for the derivative of $f$   ($| f' | \le M$)).
Well. What I want to know is, graphically and intuitively, what does this mean?
Why am I asking the question?
In this youtube video (min: $43:35$) An attempt is made to give an intuitive explanation, but it is not rigorous at all and is therefore confusing, since the lecturer treats $M$ as $f'(x)$ and not as a bound of $f'(x)$
Thank you.
 A: $
\newcommand{\ve}{\varepsilon}
$So let's go over the mean value theorem, in its usual statement, with a function $B$ to match the notation of the video.
If $B$ is continuous on $[a,b]$ and differentiable on $(a,b)$, for $a< b$, then $\exists c \in (a,b)$ where
$$\frac{B(b) - B(a)}{b-a} = B'(c)$$
That is, there is a point in the interval $(a,b)$ where the average rate of change equals the instantaneous rate of change.
If we let $\varepsilon > 0$ and $a=t$ (and we use this since we want to think of very small changes), this holds in particular for $[t,t+\ve]$:
$$\frac{B(t+\ve) - B(t)}{\ve} = B'(t)$$
With the further assumption that $B'$ is continuous on $[a,b]$, a closed interval, we may make the claim that $\exists M \in \Bbb R_{\ge 0}$ bounding $B'$, i.e.
$$|B'(t)| \le M \qquad \forall t \in [a,b]$$
Then we have
$$\left| \frac{B(t+\ve) - B(t)}{\ve} \right| = \frac{\left| B(t+\ve) - B(t) \right|}{\ve} = |B'(t)| \le |M| = M$$
Therefore
$$ \frac{\left| B(t+\ve) - B(t) \right|}{\ve} \le M \implies \left| B(t+\ve) - B(t) \right| \le \ve M$$
I imagine in the video the instructor is just assuming outright that you're using the $t$ for which this maximum $M$ is attained and then you have $A=M$.

An intuition can be gained by noticing
$$\frac{  B(t+\ve) - B(t)  }{\ve} \le M$$
(since $x \le |x|$). Namely, for a "nice" function like $B$, your average rate of change (left hand side) over an interval can never exceed the largest instantaneous rate of change over that interval.
Thinking in terms of velocity, your function can never go faster on average for an interval (the left-hand side), than its fastest speed at any given point ($M$, which bounds the derivative).
(Taking the absolute value then just focuses you on the "magnitude" of that speed, or of that change.)
This should make sense, since the mean value theorem would guarantee a different bound $M$ if the opposite were true.
A: Here's the important bit of intuition: if curve $A$ has a slope which is never greater than the slope of curve $B$, then curve $B$ produces greater height differences than curve $A$. Here, curve $A$ is the graph of $f$, its height difference over the interval $[a,b]$ being $\vert f(b)-f(a)\vert$, and curve $B$ is the graph of the function $h:x\mapsto Mx$, its height difference over the same interval being $M(b-a)$.
Why is this true? Because of the usual formulation of the mean value theorem. If the height difference in the graph of $f$ over the interval $[a,b]$ were greater than that of $Mx$, then $\frac{\vert f(b)c-f(a)\vert}{b-a}$ would be greater than $M$. But then $\vert f'\vert$ would have to assume a value greater than $M$ somewhere in the interval, which contradicts the fact that $M$ was a bound on $f'$.
The details need some more care because of the absolute values we're working with, but this is the gist of it.
A: Graphically and intuitively, under the hypotheses of the question the graph of $f$ lies in the green shaded wedge:

(Note, however, that not every graph lying in this wedge satisfies $|f'| \leq M$.)
