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.
Best Answer
$ \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.