# Intuition on Mean Value Theorem

The following is my opinion, please correct it if I'm wrong or not good explanation:

if we get $a, b$, then there must be a constant changing-rate $m$ of $(a, f(a))$ and $(b, f(b))$, if the function go this straight line, then everything point has derivative equal to m, if $f'(x)$ greater or lower than m, then it must be somewhere to lower or greater than m in order to reach the f(b), between them, it meets the m. If we get this 'feeling', then theorem is just natural and obvious : there must exist at least one x for, $f'(x)=\frac{f(b)-f(a)}{b-a}=m$

I realize if I think like this way to get the mathematical feeling behind definitions or theorems, then most of them are just natural, for example, fundamental theorem of calculus, if a function $f(x)$, we consider function value as changing-rate, $$\lim\limits_{n\rightarrow\infty}\left[\sum_{i=1}^{n}f(x_i)(x_{i}-x_{i-1})\right]$$ is just how much the original function-value changes, i.e. $\Delta F(x)=F(b)-F(a)$

But for some other things which always called rules, it seems cannot be understood directly, like the Chain rule, I could prove it by basic definition of derivative, but just cannot 'feel' it as the same way with Mean-Value theorem.

• @Xingdong: I very much applaud your desire to understand the intuition behind definitions and theorems. There are good intuitive reasons for the Chain Rule. The derivative is the local growth rate of a function. Now look at $f(g(x))$. Let $x$ grow by a small amount $h$. Then $g(x)$ grows by about $hg'(x)$. So $f(g(x))$ grows by about $(hg'(x))f'(g(x))=hg'(x)f'(g(x))$. Thus the growth rate of $f(g(x))$ at $x$ is $g'(x)f'(g(x))$. – André Nicolas Jan 11 '12 at 19:49

Generally speaking, yes, your intuition is in fact the standard way in which, for instance, I explain the Mean Value Theorem: if you think of $f(t)$ as the position function, and the slope of the line from $(a,f(a))$ to $(b,f(b))$ is the average velocity.

If the average velocity is $m$ units of distance per units of time, then we expect there to be some instant at which we were going at exactly $m$ units of distance per units of time: intuitively, we can't always be going slower than the average, nor always faster than the average, so "sometime between an instant where we are going slower than the average and an instant where we are going faster than the average, we should be going at exactly the average velocity."

Of course, getting all the details nailed down is somewhat complicated, and you have to take into account functions whose derivatives may not be continuous (for continuous derivatives, you can invoke the Intermediate Value Theorem). So, after a first flush of being sure this will definitely be true, it is suddenly not entirely so obvious that this will work for all functions that are differentiable everywhere on $(a,b)$ (and continuous on $[a,b]$), so there actually is something to the Mean Value Theorem.

Also related is Darboux's Theorem, which is the one that tells you that the property we want always holds, even if the derivative is not continuous: if $f(x)$ is differentiable on $(a,b)$ and continuous on $[a,b]$, then $f'(x)$ has the intermediate value property: if $f'_{+}(a)$ is the right derivative at $a$ ($\lim\limits_{h\to 0^+}\frac{f(a+h)-f(a)}{h}$) and $f'_{-}(b)$ is the left derivative at $b$ ($\lim\limits_{h\to 0^-}\frac{f(b+h)-f(b)}{h}$), and $k$ is a number between $f'_{+}(a)$ and $f'_{-}(b)$, then there is a point $t$, $a\lt t\lt b$, such that $f'(t)=k$. This picture can aid in intuition. from Cauchy's Generalized Mean Value Theorem. Required function. (S.A. pp 140 t5.3.5)

I parameterized the picture to question the Cauchy Generalized Mean Value Theorem, but you can ignore the $g(t)$ on the x-axis here. Ergo undo the parameterization by the agency of $g(t) = t$. Then $g(c)$ is veritably c, $g(d)$ is d.