Which value is the *mean value* in the mean value theorem: the argument, the function, or the derivative?

Question

After looking over a few statements and proofs of the real number mean value theorem of single variable calculus of single valued functions, I realized that it is not clear to me what is actually meant by mean value. I understand the proofs. I just don't know exactly what the name of the theorem means.

For the case of $f\left[x\right]$ continuous on $x\in\left[x_0,x_1\right]$ and differentiable in $x\in\left(x_0,x_1\right),$ and $c\in\left(x_0,x_1\right)$ such that

$$f^\prime\left[c\right]=\frac{f\left[x_1\right]-f\left[x_0\right]}{x_1-x_0},$$

there are three possible values which might be designated the mean value. Either $c$, $f\left[c\right]$ or $f^\prime\left[c\right].$ Which one is it? Are there good names for the other values in this context?

Answer 1

The mean value would be $\,f'(c)\,$.

This makes a lot more sense if you consider the integral form of MVT $\,f(c) = \frac {1}{b-a}\int _{a}^{b}f(x)\,dx\,$, where the RHS is indeed the "mean value" of $\,f(x)\,$ over $\,[a,b]\,$.

Answer 2

It would seem to me that $f'(c)$ is the 'mean value' in the theorem.

With some hand-waving, the "average" gradient of the graph of $f$ taken over the interval $(x_0,x_1)$ is $\frac{f\left(x_1\right)-f\left(x_0\right)}{x_1-x_0}$. Whilst the derivative $f'(c)$ may not always equal this value, it has to at some point, because that's the 'mean'.

I don't think there are particular names for the values of $c$ or $f(c)$.

Answer 3

In my opinion, the mean value denotes the mean value of $f$ on the interval $[c_0,c_1]$, i.e. the value of the fraction $$\frac{f(x_1)-f([x_0)}{x_1-x_0},$$ which is the equivalent of the average speed, if $x$ is time, and $f(x)$ is position at a given time.

In terms of kinematics, the Mean value theorem asserts that the average speed on a time interval is equal to the instantaneous speed at some time of this time interval, whence the name of the theorem.