Lately, I have been disappointed that my students don’t find the Mean Value Theorem as wonderful as I do. After all, they claim, it “makes sense”. However, it is one of the most important theorems in calculus, used to prove many of the bigger names like Fubini’s Theorem, L’Hopital’s Rule, and the Fundamental Theorem of Calculus. In addition, it is far from obvious – I doubt any of them could prove it on their own. I needed a way to reinvigorate the topic.

So, this semester, I began by giving my students a short “Warmup Quiz” with the following True/False questions:

(1) If $f'(x)=0$ for all $x$ in the domain of $f(x)$, then $f(x)$ is a constant function.

(2) If $f(x)$ is a function on the interval $[a,b]$, then there is a value $c\in [a,b]$ such that $\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}$.

(3) If $f'(x)<g'(x)$ for all $x$, then $f(x)<g(x)$.

(4) If $f'(x)>0$ for all $x$ in the domain of $f'(x)$, then $f(x)$ is an increasing function.

(5) If $f(x)$ is a function on $[a,b]$ and $f(a)=f(b)$, then there is some point $c\in [a,b]$ at which $f'(c)=0$.

(6) If $f(x)$ is an increasing function, then $f'(x)>0$ for all $x$ in the domain of $f(x)$.

(7) If $f(x)$ is a function on an interval $[a,b]$, then there is a maximum value of $f(x)$ for $x\in [a,b]$.

As soon as they finished, I asked how it went, and the students were confident. And then we went over it together…

It turns out that every statement above was false:

(1) Consider the function defined as the greatest integer less than $x$ whose domain is all non-integers. This function is non-constant, but every point on its graph is a stationary point.

(2) Consider the function $f(x)=|x|$ on the interval $[-1,1]$. There is no point on $f(x)$ at which the slope is $0$.

(3) Let $f(x)=x$ and $g(x)=2x$, so that $f(x)>g(x)$ when $x<0$.

(4) Consider the function $f(x)=-1/x$. Then $f'(x)=1/x^2>0$ for all $x$ in the domain of $f(x)$. However, $f(-1)=1$ but $f(1)=-1$, so $f(x)$ is not increasing.

(5) Use the same example as in (2).

(6) Let $f(x)=x^3$. Then $f(x)$ is increasing, yet $f'(0)=0$.

(7) Let $f(x)=\begin{cases}1/x & x>0\\0 & x=0\end{cases}$. Since $\lim_{x\to 0^+} f(x)=\infty$, there is no maximum value of $f(x)$ on the interval, say, $[0,1]$.

These True/False questions highlighted to the students just how wrong their intuition could be. At that point, I had their full attention as I stated and proved Mean Value Theorem, as well as looking at examples and corollaries the following day. The key point to emphasize is that the hypothesis of MVT is not that the function in question is differentiable, it is that the function is *differentiable on a closed interval.* That difference, while subtle, is what rules out the counterexamples I came up with, and therefore allows MVT to work.