# Only one $x_0$ in the Mean Value Theorem

The Mean Value Theorem says:
If $f:[a,b] \to \Bbb{R}$ is continuous in $[a,b]$ and differentiable in $(a,b)$ then there exists $x_0\in(a,b)$ such that $$f(b)-f(a)=(b-a)f'(x_0).$$ My professor asked us to describe all the functions which have only one "$x_0$" in the sense of the Mean Value Theorem. Does anybody know some text about this? I suspect (intuitively) that such functions haven't got inflection points.$^*$ Is this true? And if it is, how to derive it? :( Need some help!

$^*$An inflection point is a point where the function changes its concavity.

It isn't true that such functions haven't got inflection points. Consider $f(x) = \exp(-x^2)$ on any large-enough interval containing the origin. $f'(x) = 0$ if and only if $x = 0$, and $f$ has inflection points at $x = \pm 1/\sqrt{2}$. In general, $f'$ might not be differentiable, so we can't talk about inflection points.

Here's what we can say about functions $f$ on $[a, b]$ whose derivatives have a unique zero $c$ in $(a, b)$.

Theorem. Let $f:[a,b]\to\mathbb{R}$ be continuous on $[a, b]$ and differentiable on $(a, b)$ and suppose $f'$ vanishes exactly once in $(a, b)$. Then the curve $(x, f(x))$ is contained inside the parallelogram $P$ cut out by the following four lines:

1. $x = a$
2. $x = b$
3. the secant line from $(a, f(a))$ to $(b, f(b))$
4. the tangent line to $f$ at $c$.

Moreover, the curve $(x, f(x))$ will touch the boundary of $P$ precisely three times: at $x \in \{a, b, c\}$. If we let $g(x) = f(x) - l(x)$, where $l(x) = \frac{f(b)-f(a)}{b-a}(x-a) + f(a)$, then $g$ will be monotonic on $[a, c]$ and oppositely monotonic on $[c, b]$. (Phrased in terms of $f$ this is "$(x, f(x))$ will move toward its tangent line on $[a, c]$ and away on $[c, b]$".)

Proof. We use the beloved Rolle's theorem: if a differentiable function vanishes on the ends of an interval, then its derivative vanishes somewhere within.

If $g(x)$ is as in the statement of the theorem, then $g'$ has a unique zero in $(a, b)$ if and only if $f'$ does, so let's work with $g$.

Claim 1: $g$ has no zeroes in $(a, b)$ i.e. the curve $(x, f(x))$ only touches $P$'s secant-side in the corners. To wit, suppose $g$ vanishes at some $z$ in $(a, b)$. Then each of the intervals $(a, z)$, $(z, b)$ contains a zero of $g'$ by Rolle's theorem, contradicting uniqueness.

From the intermediate value theorem it follows that the sign of $g$ is constant on $(a, b)$. Without loss of generality (just multiply by $-1$ if need be), $g$ is positive on $(a, b)$.

Claim 2: $g$ is strictly increasing on $[a, c]$ i.e. the curve $(x, f(x))$ moves towards $P$'s tangent-side until it touches it. Suppose not. Let $a \le p < q \le c$ be such that $g(p) \ge g(q)$. We can use IVT again to push $q$ closer to $p$ and guarantee the equality $g(p) = g(q)$ while still maintaining $p < q$. By Rolle's theorem applied to $g(x) - g(p)$ (whose derivative is $g'$), $g'$ has a zero in $(p, q)$. Contradiction!

By symmetry, it follows that $g$ is strictly decreasing on $[c, b]$ i.e. $(x, f(x))$ moves away from $P$'s tangent-side after touching it. As a corollary, we have that the unique zero of $g'$ is the point at which $g$ attains its maximum: $g(c) > g(x)$ for all $x \in [a, b]\setminus\{c\}$.

Thus we have proven that the curve $(x, g(x))$ is bounded by the rectangle with four corners $(a, 0), (a, f(c)), (b, f(c)), (b, 0)$. The parallelogram version follows for $f$ by adding $l$.

Note that on its way across $P$ and back, $f$ has a lot of wiggle-room. This is better visualized with $g$. The only restriction is that $g$ never go flat. Have a look at some old, very-worn-down staircases!

We will assume that $a < b$ (since $a = b$ is not really practical/interesting). Now suppose that there exist $x_1,x_2 \in (a,b)$ such that $x_1 < x_2$ and $$f'(x_1) = \frac{f(b)-f(a)}{b-a} = f'(x_2)$$ If $f'$ is again differentiable in $(a,b)$ we can apply the Mean Value Theorem and obtain that there exists $x_0 \in (x_1,x_2)$ such that $$f''(x_0) = \frac{f'(x_2) - f'(x_1)}{x_2-x_1} = 0$$ Hence, if there exist more than one such 'mean value points', it follows that $f''$ vanishes at some point in $(a,b)$. Thus, if $f''(x) \neq 0$ for all $x \in (a,b)$, we have that there exists only one 'mean value point'.

Unfortunately, this is no complete answer to your professor's question. One needs to consider that $f'$ might not be differentiable in $(a,b)$ and there also might exist functions whose second derivative vanishes, but still has only one 'mean value point'.

You could also say the function is monotonically increasing/decreasing. the problem with arguing the case of the inflection points is that if the function has an inflection point $u$ really close to $a$ or $b$, the slope of $f(x) \in [a,u]$ might not ever reach $\frac{f(b)-f(a)}{b-a}$.

• But note that $f:[-1,1]\to \Bbb{R}, f(x)=x^2$ is not monotonically increasing/decreasing, but it has only one such "$x_0$". I need some properties that describe all such functions... – Ders Dec 2 '14 at 11:24

The equation $$\frac{f(b)-f(a)}{b-a}=f'(c)$$ will have only one solution if $f'$ is injective. This will happen if $f'$ is strictly monotone. In terms of $f$, this means that $f$ is strictly convex or concave. In particular $f$ does not have inflection points.