The Timpson Lyceum · Real Analysis (2020)

Lecture 15 (Real Analysis 2020)

Overview

This lecture develops the Mean Value Theorem (MVT) and the family of results that follow from it. We begin with the geometric idea — that the average slope of a function across an interval is achieved by the instantaneous slope at some interior point — and build up to it through two preliminary results: the interior extremum lemma (the derivative vanishes at a local max or min) and Rolle's Theorem. From the MVT we then derive several cornerstone facts of differential calculus: a function with zero derivative is constant, two functions with equal derivatives differ by a constant, a positive derivative forces strict increase, and the second-derivative test for local extrema. These are the tools that justify why the derivative tells us so much about a function's shape.

Key Concepts

Average slope (secant slope). For $f$ on $[a,b]$, the average rate of change from $a$ to $b$ is the slope of the secant line: $\frac{f(b)-f(a)}{b-a}.$

Mean Value Theorem. If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $x_0 \in (a,b)$ with $f'(x_0) = \frac{f(b)-f(a)}{b-a}.$ Geometrically, the tangent at $x_0$ is parallel to the secant joining the endpoints.

Interior extremum lemma. If $f:I\to\mathbb{R}$ is defined on an open interval, $f'(x_0)$ exists, and $x_0$ is a maximizer (or minimizer) of $f$, then $f'(x_0)=0$. The proof compares the one-sided difference quotients: from the left the quotient is $\le 0$ and from the right it is $\ge 0$ (or vice versa), so the common limit must be $0$.

Rolle's Theorem (Theorem 4.17). If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there exists $x_0 \in (a,b)$ with $f'(x_0)=0$. Proof: a continuous function on a closed bounded interval attains a max and a min. If both occur at the endpoints, $f$ is constant and $f'\equiv 0$; otherwise an extremum lies in $(a,b)$, where the interior extremum lemma gives $f'(x_0)=0$.

Constant functions (Lemma 4.19). A function is constant if there is some $c\in\mathbb{R}$ with $f(x)=c$ for every $x$ in its domain (any domain — it need not be an interval). For differentiable $f$ on an open interval $I$:

• $f$ is constant $\iff f'(x)=0$ for all $x\in I$.
• The forward direction is immediate; the converse uses the MVT — for any two points the secant slope equals some interior derivative, which is $0$, so the two function values agree.

Functions differing by a constant (Theorem 4.20). If $g,h:I\to\mathbb{R}$ are differentiable, then $g$ and $h$ differ by a constant $\iff g'(x)=h'(x)$ for all $x$. Apply Lemma 4.19 to $f=g-h$.

Sign of the derivative (Corollary 4.21). If $f$ is differentiable on $I$ and $f'(x)>0$ for all $x\in I$, then $f$ is strictly increasing. For $u<v$, the MVT gives $z\in(u,v)$ with $f'(z)=\dfrac{f(v)-f(u)}{v-u}>0$; since $v-u>0$, the numerator $f(v)-f(u)>0$.

Local max / local min. $x_0$ is a local maximizer of $f:D\to\mathbb{R}$ if there exists $\delta>0$ such that $|x-x_0|<\delta \Rightarrow f(x)\le f(x_0)$. Flip the inequality for a local minimizer. The interval must be open so the point itself isn't trivially extreme. Note: $x_0$ a local extremum (with $f'$ existing) forces $f'(x_0)=0$, but the converse fails — for $f(x)=x^3$, $f'(0)=0$ yet $0$ is neither a local max nor min.

Second-derivative test (Theorem 4.22). Let $f$ be twice differentiable and $f'(x_0)=0$:

• $f''(x_0)>0 \Rightarrow x_0$ is a local minimizer (concave up — "happy");
• $f''(x_0)<0 \Rightarrow x_0$ is a local maximizer (concave down — "sad");
• $f''(x_0)=0 \Rightarrow$ inconclusive.

Here $f''$ measures concavity: a positive second derivative means the slope is increasing (concave up), a negative one means the slope is decreasing (concave down).

Worked Example

Proof of the Mean Value Theorem (via Rolle's Theorem). Tilt $f$ so the endpoints sit at the same height, apply Rolle, then untilt.

Let $m = \dfrac{f(b)-f(a)}{b-a}$ be the secant slope and define the auxiliary function $g(x) = f(x) - mx$, which is continuous on $[a,b]$ and differentiable on $(a,b)$ (a difference of such functions). First verify the endpoints match:

so $g(a)=g(b)$. By Rolle's Theorem there exists $x_0\in(a,b)$ with $g'(x_0)=0$. Since $g'(x)=f'(x)-m$,

That interior point has instantaneous slope equal to the average slope — exactly the claim. $\blacksquare$

Summary

• MVT: for $f$ continuous on $[a,b]$ and differentiable on $(a,b)$, some interior $x_0$ has $f'(x_0)=\dfrac{f(b)-f(a)}{b-a}$ — the tangent there is parallel to the endpoint secant.
• It is built from the interior extremum lemma ($f'=0$ at an interior extremum) and Rolle's Theorem (the $f(a)=f(b)$ special case), and proved by subtracting off the secant slope to reduce to Rolle.
• Zero derivative $\Rightarrow$ constant, and equal derivatives $\Rightarrow$ differ by a constant — both direct MVT consequences.
• A strictly positive derivative throughout an interval forces strict increase.
• A point is a local extremum relative to an open neighborhood; if the derivative exists there it must be $0$, but $f'(x_0)=0$ alone is not enough (witness $x^3$ at $0$).
• The second-derivative test classifies a critical point by concavity: $f''>0$ gives a local min, $f''<0$ a local max, and $f''=0$ is inconclusive.