Lecture 31 - Mean Value Theorem

Recall the definition of the derivative:

derivative

Let $f : I \to R$ where $I$ is an interval (open, closed, half-open/closed), which is definitely connected. Then $f$ is differentiable at $c \in I$ if:
$lim_{x \to c} \frac{f (x) - f (c)}{x - c}$
exists. (see Functional Limit for details, and highlighting Existence of Functional Limits for showing this).

Note that:

If $f$ is differentiable at $c$ , then we define $f^{'} (c)$ to refer the value of the limit:
$lim_{x \to c} \frac{f (x) - f (c)}{x - c} = f^{'} (c)$
specifically we call this the derivative of $f$ at $c$ .
Similar to One-Sided Limits, if $c \in I$ is an endpoint of $I$ (ex: a closed interval), then this limit coincides with a one-sided limit (which we refer as a one-sided derivative):
$lim_{x \to c^{-}} \frac{f (x) - f (c)}{x - c}, lim_{x \to c^{+}} \frac{f (x) - f (c)}{x - c}$
If $f$ is differentiable at every $c \in I$ then we say $f$ is differentiable on $I$ .
$\frac{f (x) - f (c)}{x - c}$ is the slope of the secant line. Taking its limit gets the tangent line at a point.

Using this we can prove some more interesting theorems:

Darboux's Theorem

Let $f$ be differentiable on a closed interval $I = [a, b]$ and let $α \in R$ satisfy either that:

$f^{'} (a) < α < f^{'} (b)$ , or
$f^{'} (a) > α > f^{'} (b)$
Then $\exists c \in (a, b)$ such that $f^{'} (c) = α$ .

This uses the Derivative definition so make sure to know that beforehand.

Proof

The trick here is to use the Interior Extremum Theorem, over and over again. We'll have issue using it though because it works for an open interval, and not a closed interval like the Extreme Value Theorem does.

Let $g (x) = f (x) - α x$ . By the Algebraic Differentiability Theorem then $g$ is differentiable on $[a, b]$ and $\forall x \in [a, b]$ then:
$g^{'} (x) = f^{'} (x) - α$
Now for the first case, suppose that $f^{'} (a) < α < f^{'} (b)$ . Then:
$g^{'} (a) = f^{'} (a) - α < 0$
$g^{'} (b) = f^{'} (b) - α > 0$
So then $g^{'} (a) < 0 < g^{'} (b)$ . Since $g$ is differentiable on $[a, b]$ , then it is continuous on that interval. Using the Extreme Value Theorem then $g$ has an absolute minimum on $[a, b]$ , denote it $c$ .

We want $c \in (a, b)$ so it goes to show that $c \neq a, b$ . Notice that at $a$ :
$0 > g^{'} (a) = lim_{x \to a^{+}} \frac{g (x) - g (a)}{x - a}$
Now notice that $x - a > 0$ for this limit, and thus that suggests that $g (x) - g (a) < 0$ by the Limits and Order (Order Limit Theorem). So then $\exists x_{0} > a$ such that $g (x_{0}) - g (a) < 0$ , so then we found a smaller value than $g (a)$ ! This contradicts $a$ being our supposed minimum.

Similarly for $b$ :
$0 < g^{'} (b) = lim_{x \to b^{-}} \frac{g (x) - g (b)}{x - b}$
Now $x - b < 0$ and thus by the Limits and Order (Order Limit Theorem), then $\exists x_{1} < b$ such that $g (x_{1}) - g (b) < 0$ so then $b$ cannot be a minimum!

Thus $c \in (a, b)$ , and by the Interior Extremum Theorem then $g^{'} (c) = 0$ . Correspondingly then $f^{'} (c) = α$ .

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This looks oddly similar to the Mean-Value-Theorem no? Well it turns out this will help us out. But first, let's do a slightly easier theorem:

Rolle's Theorem

Let $f : [a, b] \to R$ be continuous on $[a, b]$ be differentiable at least on $(a, b)$ . If $f (a) = f (b)$ then $\exists c \in (a, b)$ such that $f^{'} (c) = 0$ .

All this comes down to is applying the Interior Extremum Theorem again.

Proof

Let $f$ be continuous on $[a, b]$ . Then by the Extreme Value Theorem then $f$ has an absolute maximum and an absolute minimum. Now if either of those are in-between $c \in (a, b)$ then apply the Interior Extremum Theorem then $f^{'} (c) = 0$ .

Now what if $c = a, b$ ? If that's the case then $f$ is constant (because the max and the min are at the same point $c$ , so the function cannot increase nor decrease at all) on $[a, b]$ which has a zero-derivative everywhere (just pick one).

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The Mean-Value Theorem will come right out:

Mean value Theorem

Let $f : [a, b] \to R$ be continuous on $[a, b]$ and differentiable on at least on $(a, b)$ . Then $\exists c \in (a, b)$ such that $f^{'} (c) = \frac{f (b) - f (a)}{b - a}$ .

Proof

Create the function $h (x)$ defined as:
$h (x) = f (a) + \frac{f (b) - f (a)}{b - a} (x - a)$
(you can think of $h$ as the secant line between $a, b$ ). Notice that $h$ is differentiable by Algebraic Differentiability Theorem, where:

$h (a) = f (a)$
$h (b) = f (b)$
$h^{'} (x) = \frac{f (b) - f (a)}{b - a}$ for all $x \in [a, b]$ .

Then for $x \in [a, b]$ define:
$g (x) = f (x) - h (x)$
Then $g (a) = 0 = g (b)$ so by Rolle's Theorem (Baby MVT) then $\exists c \in (a, b)$ such that $g^{'} (c) = 0$ , namely:
$g^{'} (c) = 0 = f^{'} (c) - h^{'} (c) = f^{'} (c) - \frac{f (b) - f (a)}{b - a} \Rightarrow f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
as desired.

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What's cool here is that for:
$f^{'} (c) = \frac{f (b) - f (a)}{b - a}$
on the RHS we only have a constant, but on the left hand side we have information about $f^{'} (c)$ . Thinking, we could try different $a, b$ values to learn more information about $f^{'}$ at points within those intervals (however, you don't know the exact value of $c$ though).

Let's apply the MVT!

Examples

Let $f$ be differentiable on the interval $I$ . If $f^{'} (x) = 0$ for all $x \in I$ then $x$ is constant. We use this idea of anti-derivation all the time, but using the Mean Value Theorem (MVT) we get that for free.

Proof

Take any $x_{1}, x_{2} \in I$ and WLOG $x_{1} < x_{2}$ . Now since $f$ is differentiable on $I$ then it satisfies the properties for Mean Value Theorem (MVT). As such then $\exists c \in [x_{1}, x_{2}]$ such that:
$f^{'} (c) = \frac{f (x_{2}) - f (x_{1})}{x_{2} - x_{1}}$
But we know that $f^{'} (c) = 0$ (per our given) so then:
$f (x_{2}) - f (x_{1}) = 0 \Rightarrow f (x_{2}) = f (x_{1})$
so since $x_{1}, x_{2}$ were arbitrary then all points of the function must be equal, so $f$ is constant.

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See this again in Derivative is Zero Implies Constant Function.