Lecture 32 - Generalized MVT

Review from last time:

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.

But in general there's a generalized version:

Theorem

If $f, g$ are continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ then $\exists c \in (a, b)$ where:
$[f (b) - f (a)] g^{'} (c) = [g (b) - g (a)] f^{'} (c)$
If $g^{'} \neq 0$ over $(a, b)$ then the conclusion can be stated as:
$\frac{f^{'} (c)}{g^{'} (c)} = \frac{f (b) - f (a)}{g (b)} - g (a)$

Proof

Apply Rolle's Theorem (Baby MVT) (or just Mean Value Theorem (MVT)) to the function $h (x) = [f (b) - f (a)] g (x) - [g (b) - g (a)] f (x)$ . This is just the linear combination of $f (x), g (x)$ . Notice that $h (a) = h (b)$ so then by Rolle's Theorem (Baby MVT) then $\exists c \in (a, b)$ such that $h^{'} (c) = 0$ . Taking the derivative gives:
$h^{'} (x) = [f (b) - f (a)] g^{'} (x) - [g (b) - g (a)] f^{'} (x)$
Thus:
$h^{'} (c) = [f (b) - f (a)] g^{'} (c) - [g (b) - g (a)] f^{'} (x)$

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Application

Remember in Calculus the following limit problem:
$lim_{x \to 1} \frac{\ln (x)}{x - 1}$
(we know we haven't formally constructed $\ln$ but just roll with it). Normally we'd use L'Hopital's Rule(s). We'd say:
$\overset{L^{'} H}{=} lim_{x \to 1} \frac{\frac{1}{x}}{1} = 1$
See the L'Hopital's Rule(s) for more on where this comes into play.

The idea here is that $x = f (t), y = g (t)$ gives some parametric curve:

That the line connecting two points $(f (a), g (a))$ and $(f (b), g (b))$ exists with the same slope somewhere else on the line. This allows for vertical slopes, by the way. Notice that we could generalize this to finitely many functions $f_{1}, f_{2}, \dots, f_{n}$ .

L'Hopital's Rule:

\frac{0}{0}

case

Let $f, g$ be continuous on an interval containing $a$ and suppose $f, g$ are differentiable on this interval with the possible exception of the point $a$ . If $lim_{x \to a} f (x) = lim_{x \to a} g (x) = 0$ and $g^{'} (x) \neq 0$ for all $x \neq a$ , then:
$lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = L \in R \Rightarrow lim_{x \to a} \frac{f (x)}{g (x)} = L$

Proof

Apply the Generalized Mean Value Theorem via the following method. For argument let $(b, c)$ be our interval.

Consider the case that $a = b$ . Let $ε > 0$ . Supposing $lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = L$ , choose $δ > 0$ such that $0 < | x - b | < δ$ implies that $| \frac{f^{'} (x)}{g^{'} (x)} - L | < \frac{ε}{2}$ . Let $x \in (b, b + δ)$ . Let $y \in (b, x)$ . By the Generalized Mean Value Theorem, then $\exists t \in (y, x)$ such that:
$f^{'} (t) (g (x) - g (y)) = g^{'} (t) (f (x) - f (y))$
and because $g (x) \neq g (y)$ and $g^{'} (t) \neq 0$ :
$\frac{f^{'} (t)}{g^{'} (t)} = \frac{f (x) - f (y)}{g (x) - g (y)}$
Now since $b < t < b + δ$ then our implication we started with gives:
$L - \frac{ε}{2} < \frac{f (x) - f (y)}{g (x) - g (y)} < L + \frac{ε}{2}$
for all $y \in (a, x)$ . Let $y \to b^{+}$ then by the Algebraic Limit Theorem for Functional Limits and Limits and Order (Order Limit Theorem) then taking a limit and using $lim_{x \to a} f (x) = 0 = lim_{x \to a} g (x)$ then:
$L - ε < \frac{f (x)}{g (x)} < L + ε$
Doing the cases for $a = c$ and $c \neq a, b$ are done very similarly.

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L'Hopital's Rule:

\frac{\infty}{\infty}

case

Suppose $f, g$ are differentiable on $(a, b)$ and that $g^{'} (x) \neq 0$ for all $x \in (a, b)$ . If $lim_{x \to a} g (x) = \infty$ (defined here) (or $- \infty$ ), then:
$lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = L \Rightarrow lim_{x \to a} \frac{f (x)}{g (x)} = L$

Notice that we only need the denominator to go to

\infty

in this case. This is different than the

\frac{0}{0}

case.

Proof

Let $ε > 0$ . Because $lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)} = L$ then $\exists δ_{1} > 0$ such that:
$| \frac{f^{'} (x)}{g^{'} (x)} - L | < \frac{ε}{2}$
for all $a < x < a + δ_{1}$ . For convenience of notation, let $t = a + δ_{1}$ and note that $t$ is fixed for the remainder of the argument.

Our functions are not defined at $a$ , but for any $x \in (a, t)$ we can apply the Generalized Mean Value Theorem on the interval $[x, t]$ to get:
$\frac{f (x) - f (t)}{g (x) - g (t)} = \frac{f^{'} (c)}{g^{'} (c)}$
for some $c \in (x, t)$ . Our choice of $t$ then implies that:
$L - \frac{ε}{2} < \frac{f (x) - f (t)}{g (x) - g (t)} < L + \frac{ε}{2}$
for all $x \in (a, t)$ .

In an effort to isolate the fraction $\frac{f (x)}{g (x)}$ , the strategy is to multiply the inequality by $\frac{g (x) - g (t)}{g (x)}$ . We need to be sure, however, that this quantity is positive, which amounts to insisting that $1 \geq \frac{g (t)}{g (x)}$ . Now, because $t$ is fixed and $lim_{x \to a} g (x) = \infty$ then we can choose $δ_{2} > 0$ so that $g (x) \geq g (t)$ for all $a < x < a + δ_{2}$ (because $t$ is fixed, so eventually we pass $g (t)$ as $g (x) \to \infty$ ). Carrying out the desired multiplication results in:
$(L - \frac{ε}{2}) (1 - \frac{g (t)}{g (x)}) < \frac{f (x) - f (t)}{g (x)} < (L + \frac{ε}{2}) (1 - \frac{g (t)}{g (x)})$
which after some manipulation becomes:
$L - \frac{ε}{2} + \frac{- L g (t) + \frac{ε}{2} g (t) + f (t)}{g (x)} < \frac{f (x)}{g (x)} < L + \frac{ε}{2} + \frac{- L g (t) - \frac{ε}{2} g (t) + f (t)}{g (x)}$
Notice the $g(x)

Definition

Given $g : A \to R$ and a Limit Point $c$ of $A$ , we say that $lim_{x \to c} g (x) = \infty$ if, $\forall M > 0$ then $\exists δ > 0$ such that when $0 < | x - c | < δ$ then $g (x) \geq M$ .
We define $lim_{x \to c} g (x) = - \infty$ in a similar way by considering only $M < 0$ instead.

We use this to prove the $\infty$ case of L'Hopital's Rule(s).s in the denominator. Again, because $t$ is fixed then $lim_{x \to a} g (x) = \infty$ . Thus, we can choose a $δ_{3}$ such that $x \in (a, a + δ_{3})$ implies that $g (x)$ is large enough to ensure that the terms:
$\frac{- L g (t) + \frac{ε}{2} g (t) + f (t)}{g (x)}, \frac{- L g (t) - \frac{ε}{2} g (t) + f (t)}{g (x)}$
are less than $\frac{ε}{2}$ in absolute value (because the numerators are fixed, and $g (x) \to \infty$ ), so using that in our original inequality and choosing $δ = min {δ_{1}, δ_{2}, δ_{3}}$ guarantees that:
$L - \frac{ε}{2} + \frac{ε}{2} < \frac{f (x)}{g (x)} < L + \frac{ε}{2} + \frac{ε}{2}$
Thus:
$| \frac{f (x)}{g (x)} - L | < ε$
for all $a < x < a + δ$ .

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!Infinity

We use this to prove the $\infty$ case of L'Hopital's Rule(s).