Properties

Definition 4.2.1  
Let $f$ be a function defined over an interval $[a,b]$. The definitions are still valid if $f$ is defined over the entire real line, but this needn't be the case.

1. $f$ attains a local minimum at a point $c$ if there is an interval around $c$ such that $f(c) \geq f(x)$ for any point in that interval.
2. $f$ attains a local maximum at a point $c$ if there is an interval around $c$ such that $f(c) \leq f(x)$ for any point in that interval.

An absolute maximum is simply the largest value a function obtains and an absolute minimum is then it smallest value. Both are usually taken over a finite interval, as over an infinite interval they will tend to be infinite rather often. Even the boringly mundane function $f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: x \mapsto x$ would have absolute maximum $+\infty$ and absolute minimum $-\infty$.

Theorem 4.2.2  
If $f$ is differentiable at a point $c$ and if $c$ is a local minimum or a local maximum for $f$, then $f'(c)=0$

Proof.
Take $c$ to be a local minimum, then $f(c+h) \geq f(c)$ for any $h$ small enough, so

$$\frac{f(c+h)-f(c)}{h}$$ (4.2.1)

is positive whenever $h$ is positive and is negative whenever $h$ is negative. This implies

$$\lim_{h \to 0^-}\frac{f(c+h)-f(c)}{h} \leq 0 \hbox{\ \ and\ \ } \lim_{h \to 0^+}\frac{f(c+h)-f(c)}{h} \geq 0$$ (4.2.2)

As $f$ is differentiable in $c$, the limit has to exists, which implies that the left and the right limit have to be equal, hence

$$f'(c)=\frac{f(c+h)-f(c)}{h}=0$$ (4.2.3)

For a local maximum the proof is pretty much the same, only the inequalities flip. $\qedsymbol$

Remark 4.2.3  
A natural question to ask would be if (4.2.2) can be reversed, in other words does the fact that $f'(c)$ is zero always imply that $c$ is a local minimum or maximum. This is not always the case. Take for instance

$$f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: x \mapsto x^3$$ (4.2.4)

Clearly, $f'(0)=0$, but on the other hand 0 is not a minimum or a maximum.

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Theorem 4.2.4 (Rolle)  
Let $f : [a,b] \to \hbox {I \hskip -5.2pt {R}}: x \mapsto f(x)$ be a differentiable function over the interval $[a,b]$ and $f(a)=f(b)$, then there exists a $c$ n the interval such that $f'(c)=0$.

Remark 4.2.5  
A typical picture for Rolle's theorem would be:

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Note that the condition $f(a)=f(b)$ is necessary. The function $f : \hbox {I \hskip -5.2pt {R}}\to \hbox {I \hskip -5.2pt {R}}: x \mapsto x$ is clearly differentiable, but the derivative $f'(x)=1$ has no zero point. The next theorem is a generalization and remedies the case where $f(a) \not= f(b)$.

Theorem 4.2.6 (Lagrange's Thm a.k.a. the Mean Value Theorem for Derivatives)  
Let $f : [a,b] \to \hbox {I \hskip -5.2pt {R}}: x \mapsto f(x)$ be a differentiable function over the interval $[a,b]$, then there exists a $c$ in the interval such that

$$f'(c) = \frac{f(b)-f(a)}{b-a}$$ (4.2.5)

Proof.
First put

$$g(x) := f(x) - \frac{f(b)-f(a)}{b-a} (x-a)$$ (4.2.6)

Note that $g$ is then again a differentiable function over $[a,b]$ with

$$g(a) := f(a) - \frac{f(b)-f(a)}{b-a} (a-a) = f(a)$$ (4.2.7)

and

$$g(b) := f(b) - \frac{f(b)-f(a)}{b-a} (b-a) = f(a)$$ (4.2.8)

So we can apply Rolle's theorem. Hence, there exists a $c$ in the interval $[a,b]$ such that $g'(c)=0$. Finally, note

$$0=g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}$$ (4.2.9)

$\qedsymbol$

Theorem 4.2.7 (de l'Hôpital's Rule)  
Suppose $f$ and $g$ are differentiable and let $\alpha$ be any point on the extended real line, i.e. $\alpha$ can be any real number or plus or minus infinity. If $g'(\alpha)\not=0$ near $\alpha$ (except possibly at $\alpha$) and either

$$\lim_{x \to \alpha}f(x)=0 \hbox{\ \ and\ \ }\lim_{x \to \alpha}g(x)=0$$ (4.2.10)

or

$$\lim_{x \to \alpha}f(x)=\pm\infty \hbox{\ \ and\ \ }\lim_{x \to \alpha}g(x)=\pm\infty$$ (4.2.11)

then

$$\lim_{x \to \alpha}\frac{f(x)}{g(x)} = \lim_{x \to \alpha}\frac{f'(x)}{g'(x)}$$ (4.2.12)

if the limit on the right side is well defined (i.e. is a real number or plus or minus infinity).

Remark 4.2.8  
Just to get a flavor for why this is true, have a look at the following picture:

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In this picture, $f(x)$ and $g(x)$ happen to have the same tangent at 0. The calculation in this situation is

$$\lim_{x \to 0} \frac{x^2+x}{\ln(x+1)} = \lim_{x \to 0} \frac{2x+1}{\frac{1}{x+1}} = \frac{1}{1} = 1$$ (4.2.13)

This little formula will often trivialize limits that are very hard to calculate:

• A standard example for $\frac{0}{0}$ is:

  $$\lim_{x \to 0}\frac{\sin(x)}{x} \overset{H}{=} \lim_{x \to 0}\frac{\cos(x)}{1} = \frac{1}{1} = 1$$ (4.2.14)

  
  

• A standard example for $\frac{\infty}{\infty}$ is:

  $$\lim_{x \to +\infty} \frac{\ln(x)}{x} \overset{H}{=} \lim_{x \to +\infty} \frac{\frac{1}{x}}{1} = \frac{0}{1} = 0$$ (4.2.15)

  
  

• Sometimes, de l'Hôpital's rule has to be applied multiple times:

  $$\lim_{x \to +\infty}x^2e^{-x} = \lim_{x \to +\infty}\frac{x^2}{e^... ...}{e^x} \overset{H}{=} \lim_{x \to +\infty}\frac{2}{e^x} = \frac{0}{+\infty} = 0$$ (4.2.16)