Other applications of the (first) derivative.

A consequence of Lagrange's theorem (in its second form 8.5) is the following well-known theorem:

Theorem 6.9.1   Let $ f$ be a function defined and differentiable on an interval $ I$ .
  1. If $ \forall x \in I$ , $ f'(x)=0$ , then $ f$ is constant on $ I$ .
  2. If $ \forall x \in I$ , $ f'(x)>0$ , then $ f$ is strictly increasing on $ I$ .
  3. If $ \forall x \in I$ , $ f'(x) \geq 0$ , then $ f$ is increasing on $ I$ .
  4. If $ \forall x \in I$ , $ f'(x)<0$ , then $ f$ is strictly decreasing on $ I$ .
  5. If $ \forall x \in I$ , $ f'(x) \leq 0$ , then $ f$ is decreasing on $ I$ .

Example 6.9.2   Take $ f(x)=x^3-3x$ ; as $ f$ is a polynomial function, it is differentiable on $ \mathbb{R}$ .

$ \forall x \in \mathbb{R}, \quad f'(x)=3x^2-3=3(x^2-1)=3(x-1)(x+1)$ .

$\displaystyle f'(x)=0$ $\displaystyle \Longleftrightarrow x= \pm 1$    
$\displaystyle f'(x)>0$ $\displaystyle \Longleftrightarrow x<-1$    or $\displaystyle x>1$    
$\displaystyle f'(x)=0$ $\displaystyle \Longleftrightarrow -1<x<1$    

Thus, $ f$ is strictly increasing on $ (- \infty , -1)$ and on $ (1,+\infty)$ and is strictly decreasing on $ (-1,1)$ . It has a maximum at 1 and a minimum at 1 (v.s. Def. 5.6).

Theorem 6.9.3   If $ f$ is differentiable at $ x_0$ and has an extremum at $ x_0$ , then $ f'(x_0) =0$ .

\begin{figure}\mbox{\subfigure{\epsfig{file=error.eps, height=1cm}} \qquad
The converse is not true!}
\end{figure}
For example, take $ f(x)=x^3$ . Then $ f$ has no extremum, but $ f'(0)=0$ . We will se later that such a point is called an inflection point (see Definition 10.8 and Figure8).

Figure 8: An inflection point.
\begin{figure}\mbox{\epsfig{file=inflection1.eps,height=4cm}}\end{figure}

Noah Dana-Picard 2007-12-28