Two global theorems.

Theorem 6.8.1 (Rolle's Theorem)   Let $ f$ be a function, continuous on $ [a,b]$ and differentiable on $ (a,b)$ . If $ f(a)=f(b)$ , then there exists at least one point $ c \in (a,b)$ such that $ f'(c)=0$ .

Figure 6: Global Theorems.
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\subfigure[Rolle's theorem]{\ep...
...ure[Lagrange's theorem]{\epsfig{file=Lagrange1.eps,height=4cm}}
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This means that, at least at one point,the graph has a tangent parallel to the $ x-$ axis (see Figure 6(a)).

Example 6.8.2 (see Figure 7)   Take $ f(x)=\sqrt{1-x^2}$ . Then $ f'(x)=\frac {-2x}{2\sqrt{1-x^2}}=\frac {-x}{\sqrt{1-x^2}}$ .

$ f'(x)=0 \Longleftrightarrow x=0$ .

Figure 7: Rolle's Theorem.
\begin{figure}\mbox{\epsfig{file=Rolle1.eps,height=4cm}}\end{figure}

Theorem 6.8.3 (Lagrange's Theorem; first form)   Let $ f$ be a function, continuous on $ [a,b]$ and differentiable on $ (a,b)$ . There exists at least one point $ c \in (a,b)$ such that $ f'(c)= \frac {f(b)-f(a)}{b-a}$ .

This means that, at least at one point,the graph has a tangent parallel to the chord $ AB$ (see Figure 6(b)).

Example 6.8.4   Take $ f(x)=x^2$ , with $ x \in [0,2]$ .

We have $ f'(x)=2x$ and $ \frac {f(2)-f(0)}{2-0}=\frac 42 = 2$ .

$ f'(x)=2 \Longleftrightarrow 2x=2 \Longleftrightarrow x=1$ .

Theorem 6.8.5 (Lagrange's Theorem; second form)   Let $ f$ be a function differentiable on a neighborhood of $ x_0$ . Then for any real number $ h$ , there exists $ \theta
\in (0,1)$ such that $ f(x)= f(x_0)+f'(x_0+\theta h)$ .

Noah Dana-Picard 2007-12-28