A trigonometric polynomial.

$ f(x)= \sin x + \frac 12 \cos 2x$ .

This function is defined on $ \mathbb{R}$ and is periodic with period equal to $ 2\pi$ ; therefore we can study it on $ [0,2 \pi ]$ .

Because of the periodicity, the function $ f$ has no limit at $ \pm \infty$ . Moreover, there is no other open endpoint of the domain and the function is continuous on $ \mathbb{R}$ , therefore we have no limit to compute.

The function is given by a trigonometric polynomial, thus it is differentiable on $ \mathbb{R}$ , whence on $ [0,2 \pi ]$ .

$\displaystyle \forall x \in [0,2 \pi ], \; f'(x)=\cos x - \sin x.$    

Where does the first derivative vanish?

$\displaystyle f'(x)=0 \Longleftrightarrow \cos x = \sin x \Longleftrightarrow x=\frac {\pi}{4} + 2k \pi, \; k \in \mathbb{Z}.$    

Figure: The graph of $ f(x)= \sin x + \frac 12 \cos 2x$ .
\begin{figure}\mbox{\epsfig{file=graph6.eps,height=6cm}}\end{figure}

Noah Dana-Picard 2007-12-28