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Curve Sketching of a Periodic Function

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z overview
We analyze the function $ f(x)=\sin(x) + \frac{1}{3}\,\sin(3x)$.

Symmetry: Since $ \sin(x)=-\sin(-x)$, the function is odd.

Periodicity: As the sine function, the function is $ 2\pi$-periodic. Hence, it is sufficient to consider the interval $ [-\pi,\pi]$.

Points of discontinuity and Poles: The function is a composition of continuous functions, and, therefore, has no points of discontinuity or poles.

Zeros: By the addition theorem, $ \sin(3x) = 3\sin(x)-4\sin^3(x)$, and it follows that

$\displaystyle f(x) = 2 \sin(x)-\frac{4}{3}\,\sin^3(x) = \sin(x)\underbrace{\left(2-\frac{4}{3}\sin^2(x)\right)}_{\neq 0} \,. $

Hence, $ f$ vanishes at 0 and $ \pm \pi$.

Extrema: The derivative

$\displaystyle f^\prime(x) = 2 \cos(x)-4\sin^2(x)\cos(x) = -2\cos(x)+4\cos^3(x) $

vanishes if $ \cos(x) = 0$ or $ \cos(x)=\pm 1/\sqrt{2}$. Consequently, the critical points of $ f$ are

$\displaystyle x=\pm \pi/2 \,,\quad x=\pm \pi/4 \,,\quad x= \pm 3\pi/4 \,. $

Since the function is periodic and defined on $ \mathbb{R}$, we do not have to consider boundary values. Thus, the type of the critical points can be determined from the sign of the second derivative

$\displaystyle f^{\prime\prime}(x) = 2 \sin x -12 \cos^2 x \, \sin x = 2 \sin x(1-6\cos^2 x) $

and by comparing function values. This yields

\begin{displaymath} \begin{array}{c\vert c\vert c\vert l} x & f(x) & f^{\prime\... ...& \sqrt{8}/3 & -2\sqrt{2}<0 & \mbox{global maximum} \end{array}\end{displaymath}

Inflection Points: The second derivative

$\displaystyle f^{\prime\prime}(x) = 2\sin x -12\cos^2 x \,\sin x=\sin x\,(6\sin^2x-5) $

vanishes if $ \sin x =0$ or $ \sin x =\pm \sqrt{5/6}$, i.e.

$\displaystyle x=0 \quad\lor\quad x=\pm \pi \quad\lor\quad x=\pm a \quad\lor\quad x = \pm(\pi-a) $

with $ a = \operatorname{arcsin}\sqrt{5/6}\approx {\tt 1.1503}$. Since the third derivative is nonzero at these points, $ f$ has inflection points at $ (0,0)$, $ (\pm \pi,0)$, $ (\pm a,\pm b)$, and $ (\pm(\pi-a),\pm b)$ with

$\displaystyle b = f(a) = \sin a\,(2-(4/3)\sin^2 a) = \sqrt{5/6}\,(2-(4/3)(5/6)) \approx {\tt0.8114} \,, $

noting that $ \sin a = \sin(\pi-a)$ and $ f(-a)=-f(a)$.

Asymptots: Since $ f$ is periodic and not constant, it has no asymptotes.

\includegraphics[width=10.4cm]{Kurvendiskussion_1_en}

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  automatisch erstellt am 22.  6. 2016