# How to graph $x\sin(x)$

Graph $x\sin(x)$

I am having problem with the period section of the graph. I can see that the function is even, and that as $x \rightarrow \pm\infty$, the amplitude $\rightarrow \infty$.

I am used to dividing $2\pi$ by the coefficient of the angle inside the $\sin$ term; however, I have notes saying that the function is not periodic. Why can't I say that the function is periodic with period $\frac{2\pi}{1} = 2\pi$? Is it the $x$ term attached? How does this term affect things and how do I solve these types of problems generally?

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I would draw the lines $y=x$ and $y=-x$ as guides. Note that at $x=0$, $\pm \pi$, $\pm 2\pi$, and so on, our function is $0$. At $x=\pi/2$, $2\pi+\pi/2$, $4\pi+\pi/2$, and so on, our function has value $x$, and at $3\pi/2$, $2\pi+3\pi/2$, and so on our function has value $-x$. that should help. It is not periodic because there is no simple repetition. – André Nicolas May 17 '12 at 2:07
I think it's useful to think of functions like $f(x)\sin(x)$ where $f(x)$ acts like an envelope or a varying amplitude, and draw $f(x)$ as Anrdé recommends. For the periodic part, $g(x)=x\sin(x)$ must satisfy $g(x+T)=g(x)$. It's easy to see this is not the case for any $T\neq0$. – azdahak May 17 '12 at 3:12

Expanding on the comment by @Andre, graphs of such nature are found (physically) in applications like Vibrations and Dynamics. Think of $\sin(x)$ as a graph exhibiting the Oscillatory motion of a particle. The usual form of such equations is $A\sin(\omega t + \alpha) \text{ where }\omega t = \theta$, A is amplitude and $\alpha$ is phase lag. We have $A\sin(\theta)$.
For an unstable particle without damping, the amplitude goes on increasing with time. So, as $t$ increases, $\omega t$ increases, $\theta$ increases and also, A increases. You have the graph for $x \sin(x)$ which looks like:.