In my attempt to sketch functions, I usually employ the following algorithm:

  • Differentiate the function
  • Find which $x$ has $\frac{dy}{dx} = 0$
  • Find which $y$ correspond to the aforementioned $x$
  • Find its limits to infinity and negative infinity (unless the function is not defined at one of those limits, like a logarithmic function)
  • Find the nature of the function between the ends of the function found using the limit nature and the turning point, and/or between turning point to turning point until you have a satisfactory framework for the function's rate of change that can model it appropriately

However, upon trying to graph:

$$y = x\sin\ 3x$$

I had some trouble.

I knew the function would be even, so I could only really concern myself with one side and draw the same for the other side, but finding which $x$ has $\frac{dy}{dx} = 0$ was an issue. $f'(x)$ here is $sin\ (3x) + 3xcos\ (3x)$. However, trying to find turning points is proving to be difficult for me, as I have:

$$0 = sin\ (3x) + 3xcos\ (3x)$$

$$sin\ (-3x) = 3xcos\ (3x)$$

However, other than $0$, I can't seem to find any other solutions for this, allow graphing out the functions tells me there're infinitely many. How do I work something like this out? I was thinking once I got a groove I could do something like $$sin\ (-3x+2\pi\ k) = 3xcos\ (3x+2\pi\ k)$$ once I worked out a solution for $x$.

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    $\begingroup$ "Uncountable" has a technical meaning that is different from how you used it here. To avoid confusion, you should just say 'infinite'. (Believe it or not, it would be technically accurate to call this a function with a countable number of turning points.) $\endgroup$ – spaceisdarkgreen Jul 8 '17 at 6:38
  • $\begingroup$ Why is that? Looking at the graph of this function it seems hard to believe. $\endgroup$ – sangstar Jul 8 '17 at 6:38
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    $\begingroup$ $\sin3x$ is between $-1$ and $1$, so $x\sin3x$ is between $-x$ and $x$. Simply "stretch" the graph of $\sin3x$ to fit between the lines $y=-x$ and $y=x$, like so: wolframalpha.com/input/… $\endgroup$ – Rahul Jul 8 '17 at 6:40
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    $\begingroup$ Countable means you can count it, that it fits into a sequence and it can be infinite. Uncountable means more than there are positive integers. For example, there are much more real numbers than positive integers. $\endgroup$ – Ennar Jul 8 '17 at 6:40
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    $\begingroup$ Re: uncountable, see en.wikipedia.org/wiki/Countable_set and compare en.wikipedia.org/wiki/Uncountable_set. But probably the best advice for now is to understand that "uncountable" is a term that has a specific technical meaning in mathematics and it's best to avoid using it in an informal sense, just like how you should say "complicated" rather than "complex" to describe things that don't involve complex numbers. $\endgroup$ – Rahul Jul 8 '17 at 6:43

Substitute $t = 3x$ to get $$\sin t + t\cos t = 0\tag{1}$$ and note that if $t_0$ is a solution to the above, then $\cos t_0 \neq 0$. Assume the contrary, i.e. $\cos t_0 = 0$. Then we have $$0 = \sin t_0 + t_0\cos t_0 = \sin t_0,$$ but $\sin t_0^2 + \cos t_0^2 = 1$, so we can't have $\sin t_0 = \cos t_0 = 0$ and we arrive at contradiction.

Now, if $t_0$ is a solution of $(1)$, since $\cos t_0\neq 0$, we can write $$0 = \sin t_0 + t_0\cos t_0 = \cos t_0(\tan t_0 + t_0)$$ which implies that $$\tan t_0 + t_0 = 0.$$ This means that any solution of $(1)$ is also a solution of $$\tan t = - t\tag{2}$$ and conversely, any solution of $(2)$ must be a solution to $(1)$.

We conclude that equations $(1)$ and $(2)$ are equivalent, i.e. have the same set of solutions. There is a great answer how to approach this kind of problem here.

  • $\begingroup$ On a side note, there are no uncountable number of turning points, only countable. $\endgroup$ – Ennar Jul 8 '17 at 6:35
  • $\begingroup$ Can your postulate that $cos\ t_0 \ne 0$ be applied equally to $sin\ t$? If not, why is that? And why does that postulate justifying dividing by cosine? I thought this would only be logistical if the RHE of the equation was nonzero. $\endgroup$ – sangstar Jul 8 '17 at 7:03
  • $\begingroup$ my apologies, I posted my comment before it was finished accidentally and had to edit it while it looked inarticulate. Which bit do you not understand in my question? $\endgroup$ – sangstar Jul 8 '17 at 7:06
  • $\begingroup$ @sangstar, does my edit answer your question? $\endgroup$ – Ennar Jul 8 '17 at 7:13
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    $\begingroup$ It absolutely does. +1 to you, good sir. $\endgroup$ – sangstar Jul 8 '17 at 7:15

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