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 $x$.
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.