Explain why graph of $f(x)=(x+1)\sin(3x)$ lies below the $x$-axis in interval $[4\pi/9,5\pi/9]$ $f(x)=(x+1)\sin(3x)$
Explain why the graph of $f$ lies below the $x$-axis for values of $x$ in the interval $[4\pi/9, 5\pi/9]$.
From what I know/understand, I'd have to look at the function in two parts: $y=x+1$ and $y=\sin(3x)$ and then construct a table of signs to figure out what's happening to the whole function.
For $y=x+1$ the function is increasing on its whole domain and the $x$ intercept of its graph is $x=-1$.
For $y=\sin(3x)$ I come a little undone. I don't know how to find the $x$ intercept to be able to have a value for my table of signs. What alternative method can I use? 
 A: first find the critical points($x$ at which $f(x)= 0$ or undefined). luckily your function $f$ is defined everywhere. because $f$ is already factored, finding the zeros are made easier. the critical numbers are given by $$(x+1)\sin 3x = 0 \to x = -1 \text{ or} \sin 3x = 0 \to x = -1, 3x = 0, \pi, 2\pi, 3\pi, \cdots $$ dividing by $3,$ the critical numbers of $f$ are
$$\cdots, -2\pi/3, -1, -\pi/3, 0, \pi/3, 2\pi/3, \cdots $$
now pick a test point in each these intervals and the sign of $f$ at that point should determine the sign for all points in the interval.
A: Hopefully you know the period of $sin(x)$. When a factor greater than $1$ (or less than $-1$) is placed inside the $sin$ function and next to the $x$, it shortens the period by that factor. One way to think of it is that the factor is "accelerating" the $x$ values so the whole function completes each cycle "more quickly". So the period for $y=sin(3x)$ is $1/3$ the size of the period of $y=sin(x)$.
You can graph $y=sin(x)$, and just relabel all the $x$ values where it crosses the $x$ axis with $1/3$ their normal value. So instead of crossing at $\pi$, it would cross at $\pi /3$ (for example). That should make it easier to chart the positive and negative values of $y=sin(3x)$.
