I just noticed today that the graph of $x \sin x$ is like a $\sin x$ trapped in $x$ and $-x$. Upon this realisation, I tried to plot some graphs by hand, others by desmos. I tried to investigate this property according to which always $\sin x$ would be trapped inside the $+f(x)$ and $-f(x)$ for a function $g(x)=f(x) \sin x$ and its shape would change in order to fit the function at varying x coordinates. But, rather than doing induction I wanted to prove that this type of property will always be valid.
I defined a function;
$$g(x) = f(x) \sin x$$
$$-f(x)\le g(x) \le f(x)$$
$$-1 \le \sin x\le 1$$
Now we can argue that $f(x)$ will act like a varying amplitude for $\sin x$ wave(/graph) and thus it should be trapped. But this is not satisfactory enough.
Thus my question is, “Is there a rigorous proof for this sort of property?”
Following are the graphs I tried to analyse the property off of:
1. $x \cdot \sin x$
2. $x^2 \sin x$
3. $x^3 \sin x$
4. $\frac{\sin x}{x^2+1}$
5. $\ln x \cdot \sin x$
6. $(3x^2-2x^3) \sin x$
7. $((1-x^{\frac{2}3})^{\frac{3}2}) \sin x$
8. $\sqrt{(x-1)(x-2)(x-3)} \cdot \sin x $
9. $x \sqrt{\frac{x+5}{x-5}} \cdot \sin x $
Notice that $\sin(x)$ is bounded, indeed $|\sin(x)|\leq 1$. This gives us: $$|f(x)\sin(x)|\leq|f(x)|\cdot|\sin(x)|\leq|f(x)|\cdot 1=|f(x)|.$$ In other words, an element of $g(x)$ can never go above the graph of $f(x)$ or below the graph of $-f(x)$. Can you try to generalize this for a function $h(x)$ such that $|h(x)|\leq n$ for $n$ a natural number? What would happen to the graph of $h(x)f(x)$?