Let $x=e^{-3t}(3 \cos(4t)+\frac{9}{4} \sin(4t)), t\geq 0$. Prove that $\lvert x \rvert \leq 3$.
Now, using a graphical approach, the above result is clear. However, is there a way to prove the above upper bound using algebraic methods? I tried doing the obvious, like finding the first derivative, but this method is not sufficient since it only proves that $t=0$ yields a $\textbf{local maximum}$, and this function is not monotone decreasing over the given interval.
On
$$x(t)=e^{-3t}[3 \cos 4t-9\frac{\sin (4t)}{4}, t\ge 0$$ $$\frac{dx}{dt}==\frac{75}{4}e^{-3t} \sin (4t), `~x''(t)=-\frac{75}{4}e^{-3t}[4 \cos 4z-3 \sin 4z]$$ When $4t=n\pi, n=0,1,2,3..$ there will be max of min. As $x''(0)<0$ there will be a max at $x=0$. Due to exponential fall-off of $x(t)$ other max/min will be of lesser absolute height. So $$x(t) \le x(0)=3.$$ Next at $t=\pi/4$ there will be the first min as $x'(\pi/4)=0, x''(\pi/4)=75e^{-3\pi/4}$ and $x_{min}=x(\pi/4)=-3 \le 3e^{-3\pi/4}$ So, $x(t)$ will lie between these the first max and the first min, i.r. $$-3e^{=3\pi/4} \le x(t) \le 3 \implies |x(t)|\le 3.$$
One may also write $$x(t)=\frac{\sqrt{15}}{4} e^{-3t} \sin [4t+\tan^{-1}(4/3)] \implies |x(t)|=\frac{15}{4} e^{-3t} |\sin [4t+\tan^{-1}(4/3)]$ \implies x(0)=\frac{15}{4}e^{-3t} \sin (\tan^{-1}(4/3)=3.$$ Therefore $$|x(t)| \le 3$$
Notice that $x(0)=3+9/4>3$ so the statement is not accurate. But since $e^{-t}$ decreases faster than any polynomial, let alone a bounded function, then there exists $T>0$ such that $|x(t)|<1$ for all $t>T$.