Calculus one: how to graph this equation

Question

How would you graph this equation $f(x) = e^{-x}\sin(x)$ over the domain of $[-\pi,\pi]$? Like what are the steps to do the math? I do not just want the answer. I want to understand the material as well. I am assuming since this is calculus one you need to find the derivative but what do you do after that? Thank you so much for your advice and expertise.

Answer 1

Considering the function $$f(x) = e^{-x}\sin(x)$$ there are a few things we can notice first $$f(-\pi)=0 \qquad f(0)=0 \qquad f(\pi)=0$$ On the other side, and this would help $$f(-x)=-\frac 1 {f(x)}$$ Now, considering $$f'(x) = e^{-x}(\cos(x)-\sin(x))$$ there is a maximum at $x=\frac \pi 4$ and a minimum at $x=-\frac {3\pi} 4$.

Let us try to evaluate $$f(-\frac {3\pi} 4)=-\frac{e^{3 \pi /4}}{\sqrt{2}}$$ We know that $e=2.718$; let us use $3$ instead. $\frac {3\pi} 4 \approx 2.36$; let us use $2$; this makes $e^{3 \pi /4}\approx 3^2=9$ and being lazy, use $\sqrt{2} \approx 1.5$. Combining all of that would give $$f(-\frac {3\pi} 4)\approx -6\implies f(\frac {\pi} 4)\approx \frac 16$$ Now, try to draw a smooth curve going through these five points.

Answer 2

As $x$ goes from $0$ to $\pi,$ $\sin x$ goes from $0$ up to $1$ and back down.

Now look at what $x\mapsto e^{-x}$ looks like. The number $e$ is more than $2,$ so every time $x$ increases by $1,$ $e^{-x}$ gets cut down to less than half what it was. So $e^{-1}\sin 1$ is less than half of $\sin1,$ and $e^{-2}\sin2$ is less than half of $e^{-1}\sin2$ which is less than half of $\sin 2$, and $e^{-3}\sin 3$ is less than half of half of half of $\sin 3.$

Likewise every time $x$ decreases by $1$ unit, as when $x$ goes from $0$ to $-1$, $e^{-x}$ gets more than twice as big as it was.

Answer 3

You probably know the graph of $y = A \sin x$: $A$ stretches/compresses the graph of $y = \sin x$ to lie between $A$ and $-A$ on the $y$-axis. The amplitude is changed from $1$ to $|A|$.

For simple functions $A(x)$, ones that don't cross the $x$-axis, like $e^{-x}$, sketch the graphs of $y = A(x)$ and $y = - A(x)$. $y = A(x) \sin x$ oscillates between the two graphs. You can think of $A(x)$ as causing a varying amplitude.

Answer 4

Start by graphing $e^{-x}$ and $-e^{-x}$ on $[-\pi,\pi]$ as best you can. Then mark the points $$(\pi/2,e^{-\pi/2}),(-\pi/2,-e^{-\pi/2}),(0,0),(\pi,0),(-\pi,0).$$ Then draw a "sinusoid" going through these points. This gives you a rough idea of what the graph of $f(x)$ will look like.