find the critical point(s) of $$e ^{-x}\sin x$$ on (-pi,pi)

Question

find the critical point(s) of $$e ^{-x}\sin x$$ on         (-pi,pi)

zehanz · Answer

You differentiate and solve $f'(x)=0$.
You need the Product Rule and the Chain Rule.

Let me help you by taking the first steps:
$$f(x)=e^{-x}\sin x \Rightarrow f'(x)=e^{-x}\cdot -1\cdot \sin x +e^{-x}\cos x$$If you factor out $e^{-x}$, $f'$ looks a lot better.

Now it is time for you to solve $f'(x)=0$...

anonymous · Answer

i got that far but now i'm having trouble solving for f'(x)=0

zehanz · Answer

Solve: $e^{-x}(\cos x-\sin x)=0 \Leftrightarrow \cos x-\sin x=0$ 
(because $e^{-x}>0$ for every x.

So you can rewrite it as cos x = sin x.
What are the solutions on (-pi, pi)?

anonymous · Answer

that would be your crit point? so $$\pi/4, 3\pi/4$$

zehanz · Answer

I go along with π/4, but I think the other one is -3π/4 (see image

anonymous · Answer

If the domain is -pi,pi is that the whole unit circle?

zehanz · Answer

In your question, the domain is (-pi,pi), so that's all of the unit circle, minus the point (-1,0), but that is of no consequence here.

So these are the critical values.
The critical points are the corresponding points on the graph of f.

anonymous · Answer

but then wouldn't there be four critical points? one in each quadrant

zehanz · Answer

No, because in the second quadrant, sin x > 0 and cos x < 0, so they never are equal there.
In the fourth quadrant it is just the other way round: cos is pos, sin is neg, so not equal...

anonymous · Answer

Oh you're right lol good call

zehanz · Answer

Here is a drawing, showing the two critical points: