Question

if f(x)=e^( sin x), how many zeros does f'(x) have on the interval [ 0, 2pi] ?

I used chain rule to find the first derivative and set it equal to zero, but I'm not sure where to go from there.

Help ..please!

Answer 1

Well, if you got f' and set it to 0, you have a periodic answer. You need to look at that solution in the domain that was supplied of 0 to 2pi.

Answer 2

Dude your く isn't showing!

Answer 3

From what you said, you have \[f(x)=e^{\sin x} \rightarrow f'(x)=(\cos x )e^{\sin x}\] Well, that means it is 0 when either factor is 0. So on [0, 2pi] when is either cos x or e^(sin x) 0?

Answer 4

@wio Yah, I need to re-edit so the number does not kill it. Nihon go ga wakarimasuka

Answer 5

少し分かりますが、単語が苦手です。

Answer 6

Basically, since \(e^{g(x)}=0\) is never the case, you only need to look at \(g'(x)\) in this case.

Answer 7

Yah, which makes it simpler because he only has cos x to deal with.

漢字は難しいでしょ

Answer 8

Oh! ok, got it. so then, there's two zeros in the interval.

Answer 9

If that is inclusive, rather than [0, 2pi), then...

Answer 10

ひらがなをつかってほしかったら、そうしますが、漢字ができるのでぼくはすきです。

Answer 11

Actually, yah. I was thinking sin x for a sec. cos x will only be when you are at the top and the bottom. But it does point out where you need to be careful.

Answer 12

Thanks.

Answer 13

@ Wio ..I don't actually understand what you are saying.

Answer 14

He and I are speaking in Japanese, completely unrelated to your problem. LOL

Answer 15

I figured it was something like that.

Answer 16

lol

Answer 17

He liked my icon, which is Jaopanese for Mathematics, and we have traded a bit on how it can be a challenege to learn, or basically that.

Answer 18

And half of that is subtext rather than text because you would only comment on the ku being missing if you knew it and liked it.