Question

try this again I am looking for the critical points on the Eq. (e^x + e^-x)/2

Answer 1

\[\Large \frac{d}{dx}\frac{1}{2}\left(e^x+e^{-x}\right)\quad=\quad?\]Have you tried taking the derivative yet? :)

Answer 2

That is where I am stuck. it seems that I get e^x/e^x? =1?

Answer 3

Mmmm I'm not quite sure what you did there...

Answer 4

\[\Large \frac{d}{dx}\frac{1}{2}\left(e^x+e^{-x}\right)\quad=\quad\frac{1}{2}\left(e^x-e^{-x}\right)\]So this is the derivative, yes?

Answer 5

no, the derivative is simply 0 because e^x/e^x? =1 and the derivative of 1 is 0.

Answer 6

The D' of e^x is x so I figured that e^-x = 1/e^x
?
ok I can see that, but how do you solve for x. Zimmah I thin that is what I wanted but is that a critical point?

Answer 7

Irdihawk, I remember you posting this question before and having tried to use the `quotient rule` on it.

You `can` use the quotient rule, but it's just a bunch of extra work taking that path.

Answer 8

\[\Large e^x+e^{-x}\quad\ne\quad \frac{e^x}{e^x}\]Is this what you were trying to do?

Answer 9

yes I went back and re read about e^x that is another but wrong choice, I can see that

Answer 10

Actually i was wrong, the question was not e^x*e^-x but e^x+e^-x so it's different.

Answer 11

hawk you seem kinda confused :c lemme just list the steps and see if this helps or not.

Answer 12

t seems that Ln is rule that needs be used. because e^x D/dx =e^x?

Answer 13

Taking the derivative gives us:\[\Large \frac{d}{dx}\frac{1}{2}\left(e^x+e^{-x}\right)\quad=\quad\frac{1}{2}\left(e^x-e^{-x}\right)\]To find critical points we set this equal to zero:\[\Large \frac{1}{2}\left(e^x-e^{-x}\right)\quad=\quad 0\]Multiply both sides by 2,\[\Large e^x-e^{-x}\quad=\quad 0\]

Answer 14

\[\Large e^x\quad=\quad e^{-x}\]

Answer 15

Ya you can use natural log I suppose, that works out quite nicely.\[\Large \frac{e^x}{e^{-x}}\quad=\quad 1\qquad\to\qquad e^{2x}\quad=\quad 1\]

Answer 16

\[\Large \ln e^{2x}\quad=\quad \ln 1\qquad\to\qquad 2x \ln e\quad=\quad \ln 1\]

Answer 17

It's quite obvious from \[e ^{x}=e ^{-x} \] that x=0

Answer 18

I dunno how obvious it is :)
Lot of people find exponentials confusing :3

Answer 19

The answer is 0, so I think that what you have shown me is right.

thanks, you make it look so simple! at 56 it is not as easy as it used yo be thanks again