HELP PLEASE stationary values

AM4R

Posted 08 November 2006 - 08:04 PM

Can someone please please try and help me with this question thanks.

Find the st values and determine each stationary value (min or max)

y=e^-x(sinx)

When i differentiate I get -e^-x = 0.

What do I do when I get to this stage??

Thanks

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Discogirl17

Posted 08 November 2006 - 08:20 PM

Use logs of base e on both sides and u'll end up with x= log of base e to the 0. I think!

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AM4R

Posted 08 November 2006 - 08:30 PM

Thanks for the help but how do I do that?

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Steve

Posted 08 November 2006 - 09:03 PM

That's a good question! You can't - think about the graph of y=e^x

, this is never zero.

You've not differentiated correctly, I get $\frac{dy}{dx}=e^{-x}(\cos x-\sin x)$ .

Does this help?

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AM4R

Posted 08 November 2006 - 09:57 PM

yup your right, but where do I go from there? and how do I factorise it?

Thanks

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Discogirl17

Posted 09 November 2006 - 11:17 AM

Ah yes he's right. You must now use logs of base e. So you get log to the base e of e to the power of -x, this simply becomes -x since log of e to the e is 1. So you have -x times (cos x- sinx) which becomes -xcosx+xsinx. I'm sure you (using the standard formulas given in past papers and textbooks etc)can rearrange this to form the answer.

P.S I may be wrong, maybe its best to PM Dave, he's the maths wizard! Maths isn't exactly my forte!

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Steve

Posted 09 November 2006 - 07:36 PM

Sorry Discogirl, but that's completely wrong!

We want to solve

$\begin{align*}\frac{dy}{dx}&=0\\ e^{-x}(\cos x-\sin x)&=0\end{align*}$

So, just in the usual way, $e^{-x}=0$ or $\cos x-\sin x=0$ . The first equation has no solutions, but it remains to find solutions to the second one, i.e. $\cos x=\sin x$ .

This is something you'll have done at Higher.