# stationary points

• May 15th 2009, 02:32 PM
slaypullingcat
stationary points
a) given y=e^-x cos x I found y'=-e^-x(sin x + cos x) and
y"=2e^-x(sin x). I checked and they are right.

b) find stationary points between -pi<x<pi and determine their nature.
I can do this with simple quadratics but this function is beyond me at present.

c) Find the points of inflection for -pi < x < pi

Thank you for any help
• May 15th 2009, 02:52 PM
skeeter
Quote:

Originally Posted by slaypullingcat
a) given y=e^-x cos x I found y'=-e^-x(sin x + cos x) and
y"=2e^-x(sin x). I checked and they are right.

b) find stationary points between -pi<x<pi and determine their nature.
I can do this with simple quadratics but this function is beyond me at present.

$\textcolor{red}{-e^{-x}(\sin{x} + \cos{x}) = 0}$

$\textcolor{red}{-e^{-x}}$ is always negative ... that means
$\textcolor{red}{\sin{x} = -\cos{x}}$ ... now, for what two values of x in the defined interval is that true?

c) Find the points of inflection for -pi < x < pi

$\textcolor{red}{2e^{-x}\sin{x} = 0}$

same idea ... the only place where y'' = 0 in the given interval is at the endpoints and x = 0 .

.
• May 15th 2009, 04:29 PM
slaypullingcat
sin 3pi/4=-cos 3pi/4 and sin pi/4=-cos pi/4 ? Is that correct?

If so I just sub both values of x back into y to find the y coordinate?
• May 16th 2009, 06:00 AM
skeeter
Quote:

Originally Posted by slaypullingcat
sin 3pi/4=-cos 3pi/4 (yes) and sin pi/4=-cos pi/4 (no)

$\sin\left(-\frac{\pi}{4}\right) = -\cos\left(-\frac{\pi}{4}\right)$