how do you find extrema and points of inflection of y=xe^(-x) and for y=lnx/x

Question

anonymous · Answer

Well extrema can occur where the first derivative is zero or undefined. The first one you need to do integration by parts
The second is a usub.
Then points of inflection can occur where the second derivative is zero or undefined.

anonymous · Answer

can u show the steps for the 1st one? for the extrema and points..id rlly appreciate it! ill try 2 attempt the seond prob:)

anonymous · Answer

y'=e^(-x)-x*e^(-x)
y'=0, then x=1

anonymous · Answer

y''=-e^(-x)-+xe^(-x)-e(-x)
y''=0, then x=2

find corresponding y (both cases)

anonymous · Answer

derv of xe^(-x) is x-e^-x+e^-x thou right? usin the product rule

anonymous · Answer

I love how I was talking about integration xDDDDD I misread.

anonymous · Answer

$$y'=e ^{-x}-x*e ^{-x}$$

anonymous · Answer

sry..i kno this is a stupid que...@inik..is ur deriv jus  anothr way u wrote wat i got..jus makin sure im doin it right too

anonymous · Answer

did you get the above that I posted?

anonymous · Answer

$$y''=-2e ^{-x}+x*e ^{-x}$$
get it = 0

anonymous · Answer

how do u find second derivatives?

anonymous · Answer

how about the second one...?
can you type first derivative?

anonymous · Answer

take a derivative of the first one
see above

anonymous · Answer

1st deriv is $$e ^{-x}-xe ^{-x}$$

anonymous · Answer

?

anonymous · Answer

ok
$$y''=-e ^{-x}-e ^{-x}+x*e ^{-x}$$
see it?

anonymous · Answer

y is it -e ^{-x}-e^{-x}?

anonymous · Answer

you take derivative of each term

anonymous · Answer

oops..sorry..i got it now!

anonymous · Answer

$$(e ^{-x)})'=-e ^{-x}$$
$$(xe ^{-x})'=e ^{-x}-xe ^{-x}$$

anonymous · Answer

Good! Are you OK to finish it?

anonymous · Answer

not rlly:(..watz the next step?

anonymous · Answer

let's finish with first equation.
now, that you have 1st & second derivatives, =0 & find x of critical point(s) & inflection ; plug-in in origial function & find y...

I need to log-out for few min... will be back shortly :(

anonymous · Answer

ohhok..ill b workin on it! thank u so much 4 ur patience!:))

anonymous · Answer

im back!
how are you doing?

anonymous · Answer

still workin on it..returned 2 a couple old problems..ill post wat i get when im done!:) thank u so much thou

anonymous · Answer

watz ur name?

anonymous · Answer

Good Luck...
posted it as a new one if there is questions :)
Signing off
Paul

anonymous · Answer

ohok..thank paul! i will