Mathematics 72 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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:)

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

OpenStudy (anonymous):

\[y'=e ^{-x}-x*e ^{-x}\]

OpenStudy (anonymous):

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

OpenStudy (anonymous):

did you get the above that I posted?

OpenStudy (anonymous):

\[y''=-2e ^{-x}+x*e ^{-x}\] get it = 0

OpenStudy (anonymous):

how do u find second derivatives?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

take a derivative of the first one see above

OpenStudy (anonymous):

1st deriv is \[e ^{-x}-xe ^{-x}\]

OpenStudy (anonymous):

?

OpenStudy (anonymous):

ok \[y''=-e ^{-x}-e ^{-x}+x*e ^{-x}\] see it?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

you take derivative of each term

OpenStudy (anonymous):

oops..sorry..i got it now!

OpenStudy (anonymous):

\[(e ^{-x)})'=-e ^{-x}\] \[(xe ^{-x})'=e ^{-x}-xe ^{-x}\]

OpenStudy (anonymous):

Good! Are you OK to finish it?

OpenStudy (anonymous):

not rlly:(..watz the next step?

OpenStudy (anonymous):

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 :(

OpenStudy (anonymous):

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

OpenStudy (anonymous):

im back! how are you doing?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

watz ur name?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ohok..thank paul! i will

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