Question

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

Answer 1

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.

Answer 2

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:)

Answer 3

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

Answer 4

y''=-e^(-x)-+xe^(-x)-e(-x)
y''=0, then x=2

find corresponding y (both cases)

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

did you get the above that I posted?

Answer 10

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

Answer 11

how do u find second derivatives?

Answer 12

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

Answer 13

take a derivative of the first one
see above

Answer 14

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

Answer 15

?

Answer 16

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

Answer 17

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

Answer 18

you take derivative of each term

Answer 19

oops..sorry..i got it now!

Answer 20

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

Answer 21

Good! Are you OK to finish it?

Answer 22

not rlly:(..watz the next step?

Answer 23

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 :(

Answer 24

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

Answer 25

im back!
how are you doing?

Answer 26

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

Answer 27

watz ur name?

Answer 28

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

Answer 29

ohok..thank paul! i will