use the 2nd derivative test to find the local extrema for the function

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use the 2nd derivative test to find the local extrema for the function

Mathematics
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|dw:1352126604670:dw|
use the product rule

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i got that
|dw:1352126927234:dw|
set =0 and find critical points
u got it wrong!!!
!!!
|dw:1352127080538:dw|
|dw:1352127105772:dw|
\[ \large y'=e^x+xe^x=(1+x)e^x \]
dont factor the first one yet
so \[ \large y''=e^x+(1+x)e^x=(2+x)e^x \]
|dw:1352127282428:dw|
okay after that
i dont get that part
solve y'=0
what is e^x=0
\[ \large 0=y'=(1+x)e^x \]
so x=-1
wait how did u get 1 it is (x+2)e^x right so e^x=0 (x+2)=0
no. u r looking for extrema.
i know i am finding the critical points by setting=0 and then using sign chart i will find it
yes. the only critical point is x=-1
u r using first derivative right
then \[ \large y''(-1)=(2+(-1))e^{-1}=(1)e^{-1}>0 \]
yes i'm using the first derivative
but question asks for second derivative
so y'' is positive when x=-1. what does this tell u??

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