ksaimouli
  • ksaimouli
use the 2nd derivative test to find the local extrema for the function
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ksaimouli
  • ksaimouli
|dw:1352126604670:dw|
ksaimouli
  • ksaimouli
@Aperogalics
helder_edwin
  • helder_edwin
use the product rule

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

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