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



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ksaimouli
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dw:1352126604670:dw

ksaimouli
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@Aperogalics

helder_edwin
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use the product rule

ksaimouli
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i got that

ksaimouli
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dw:1352126927234:dw

ksaimouli
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set =0 and find critical points

helder_edwin
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u got it wrong!!!

ksaimouli
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!!!

ksaimouli
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dw:1352127080538:dw

ksaimouli
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dw:1352127105772:dw

helder_edwin
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\[ \large y'=e^x+xe^x=(1+x)e^x \]

ksaimouli
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dont factor the first one yet

helder_edwin
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so
\[ \large y''=e^x+(1+x)e^x=(2+x)e^x \]

ksaimouli
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dw:1352127282428:dw

ksaimouli
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okay after that

ksaimouli
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i dont get that part

helder_edwin
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solve y'=0

ksaimouli
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what is e^x=0

helder_edwin
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\[ \large 0=y'=(1+x)e^x \]

helder_edwin
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so x=1

ksaimouli
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wait how did u get 1 it is (x+2)e^x right so e^x=0 (x+2)=0

helder_edwin
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no. u r looking for extrema.

ksaimouli
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i know i am finding the critical points by setting=0 and then using sign chart i will find it

helder_edwin
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yes. the only critical point is x=1

ksaimouli
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u r using first derivative right

helder_edwin
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then
\[ \large y''(1)=(2+(1))e^{1}=(1)e^{1}>0 \]

helder_edwin
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yes i'm using the first derivative

ksaimouli
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but question asks for second derivative

helder_edwin
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so y'' is positive when x=1.
what does this tell u??