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ksaimouli
Group Title
use the 2nd derivative test to find the local extrema for the function
 one year ago
 one year ago
ksaimouli Group Title
use the 2nd derivative test to find the local extrema for the function
 one year ago
 one year ago

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ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
dw:1352126604670:dw
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
@Aperogalics
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
use the product rule
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
i got that
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
dw:1352126927234:dw
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
set =0 and find critical points
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
u got it wrong!!!
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
dw:1352127080538:dw
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
dw:1352127105772:dw
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
\[ \large y'=e^x+xe^x=(1+x)e^x \]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
dont factor the first one yet
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
so \[ \large y''=e^x+(1+x)e^x=(2+x)e^x \]
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
dw:1352127282428:dw
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
okay after that
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
i dont get that part
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
solve y'=0
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
what is e^x=0
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
\[ \large 0=y'=(1+x)e^x \]
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
so x=1
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
wait how did u get 1 it is (x+2)e^x right so e^x=0 (x+2)=0
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
no. u r looking for extrema.
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
i know i am finding the critical points by setting=0 and then using sign chart i will find it
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
yes. the only critical point is x=1
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
u r using first derivative right
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
then \[ \large y''(1)=(2+(1))e^{1}=(1)e^{1}>0 \]
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
yes i'm using the first derivative
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.1
but question asks for second derivative
 one year ago

helder_edwin Group TitleBest ResponseYou've already chosen the best response.1
so y'' is positive when x=1. what does this tell u??
 one year ago
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