## ksaimouli 3 years ago use the 2nd derivative test to find the local extrema for the function

1. ksaimouli

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2. ksaimouli

@Aperogalics

3. helder_edwin

use the product rule

4. ksaimouli

i got that

5. ksaimouli

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6. ksaimouli

set =0 and find critical points

7. helder_edwin

u got it wrong!!!

8. ksaimouli

!!!

9. ksaimouli

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10. ksaimouli

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11. helder_edwin

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

12. ksaimouli

dont factor the first one yet

13. helder_edwin

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

14. ksaimouli

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15. ksaimouli

okay after that

16. ksaimouli

i dont get that part

17. helder_edwin

solve y'=0

18. ksaimouli

what is e^x=0

19. helder_edwin

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

20. helder_edwin

so x=-1

21. ksaimouli

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

22. helder_edwin

no. u r looking for extrema.

23. ksaimouli

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

24. helder_edwin

yes. the only critical point is x=-1

25. ksaimouli

u r using first derivative right

26. helder_edwin

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

27. helder_edwin

yes i'm using the first derivative

28. ksaimouli

but question asks for second derivative

29. helder_edwin

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