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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

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  1. ksaimouli Group Title
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    |dw:1352126604670:dw|

    • one year ago
  2. ksaimouli Group Title
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    @Aperogalics

    • one year ago
  3. helder_edwin Group Title
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    use the product rule

    • one year ago
  4. ksaimouli Group Title
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    i got that

    • one year ago
  5. ksaimouli Group Title
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    |dw:1352126927234:dw|

    • one year ago
  6. ksaimouli Group Title
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    set =0 and find critical points

    • one year ago
  7. helder_edwin Group Title
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    u got it wrong!!!

    • one year ago
  8. ksaimouli Group Title
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    !!!

    • one year ago
  9. ksaimouli Group Title
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    |dw:1352127080538:dw|

    • one year ago
  10. ksaimouli Group Title
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    |dw:1352127105772:dw|

    • one year ago
  11. helder_edwin Group Title
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    \[ \large y'=e^x+xe^x=(1+x)e^x \]

    • one year ago
  12. ksaimouli Group Title
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    dont factor the first one yet

    • one year ago
  13. helder_edwin Group Title
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    so \[ \large y''=e^x+(1+x)e^x=(2+x)e^x \]

    • one year ago
  14. ksaimouli Group Title
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    |dw:1352127282428:dw|

    • one year ago
  15. ksaimouli Group Title
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    okay after that

    • one year ago
  16. ksaimouli Group Title
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    i dont get that part

    • one year ago
  17. helder_edwin Group Title
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    solve y'=0

    • one year ago
  18. ksaimouli Group Title
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    what is e^x=0

    • one year ago
  19. helder_edwin Group Title
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    \[ \large 0=y'=(1+x)e^x \]

    • one year ago
  20. helder_edwin Group Title
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    so x=-1

    • one year ago
  21. ksaimouli Group Title
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    wait how did u get 1 it is (x+2)e^x right so e^x=0 (x+2)=0

    • one year ago
  22. helder_edwin Group Title
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    no. u r looking for extrema.

    • one year ago
  23. ksaimouli Group Title
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    i know i am finding the critical points by setting=0 and then using sign chart i will find it

    • one year ago
  24. helder_edwin Group Title
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    yes. the only critical point is x=-1

    • one year ago
  25. ksaimouli Group Title
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    u r using first derivative right

    • one year ago
  26. helder_edwin Group Title
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    then \[ \large y''(-1)=(2+(-1))e^{-1}=(1)e^{-1}>0 \]

    • one year ago
  27. helder_edwin Group Title
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    yes i'm using the first derivative

    • one year ago
  28. ksaimouli Group Title
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    but question asks for second derivative

    • one year ago
  29. helder_edwin Group Title
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    so y'' is positive when x=-1. what does this tell u??

    • one year ago
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