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ksaimouli

  • 3 years ago

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

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  1. ksaimouli
    • 3 years ago
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    |dw:1352126604670:dw|

  2. ksaimouli
    • 3 years ago
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    @Aperogalics

  3. helder_edwin
    • 3 years ago
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    use the product rule

  4. ksaimouli
    • 3 years ago
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    i got that

  5. ksaimouli
    • 3 years ago
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    |dw:1352126927234:dw|

  6. ksaimouli
    • 3 years ago
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    set =0 and find critical points

  7. helder_edwin
    • 3 years ago
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    u got it wrong!!!

  8. ksaimouli
    • 3 years ago
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    !!!

  9. ksaimouli
    • 3 years ago
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    |dw:1352127080538:dw|

  10. ksaimouli
    • 3 years ago
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    |dw:1352127105772:dw|

  11. helder_edwin
    • 3 years ago
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    \[ \large y'=e^x+xe^x=(1+x)e^x \]

  12. ksaimouli
    • 3 years ago
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    dont factor the first one yet

  13. helder_edwin
    • 3 years ago
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    so \[ \large y''=e^x+(1+x)e^x=(2+x)e^x \]

  14. ksaimouli
    • 3 years ago
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    |dw:1352127282428:dw|

  15. ksaimouli
    • 3 years ago
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    okay after that

  16. ksaimouli
    • 3 years ago
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    i dont get that part

  17. helder_edwin
    • 3 years ago
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    solve y'=0

  18. ksaimouli
    • 3 years ago
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    what is e^x=0

  19. helder_edwin
    • 3 years ago
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    \[ \large 0=y'=(1+x)e^x \]

  20. helder_edwin
    • 3 years ago
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    so x=-1

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

  22. helder_edwin
    • 3 years ago
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    no. u r looking for extrema.

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

  24. helder_edwin
    • 3 years ago
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    yes. the only critical point is x=-1

  25. ksaimouli
    • 3 years ago
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    u r using first derivative right

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

  27. helder_edwin
    • 3 years ago
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    yes i'm using the first derivative

  28. ksaimouli
    • 3 years ago
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    but question asks for second derivative

  29. helder_edwin
    • 3 years ago
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    so y'' is positive when x=-1. what does this tell u??

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