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anonymous

  • 5 years ago

suppose f(x)= x^4 +ax^2. What is the value of a if f has a local min at x=2?

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  1. anonymous
    • 5 years ago
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    suppose . [f(x)=x^{4}+ax ^{2.}\] What is the value of a if f has a local min at x=2?

  2. anonymous
    • 5 years ago
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    im not sure if im supposed to find the derivative then set x = 2 then solve for a or something else..

  3. anonymous
    • 5 years ago
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    If there is a local min at x=2, then f'(2)=0.

  4. anonymous
    • 5 years ago
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    If it has a local min, then we know the derivative at that point must be 0. So take the derivative, set x = 2, and set the derivative =0, then solve for a.

  5. anonymous
    • 5 years ago
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    Which is essentially what Xavier said, but with more words ;p

  6. anonymous
    • 5 years ago
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    okay, i did that.. the answer is -8 but i got -4

  7. anonymous
    • 5 years ago
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    Did you get 4x^3+2ax=0 as the derivative?

  8. anonymous
    • 5 years ago
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    yes

  9. anonymous
    • 5 years ago
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    Substitute x=2 4(2)^2+4a=0

  10. anonymous
    • 5 years ago
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    4(2)^3+4a=0 I mean

  11. anonymous
    • 5 years ago
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    it would be 4(2)^3

  12. anonymous
    • 5 years ago
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    yeah

  13. anonymous
    • 5 years ago
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    4a=-32

  14. anonymous
    • 5 years ago
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    ohhh. okay. wow.

  15. anonymous
    • 5 years ago
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    thank soo much. im stupid..

  16. anonymous
    • 5 years ago
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    Everyone has those moments :P

  17. amistre64
    • 5 years ago
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    tag!!

  18. amistre64
    • 5 years ago
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    tag

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