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iluvvyyhu

  • one year ago

Can someone explain to me how i would solve this problem thanks <3 "Estimate graphically the local maximum and local minimum of f(x)=((1/4)x^5)+(x^2)-4x"?

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  1. iluvvyyhu
    • one year ago
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    o-o i have a graphing calculator since the review worksheet said i needed one x.x but im not sure how to use it so ya... fail x.x

  2. JamesWolf
    • one year ago
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    the local maximum is when the rate of change between x and y is 0|dw:1358379289305:dw|

  3. JamesWolf
    • one year ago
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    the rate of change between x and y is found by taking the derivative

  4. JamesWolf
    • one year ago
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    then you can find where the maximum and minimum are by solving \[\frac{ dy }{ dx } = 0\]

  5. iluvvyyhu
    • one year ago
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    im sorry but im still confused x.x what would i plug into that formula? ^-^;

  6. JamesWolf
    • one year ago
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    Do you know how to find the derivative of the formula in the question?

  7. iluvvyyhu
    • one year ago
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    no :(

  8. JamesWolf
    • one year ago
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    is this homework? what topic are you doing at the moment?

  9. iluvvyyhu
    • one year ago
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    its a segment review for my exam x.x and its pre-calc

  10. iluvvyyhu
    • one year ago
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    i switched schools so some of the stuff i dont know x.x

  11. JamesWolf
    • one year ago
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    yeah your going to need to put the function into your graph and just see where the maximum and minimum are, and guess

  12. JamesWolf
    • one year ago
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    have a look on google for an instruction book for your calc

  13. iluvvyyhu
    • one year ago
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    okay:/ thanks though <3

  14. JamesWolf
    • one year ago
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    if you cant use your calculator go here http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html

  15. JamesWolf
    • one year ago
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    copy this formula ((x^5)/4) + (x^2) - 4x its your one but arranged in a way this calc likes it

  16. JamesWolf
    • one year ago
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    and paste it after y1 = then press graph

  17. iluvvyyhu
    • one year ago
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    thanks!(:

  18. JamesWolf
    • one year ago
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    :)

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