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anonymous

  • 5 years ago

what is the maximums and minimums of the equation y=2.657x^3 + -1.48x^2 + .2819x + 7.7E-14?

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  1. anonymous
    • 5 years ago
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    thats kinda complicated to figure out you should go to http://www.wolframalpha.com

  2. anonymous
    • 5 years ago
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    your welcome

  3. anonymous
    • 5 years ago
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    if this is a y or f(x) then differentiate it then equaled it to zero and find its roots, that rot will be your max or min point..just plug it into your original equation....

  4. anonymous
    • 5 years ago
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    Y'=7.7971X^2 -2.96X +.2819.....NOW THIS IS A QUADRATIC EQUATION,.YOU CAN SOLVE FOR THE ROOTS NOW

  5. anonymous
    • 5 years ago
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    this is supposed to be a quartic...

  6. anonymous
    • 5 years ago
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    no,,, quartic eq starts at say x^4,....cubic is x^3.... quadratic is x^2

  7. anonymous
    • 5 years ago
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    what do i do after i derive the equation to find the minimums and maximums?

  8. anonymous
    • 5 years ago
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    this does start at x^4...... -1.28x4 + -33.36x3 + -323.41x2 + -1384.63x + -2206.60

  9. anonymous
    • 5 years ago
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    no it doesn't

  10. anonymous
    • 5 years ago
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    it's 2.657x^3.....

  11. anonymous
    • 5 years ago
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    hehe look up at your prob here atart at y=x^3 it a cubic eq rt?

  12. anonymous
    • 5 years ago
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    wait who are you talking to?

  13. anonymous
    • 5 years ago
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    after you done the derivative, the derivative is now a quadratic y'= ...x^2

  14. anonymous
    • 5 years ago
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    then how do you find the minimums and maximums

  15. anonymous
    • 5 years ago
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    ok ..you equate the derivative to zero and find its roots

  16. anonymous
    • 5 years ago
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    that roots will become you max or min of your original equation

  17. anonymous
    • 5 years ago
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    what is the answer after equating the derivative

  18. anonymous
    • 5 years ago
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    ok, to find the roots, do you know the quadratic formila?

  19. anonymous
    • 5 years ago
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    x=(-b+\[\sqrt{?}\]b2-4ac)2a

  20. anonymous
    • 5 years ago
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    im sorry my pc equation writer does not write well here you a=7.797.....b=-2.96...c= .2819

  21. anonymous
    • 5 years ago
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    x=(2.96+-\[\sqrt{?}\]-.0303)/15.594

  22. anonymous
    • 5 years ago
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    did you get it? the roots are imaginaries...do you know what is imaginary number?

  23. anonymous
    • 5 years ago
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    obviously... a number that's imaginary like i for example or e

  24. anonymous
    • 5 years ago
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    im just asking what are the points (max and min) for my equation

  25. anonymous
    • 5 years ago
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    no...any square root of a negative number are imaginary number like square root of -1, sqrot of -2, squroot of - numbers

  26. anonymous
    • 5 years ago
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    ok here the simplest way of geting the derivative roots is to graph the derivative and approximate its roots..did you get what i mean?

  27. anonymous
    • 5 years ago
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    Y'=7.7971X^2 -2.96X +.2819.....graph this and find its root or zeroes

  28. anonymous
    • 5 years ago
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    here one root is 0.2819...you plug it in your orig equation y=2.657(.2819)^3 + -1.48(.2819)^2 + .2819(.2819)+ 7.7E-14=?

  29. anonymous
    • 5 years ago
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    min y=0.02137744667

  30. anonymous
    • 5 years ago
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    i graph it rt now and found the root x=0.8948 now you can plug it in to your orig prob.

  31. anonymous
    • 5 years ago
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    your max or min y=0.9708296575

  32. anonymous
    • 5 years ago
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    sorry never mind that root x=.2819 or min y=.02137744667....thats a mistake,, graph it and found that the max or min y=.9708296575

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