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anonymous

  • 5 years ago

find the local min and max of f(x) x^4-2x^2+3

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  1. anonymous
    • 5 years ago
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    Graph it. Find the points that are second highest and second lowest.

  2. anonymous
    • 5 years ago
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    Not that way i wish it was that simple. My professor wants us to find the first derivative which i did but i dont know what to do next

  3. amistre64
    • 5 years ago
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    the rules for this problem are simple; to derive it; take the value of the exponent, multiply it to coeefficeint of its x and subtract 1 from the exponent and put it back up there.....

  4. anonymous
    • 5 years ago
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    you so need video chat i need you as a personal tutor to succeed amistre64

  5. amistre64
    • 5 years ago
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    f(x) = x^4-2x^2+3 f'(x) = 4x^3 -4x + 0

  6. amistre64
    • 5 years ago
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    lol.... id be distracted by all the cheesecake :)

  7. anonymous
    • 5 years ago
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    i got that part then i cant factor it so then what....cheescake????

  8. amistre64
    • 5 years ago
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    oh but you can my dear...you can.. 4x^3 -4x ..... we have things in common to take out..

  9. amistre64
    • 5 years ago
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    I see a 4 and a little x in common to factor out

  10. anonymous
    • 5 years ago
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    yea i did that like 4(x^3-x)

  11. amistre64
    • 5 years ago
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    4x(x^2 -1) is factored then notice the difference of squares and factor it again: 4x(x-1)(x+1)

  12. anonymous
    • 5 years ago
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    oh so i messed up because i didnt bring ot the 4x instead of 4

  13. amistre64
    • 5 years ago
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    the "difference of squares" is something you just need to be able to spot.... and yeah, take that little x out there its clogging up the works

  14. amistre64
    • 5 years ago
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    most likely if you see a binomial....thats a 2 term equation that begins with a "square" and a subtraction in it, look to see if the other number is a square like: (4x^2 - 9). this can be factored to (2x-3)(2x+3)

  15. anonymous
    • 5 years ago
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    amstre, can you please help theycallmekelly when you're done here? She asked for help and I have to go.

  16. amistre64
    • 5 years ago
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    I can :) have fun....

  17. anonymous
    • 5 years ago
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    thanks

  18. anonymous
    • 5 years ago
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    okay i get ya so i know how i could get my c# if it was 4(2x-3)(2x+3) but but now its 4x(x-1)(x+1) do i do the same thing i would do for the first equation

  19. amistre64
    • 5 years ago
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    in this factored form, you can easily see your values of "x" that make it go to zero. each (..) needs a specific value to go to zero, can yo ufind them?

  20. anonymous
    • 5 years ago
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    i know but say the number isnt that easy would i do the same thing

  21. amistre64
    • 5 years ago
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    if the number isnt that easy, then you would try a trial and error manner to determine if it was factorable

  22. amistre64
    • 5 years ago
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    we can give ourselves a "pool" of options to narrow down or trial and error process :) by using the factors of the last # -------------------------- the factor for the first #

  23. anonymous
    • 5 years ago
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    okay while doing this problem i get a local min at 2 but i didnt determine my local max which is f(0)=3 but i didnt get a 0 for my c# why?

  24. amistre64
    • 5 years ago
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    your criticals are what: 0, 1, -1 these can be either mins maxs or inflections. use the 2nd derivative to test all possible zeros :)

  25. anonymous
    • 5 years ago
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    i am about to send you an email give me a min so you can see what i do right and or wrong is that okay

  26. amistre64
    • 5 years ago
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    okey dokey...

  27. anonymous
    • 5 years ago
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    just sent it it was two problem one of them are completed altready

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