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anonymous

  • 5 years ago

Give an example of a function, g(x), that has a local maximum at (-3; 3) and a local minimum at (3, -3).

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  1. amistre64
    • 5 years ago
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    g'(3) = -3 eh...

  2. amistre64
    • 5 years ago
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    its a cubic at its basic

  3. amistre64
    • 5 years ago
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    centered at the the origin

  4. amistre64
    • 5 years ago
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    g'(3) = 0 and g'(-3) = 0

  5. amistre64
    • 5 years ago
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    x^3 derived is 3x^2 3(3)^2 = 0 27 = 0 27 - 27 = 0

  6. amistre64
    • 5 years ago
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    3x^2 - 27 = g' x^3 - 27x +C = function of g(x); just gotta calibrate for (3,-3)

  7. amistre64
    • 5 years ago
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    -3 = (3)^3 -27(3) + C -3 -17 + 27(3) = C C = 61 g(x) = x^2 -27x + 61 perhaps?

  8. amistre64
    • 5 years ago
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    g(x) = x^3 -27x + 61 that is lol

  9. anonymous
    • 5 years ago
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    Thanks a lot. I really appreciate it.

  10. amistre64
    • 5 years ago
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    it might be wrong; i havent checked it good wnough yet :)

  11. amistre64
    • 5 years ago
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    C = 51 maybe...

  12. anonymous
    • 5 years ago
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    Hmm, when I graph it it the maximum and minimum isn't (-3, 3) and (3, -3) let me try with C=51

  13. amistre64
    • 5 years ago
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    i see it too; :) gotta figure out where I looked at it wrong

  14. amistre64
    • 5 years ago
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    g' = 3x^2 - 27 = 0 at -3 and 3 so the derivative is good

  15. amistre64
    • 5 years ago
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    g(x) = x^3 - 27x + C if we integrate it...

  16. amistre64
    • 5 years ago
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    (x=-3,y=3) should fit it and (x=3,y=-3) should fit it, so we solve for C with these parameters

  17. amistre64
    • 5 years ago
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    -3 = (3)^3 - 27(3) + C and 3 = (-3)^3 - 27(-3) + C -3 - (3)^3 + 27(3) = C OR 3 - (-3)^3 + 27(-3) = C

  18. amistre64
    • 5 years ago
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    C = 51 or C = -51 which doesnt work out to well in teh end

  19. amistre64
    • 5 years ago
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    g(0) would also have to equal 0

  20. amistre64
    • 5 years ago
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    C = 0

  21. anonymous
    • 5 years ago
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    I just graphed it again and the minimum is at (3, -3) but not the Maximum

  22. anonymous
    • 5 years ago
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    the maximum is not at (3, -3)

  23. amistre64
    • 5 years ago
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    g(x) = x^3 - 27x ... thats it

  24. amistre64
    • 5 years ago
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    ack!!... that aint it either.. lol

  25. amistre64
    • 5 years ago
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    soooo close

  26. anonymous
    • 5 years ago
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    Haha, it's ok, thanks a lot for the help though. But I got a little help from someone but have no idea how hey got it, maybe you can help elaborate this. For this problem they started from this: (x+3)(x-3) g'(x)= X^2 -9 = 1/3x^3 -9 ???? Have no idea as to how they got here for this problem.

  27. amistre64
    • 5 years ago
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    i got some idea, but have to look at it some more

  28. amistre64
    • 5 years ago
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    g'(x) has roots at 3 and -3 so g'(x) = x^2 - 9 if we combine them; which is good; then integrate up to x^3 --- - 9x tho 3

  29. amistre64
    • 5 years ago
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    (1/3)(x^3) - 9x + C is the family of curves for it....

  30. amistre64
    • 5 years ago
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    C = -15 then i think (1/3)x^3 - 9x -15 ; lets check it

  31. anonymous
    • 5 years ago
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    Now the max is right but not the min...

  32. anonymous
    • 5 years ago
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    I did this very quickly; \(f(x)=x^3-27x-51\). Check it!!

  33. amistre64
    • 5 years ago
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    did that first lol

  34. amistre64
    • 5 years ago
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    it works for one , but not the other

  35. amistre64
    • 5 years ago
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    g(0) = 0 has to be part of te graph

  36. amistre64
    • 5 years ago
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    so adding somethig to it aint gonna help; we gotta find the right 'slope'

  37. amistre64
    • 5 years ago
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    M(x^3 - Bx) is what we need

  38. amistre64
    • 5 years ago
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    Dx(Mx^3 - MBx) 3Mx^2 - MB M(3x^2 - B) = 0 when x = 3 or -3

  39. amistre64
    • 5 years ago
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    Maybe include more powers of x?

  40. amistre64
    • 5 years ago
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    or derive again for inflection...

  41. amistre64
    • 5 years ago
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    6Mx = 0 when x = 0 lol ... that dint work out to well

  42. anonymous
    • 5 years ago
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    haha, thanks, i really appreciate the help

  43. amistre64
    • 5 years ago
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    k(-3)^3 = 3 -27k = 3 k = -1/9 maybe? x^3 ---- ; 3x^2/9 = x^2/3 i wonder 9

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