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burhan101

Determine and classify all critical points of each function on the interval -4<x<4 f(x) = x⁴-4x³

  • 10 months ago
  • 10 months ago

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  1. johnweldon1993
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    Critical points are found via taking the first derivative of the function...and setting it = to 0 So what is \[\frac{ d }{ dx }x^4-4x^3\]

    • 10 months ago
  2. burhan101
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    0=4x²(x-3)

    • 10 months ago
  3. johnweldon1993
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    One step ahead of me...so you will have 2 critical points.. being...?

    • 10 months ago
  4. burhan101
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    0 and 3 ?

    • 10 months ago
  5. johnweldon1993
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    That would be correct...Now do you have to classify if they are a max or a min or an inflection point?

    • 10 months ago
  6. burhan101
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    How do i find that out?

    • 10 months ago
  7. Loser66
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    plug them into the original function and calculate its value. f(0) =? and f(3)=? compare them to give out the answer

    • 10 months ago
  8. burhan101
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    (3, -27)

    • 10 months ago
  9. Loser66
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    ok, between them, which one is bigger? the bigger is local max, the smaller is local min. right?

    • 10 months ago
  10. burhan101
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    yes

    • 10 months ago
  11. burhan101
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    what do you mean which one :S

    • 10 months ago
  12. burhan101
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    like between (x,y)

    • 10 months ago
  13. Loser66
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    for your problem, so far you have 4 critical points, -4, 0, 3, 4, just plug them into the original function and make conclusion. done.

    • 10 months ago
  14. burhan101
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    where did -4 and 4 come from ?

    • 10 months ago
  15. oldrin.bataku
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    $$f(x) = x^4-4x^3\\f'(x)=4x^3-12x^2=4x^2(x-3)\\4x^2(x-3)=0\implies4x^2=0,x-3=0$$... so we conclude \(x=0\) and \(x=3\) are our critical points. Both lie in our interval so classify them.

    • 10 months ago
  16. oldrin.bataku
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    @burhan101 they're our endpoints, which you'd want to test when finding *absolute* extrema. @Loser66 had a minor misunderstanding

    • 10 months ago
  17. burhan101
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    @oldrin.bataku what do i do with these points now?

    • 10 months ago
  18. burhan101
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    plug them into f(x)?

    • 10 months ago
  19. oldrin.bataku
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    @burhan101 no you want to check the second derivative:$$f''(x)=12x^2-24x=12x(x-2)\\f''(0)=0\\f''(3)=12(3)=36>0$$... so our derivative is increasing near \(x=3\) meaning \(x=3\) is a relative minimum; since our derivative is neither increasing nor decreasing near \(x=0\) we find it's neither! (hence the even multiplicity)

    • 10 months ago
  20. burhan101
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    why are we plugging in 3 ?

    • 10 months ago
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