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anonymous

  • 5 years ago

Locate all inflection points of:

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  1. anonymous
    • 5 years ago
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    well lets see this dont make sence so i cant help sorry i wish i could

  2. anonymous
    • 5 years ago
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    What's the function you're looking at?

  3. anonymous
    • 5 years ago
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    \[ f(x) = x ^{4} + 6x ^{3} - 24x ^{2} + 26\] I get the answer of x=1 and x=-4

  4. anonymous
    • 5 years ago
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    This is a calculus class?

  5. anonymous
    • 5 years ago
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    yes

  6. anonymous
    • 5 years ago
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    Oh good. Didn't want to try to factor that ;p

  7. anonymous
    • 5 years ago
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    i did not know what she wanted because im in the 6th grade haha

  8. anonymous
    • 5 years ago
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    \[f'(x) = 4x^3 + 18x^2-48x = 0\] I'm getting \(x \in \{1.8807, 0, -6.3807\} \) for the critical points.

  9. anonymous
    • 5 years ago
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    this looks so confusing hehe

  10. anonymous
    • 5 years ago
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    And of those, the only inflection point is the 0.

  11. anonymous
    • 5 years ago
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    but i think ill get it soon when i get in a different grade

  12. anonymous
    • 5 years ago
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    It's ok moon, keep practicing and you'll get to calculus sooner than you think.

  13. anonymous
    • 5 years ago
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    did you take the 2nd derivative?

  14. anonymous
    • 5 years ago
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    ok thanks polpak

  15. anonymous
    • 5 years ago
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    i became a fan of you

  16. anonymous
    • 5 years ago
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    \[f''(x) = 12x^2+36x - 48 = 0\] \[\implies x^2 + 3x - 4 = 0 \implies x \in \{1,-4\}\] So actually 0 isn't an inflection point either.

  17. anonymous
    • 5 years ago
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    Since none of the critical points are also points at which f'' is 0 you don't have any points of inflection.

  18. anonymous
    • 5 years ago
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    Does that make sense?

  19. anonymous
    • 5 years ago
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    from what i am reading in my book the inflection points is a point where the concavity changes. when i test some points i come up with\[x <-4 = positive \]\ \[x > 1 = pos\]\[-4 < x < 1 = negative \] so it looks like it changes from positive at -4 to negative then postive again at 1

  20. anonymous
    • 5 years ago
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    That is what I get, too.

  21. anonymous
    • 5 years ago
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    Yes that's true. Technically they are inflection points. I just typically think of inflection points as in the stationary kind. Those are non-stationary points of inflection because while the concavity does change, the tangent to the function is non-zero.

  22. anonymous
    • 5 years ago
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    ok for a question on is problem I need to"show inflections point(s) and explain in complete sentence why these number are inflection points and not critical points".

  23. anonymous
    • 5 years ago
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    So I am kinds confussed with this

  24. anonymous
    • 5 years ago
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    "kinda"

  25. anonymous
    • 5 years ago
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    They are inflection points because the curvature changes (goes from positive to negative and then negative to positive). They are not critical points because the slope of the tangent to the curve at those points is not 0.

  26. anonymous
    • 5 years ago
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    i see now. thanks!!

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