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mathcalculus Group Title

Locate all critical points of s(t) = (2 t - 3)^(2) (8 - 2 t)^(3)?

  • 11 months ago
  • 11 months ago

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  1. Paynesdad Group Title
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    Ok is this a calc problem?

    • 11 months ago
  2. mathcalculus Group Title
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    yes

    • 11 months ago
  3. mathcalculus Group Title
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    @Paynesdad

    • 11 months ago
  4. Paynesdad Group Title
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    SO your critical points are max, min, and points of inflection, right?

    • 11 months ago
  5. mathcalculus Group Title
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    critical points are x

    • 11 months ago
  6. mathcalculus Group Title
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    we have to solve for x... to find the y's and that gives us our point

    • 11 months ago
  7. Paynesdad Group Title
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    Right but when the question says critical points...In calculus that is usually the maximum, minimum an/or points of inflection.

    • 11 months ago
  8. SithsAndGiggles Group Title
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    \[s(t)=(2t-3)^2(8-2t)^3\\ \Rightarrow~s'(t)=4(2t-3)(8-2t)^3-6(2t-3)^2(8-2t)^2\\ \Rightarrow~s'(t)=(2t-3)(8-2t)^2\left[4(8-2t)-6(2t-3)\right]\\ \Rightarrow~s'(t)=(2t-3)(8-2t)^2(50-20t)\] It should be fairly easy to find when \(s'(t)=0\).

    • 11 months ago
  9. Paynesdad Group Title
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    That would locate your max and minimums but not your points of inflection. @SithsAndGiggles

    • 11 months ago
  10. SithsAndGiggles Group Title
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    @Paynesdad, usually when a problem asks for critical points, it's referring to the critical points of \(f(x)\). Points of inflection for \(f(x)\) are critical points of \(f'(x)\), not \(f(x)\), so I would think that's the assumption to be used here.

    • 11 months ago
  11. mathcalculus Group Title
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    @SithsAndGiggles we used the product AND chain rule right?

    • 11 months ago
  12. Paynesdad Group Title
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    Yep

    • 11 months ago
  13. SithsAndGiggles Group Title
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    Product, chain, power (not necessarily in that order).

    • 11 months ago
  14. SithsAndGiggles Group Title
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    @Paynesdad, but that doesn't mean critical points can't be inflection points. Depends on the function.

    • 11 months ago
  15. Paynesdad Group Title
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    Correct .. I just wanted to know if she would need to go on and get the 2nd derivative to find the points of inflection as well or just the first derivative critical points.

    • 11 months ago
  16. mathcalculus Group Title
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    @Paynesdad oh sorry, no i was only looking for the critical points.

    • 11 months ago
  17. Paynesdad Group Title
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    Sounds like you have a good handle on this problem...Nice work @SithsAndGiggles and @mathcalculus

    • 11 months ago
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