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mathcalculus

HELP please. Locate all critical points : (The answers are to be points.

  • 8 months ago
  • 8 months ago

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  1. mathcalculus
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    • 8 months ago
  2. tkhunny
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    Have you considered the 1st and 2nd derivatives?

    • 8 months ago
  3. mathcalculus
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    okay so i did this: s'(T)= 4(t-3)^3 (t+2)^5+(t-3)^4 5(t+2)^4(1)

    • 8 months ago
  4. tkhunny
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    Very good, excepting the upper case "T" on the left hand side. It is generally considered bad form to change variables in the middle of the equation. Okay, what does that 1st derivative tell us?

    • 8 months ago
  5. mathcalculus
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    wait: it is 4(t-3)^3(1)(t-2)^5+(t-3)^4 5(t-2)(1)

    • 8 months ago
  6. mathcalculus
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    i am going to use product rule and chain rule.

    • 8 months ago
  7. tkhunny
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    It is not necessary to carry the (1). It is understood or unnecessary. Okay, you have managed a correct 1st derivative twice. Now, what does that tell us?

    • 8 months ago
  8. mathcalculus
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    then... 4(t-3)^3 (t-2)^5 +5(t-3)^4 (t-2)

    • 8 months ago
  9. mathcalculus
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    yes but i still write it to be sure.. :/

    • 8 months ago
  10. tkhunny
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    To Luis_Rivera's "point", I think the intent is to report (a,b) rather than x = a.

    • 8 months ago
  11. mathcalculus
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    now where im stuck is on the factoring part...

    • 8 months ago
  12. mathcalculus
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    well we know we have to find the points.... in order to do that we must find x, so both are right.

    • 8 months ago
  13. mathcalculus
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    @tkhunny now how do i move on after this?

    • 8 months ago
  14. tkhunny
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    That's fine. I have no objection to carrying the (1) along if it helps you understand what you are doing. Time to break out your algebra skills. 4(t-3)^3 (t+2)^5 +5(t-3)^4 (t+2)^4 Common (t-3)^3 (t-3)^3[ 4(t+2)^5 +5(t-3)(t+2)^4] Common (t+2)^4 s'(t) = (t-3)^3 (t+2)^4 [4(t+2) +5(t-3)] Okay, now what? Note: I'm not going to do that for you on the 2nd Derivative.

    • 8 months ago
  15. tkhunny
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    1) There is no (t-2), it is (t+2). Please be more careful. 2) The common factor was (t+2)^4. This leaves one (t+2) in one term, since it started with (t+2)^5

    • 8 months ago
  16. mathcalculus
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    ok

    • 8 months ago
  17. tkhunny
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    Right now, we're just playing with symbols. It is important to be able to do that, but it doesn't help much if we don't know what to do with all the nice little symbols when we are done playing with them. I ask again. What does the 1st derivative tell us?

    • 8 months ago
  18. mathcalculus
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    anyone like to help? let me hagning up thereee;/

    • 8 months ago
  19. mathmate
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    @mathcalculus when you work with a master, you would have to be attentive of what the question is about. If you don't understand the question, ask and ask again what he want to know from you. You are lucky to work with @tkhunny for the past half-an hour. If I have an advice at this point, make sure you understand the definition of "critical point" before proceeding, so you know what you're looking for!

    • 8 months ago
  20. mathcalculus
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    i've ACTUALLY been waiting..

    • 8 months ago
  21. mathcalculus
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    i do know the definition for critical point, what i needed help was on the factoring form..

    • 8 months ago
  22. mathcalculus
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    part*

    • 8 months ago
  23. tkhunny
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    You haven't answered my question, so I was beginning to wonder if you were still working on it.

    • 8 months ago
  24. mathcalculus
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    i calculated wrong. i needed to see where i made my mistake

    • 8 months ago
  25. mathcalculus
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    @tkhunny yes i was working on it.

    • 8 months ago
  26. mathcalculus
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    @tkhunny that determines the critical points. the derivative can;t = to zero.

    • 8 months ago
  27. tkhunny
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    Fair enough. It helps if you tell me, so I don't give up on you. If s'(t) = 0, what does that tell us? (t-3)^3 = 0 at t = 3 (t+2)^4 = 0 at t = -2 [4(t+2) +5(t-3)] = 0 at t = 7/9 So, it appears s'(t) CAN be zero. What do we know when it is?

    • 8 months ago
  28. mathcalculus
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    min or max?

    • 8 months ago
  29. tkhunny
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    Maybe. Not if it is superseded by an inflection point. We must keep track. No conclusions until all the evidence is in.

    • 8 months ago
  30. tkhunny
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    gtg Up to pdpilot326 from here.

    • 8 months ago
  31. pgpilot326
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    not necessarily y = x^3 has a derivative f'(0) = 0 but it's not a min nor max.

    • 8 months ago
  32. pgpilot326
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    check this... http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints.aspx

    • 8 months ago
  33. mathcalculus
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    @pgpilot is it okay to move on from where i left off?

    • 8 months ago
  34. pgpilot326
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    yep

    • 8 months ago
  35. mathcalculus
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    s'(t) = (t-3)^3 (t+2)^4 [4(t+2) +5(t-3)]

    • 8 months ago
  36. mathcalculus
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    @pgpilot326 thank you, i think I am understanding this better but i hate being asked questions like i'm some sort of mathematician..

    • 8 months ago
  37. mathcalculus
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    the problem i has was factoring it.. now to find the critical points i know i have to look for x by setting it to zero after i find the derivative.

    • 8 months ago
  38. pgpilot326
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    simplify in the brackets to better see the zero... \[[4\left( t+2 \right)+5\left( t-3 \right)]=4t+8+5t-15=9t-7\] now you have all factors so any of them can be zero... \[s'\left( t \right)=\left( t-3 \right)^{3}\left( t+2 \right)^{4}\left( 9t-7 \right)\] so s'(t) = 0 if t = ?

    • 8 months ago
  39. mathcalculus
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    t=3 t=-2 t=7/9

    • 8 months ago
  40. mathcalculus
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    from there plug into original function and find the y's to each x

    • 8 months ago
  41. mathcalculus
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    right?

    • 8 months ago
  42. pgpilot326
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    yes, you can also check the 2nd derivative or use a number line with your critical points and check to see which intervals are + or -. this will tell you if you have a max, min or neither.

    • 8 months ago
  43. pgpilot326
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    \[s′(t)=(t−3)^{3}(t+2)^{4}(9t−7)\] |dw:1375838972632:dw| so you should have a rel. max at 7/9 and a rel. min at 3

    • 8 months ago
  44. mathcalculus
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    but for this questions they want the points...

    • 8 months ago
  45. mathcalculus
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    i used the critical numbers and plugged into the orginal equation and i find it wrong.. it' seems wrong.

    • 8 months ago
  46. pgpilot326
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    i understand... just giving you a bit extra.

    • 8 months ago
  47. mathcalculus
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    (3,0),(-2,0),(-7/9,4033.09)

    • 8 months ago
  48. mathcalculus
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    oh well thankyou for that.

    • 8 months ago
  49. pgpilot326
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    it's not -7/9, it's 7/9...

    • 8 months ago
  50. mathcalculus
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    7.9**

    • 8 months ago
  51. mathcalculus
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    ignore the -

    • 8 months ago
  52. pgpilot326
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    no, 7/9 = 0.7777777777 to infinite and beyond...

    • 8 months ago
  53. mathcalculus
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    O_o

    • 8 months ago
  54. pgpilot326
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    oh, well then that's all good. why does it seem wrong?

    • 8 months ago
  55. mathcalculus
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    don't i use this critical number and plug it in?

    • 8 months ago
  56. pgpilot326
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    yes, you want s(7/9)

    • 8 months ago
  57. mathcalculus
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    oh okay got cha!

    • 8 months ago
  58. tkhunny
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    So, we're not going to pursue the points of inflection?

    • 8 months ago
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