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mathcalculus

  • 2 years ago

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

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

  3. mathcalculus
    • 2 years ago
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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)

  4. tkhunny
    • 2 years ago
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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?

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

  6. mathcalculus
    • 2 years ago
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    i am going to use product rule and chain rule.

  7. tkhunny
    • 2 years ago
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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. mathcalculus
    • 2 years ago
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    then... 4(t-3)^3 (t-2)^5 +5(t-3)^4 (t-2)

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

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

  11. mathcalculus
    • 2 years ago
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    now where im stuck is on the factoring part...

  12. mathcalculus
    • 2 years ago
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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.

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

  14. tkhunny
    • 2 years ago
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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.

  15. tkhunny
    • 2 years ago
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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

  16. mathcalculus
    • 2 years ago
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    ok

  17. tkhunny
    • 2 years ago
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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?

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

  19. mathmate
    • 2 years ago
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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!

  20. mathcalculus
    • 2 years ago
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    i've ACTUALLY been waiting..

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

  22. mathcalculus
    • 2 years ago
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    part*

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

  24. mathcalculus
    • 2 years ago
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    i calculated wrong. i needed to see where i made my mistake

  25. mathcalculus
    • 2 years ago
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    @tkhunny yes i was working on it.

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

  27. tkhunny
    • 2 years ago
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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?

  28. mathcalculus
    • 2 years ago
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    min or max?

  29. tkhunny
    • 2 years ago
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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.

  30. tkhunny
    • 2 years ago
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    gtg Up to pdpilot326 from here.

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

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

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

  34. pgpilot326
    • 2 years ago
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    yep

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

  36. mathcalculus
    • 2 years ago
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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..

  37. mathcalculus
    • 2 years ago
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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.

  38. pgpilot326
    • 2 years ago
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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 = ?

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

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

  41. mathcalculus
    • 2 years ago
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    right?

  42. pgpilot326
    • 2 years ago
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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.

  43. pgpilot326
    • 2 years ago
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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

  44. mathcalculus
    • 2 years ago
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    but for this questions they want the points...

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

  46. pgpilot326
    • 2 years ago
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    i understand... just giving you a bit extra.

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

  48. mathcalculus
    • 2 years ago
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    oh well thankyou for that.

  49. pgpilot326
    • 2 years ago
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    it's not -7/9, it's 7/9...

  50. mathcalculus
    • 2 years ago
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    7.9**

  51. mathcalculus
    • 2 years ago
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    ignore the -

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

  53. mathcalculus
    • 2 years ago
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    O_o

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

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

  56. pgpilot326
    • 2 years ago
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    yes, you want s(7/9)

  57. mathcalculus
    • 2 years ago
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    oh okay got cha!

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

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