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mathcalculus

  • one year ago

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

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

  3. mathcalculus
    • one year 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
    • one year 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
    • one year 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
    • one year ago
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    i am going to use product rule and chain rule.

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

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

  10. tkhunny
    • one year 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
    • one year ago
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    now where im stuck is on the factoring part...

  12. mathcalculus
    • one year 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
    • one year ago
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    @tkhunny now how do i move on after this?

  14. tkhunny
    • one year 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
    • one year 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
    • one year ago
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    ok

  17. tkhunny
    • one year 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
    • one year ago
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    anyone like to help? let me hagning up thereee;/

  19. mathmate
    • one year 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
    • one year ago
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    i've ACTUALLY been waiting..

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

  22. mathcalculus
    • one year ago
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    part*

  23. tkhunny
    • one year 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
    • one year ago
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    i calculated wrong. i needed to see where i made my mistake

  25. mathcalculus
    • one year ago
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    @tkhunny yes i was working on it.

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

  27. tkhunny
    • one year 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
    • one year ago
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    min or max?

  29. tkhunny
    • one year 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
    • one year ago
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    gtg Up to pdpilot326 from here.

  31. pgpilot326
    • one year 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
    • one year ago
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    check this... http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints.aspx

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

  34. pgpilot326
    • one year ago
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    yep

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

  36. mathcalculus
    • one year 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
    • one year 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
    • one year 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
    • one year ago
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    t=3 t=-2 t=7/9

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

  41. mathcalculus
    • one year ago
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    right?

  42. pgpilot326
    • one year 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
    • one year 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
    • one year ago
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    but for this questions they want the points...

  45. mathcalculus
    • one year 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
    • one year ago
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    i understand... just giving you a bit extra.

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

  48. mathcalculus
    • one year ago
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    oh well thankyou for that.

  49. pgpilot326
    • one year ago
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    it's not -7/9, it's 7/9...

  50. mathcalculus
    • one year ago
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    7.9**

  51. mathcalculus
    • one year ago
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    ignore the -

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

  53. mathcalculus
    • one year ago
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    O_o

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

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

  56. pgpilot326
    • one year ago
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    yes, you want s(7/9)

  57. mathcalculus
    • one year ago
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    oh okay got cha!

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

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