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Ephilo

can some one graph this please?? ill award and fan ! it is an upside-down parabola with a vertex at x=1 (third charateristic) and zeros at x=0 and x=2 (first charateristic) and the second says that the slope of the tangent line is positive when x<1 so the graph rises from left to right until it gets to x=1 and the fourth says that the slope is negative meaning it falls from left to right after x=1 making the graph an upside-down parabola

  • one year ago
  • one year ago

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  1. ovenmitt12
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    The vertex is (1,0)? and it has two other zeros? Or is it just a plain old parabola with the vertex at (1,1)?

    • one year ago
  2. Ephilo
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    @ovenmitt12 the original question is this Sketch the graph of a function f(x) having the following characteristics...? f (0) = f (2) = 0 f ' (x) > 0 if x < 1 f ' (1) = 0 f ' (x) < 0 if x > 1 f " (x) < 0

    • one year ago
  3. Ephilo
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    i thought that would be what the graph would look like

    • one year ago
  4. ovenmitt12
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    I think you have it.... It would be a parabola with: vertex (1,1) P(0,0) and P(2,0) and with f''(x) < 0 it would be concave down....

    • one year ago
  5. ovenmitt12
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    My turn... What are the zeros of x^3-4x+1?

    • one year ago
  6. Ephilo
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    i understand it i just don't write get how im suppose to graph that :(

    • one year ago
  7. ovenmitt12
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    I think you just have to graph the original function that would have been a parabola... I don't think you would have to graph any of the derivative or the second deriv.

    • one year ago
  8. Ephilo
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    can you draw that out please,because what the graph is suppose to look like is were im lost

    • one year ago
  9. ovenmitt12
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    |dw:1363820293103:dw|

    • one year ago
  10. ovenmitt12
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    lol sorry this is my first time, I didn't see the draw function.

    • one year ago
  11. SandeepReddy
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    |dw:1363820649485:dw|

    • one year ago
  12. ovenmitt12
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    Look at that... straight lines lol. I should have used those

    • one year ago
  13. SandeepReddy
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    lol!

    • one year ago
  14. Ephilo
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    lolololol!

    • one year ago
  15. Ephilo
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    thats funny lol but thanks it really helped!! :P :D

    • one year ago
  16. ovenmitt12
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    |dw:1363820563396:dw|

    • one year ago
  17. Ephilo
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    @ovenmitt12 thanks so much !

    • one year ago
  18. SandeepReddy
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    |dw:1363821010459:dw|

    • one year ago
  19. Ephilo
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    @ovenmitt12 can you help me with another question please?

    • one year ago
  20. ovenmitt12
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    Only if you answer mine lol

    • one year ago
  21. genius12
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    The equation becomes: f(x) = -x(x - 2) = -x^2 + 2x If the tangents to graph are positive for x < 1, then that just means the parabola is increasing from (-infinity, 1). For it to be increasing over that interval, it needs to be upside down, and so I placed a negative sign in front of the leading coefficient, which reflects it in the x-axis. Secondly, if the graph has a vertex at x = 1, then that implies this: f'(1) = 0 --> Mean the value of the first derivative at x = 1 is 0. Let's see if that's the case. f'(x) = -2x + 2 --> f'(1) = -2(1) + 2 = 0 --> It works! But we need to know what the y-value of the vertex is at x = 1. The derivative tells us that there definitely is a vertex at x = 1, now to find the y-value of this vertex, we just plug in x = 1 in to our function and get the value: f(1) = -(1)(1 - 2) = -(1)(-1) = 1 --> So the vertex is at (1, 1) So this equations gives us that there is definitely a vertex at x = 1, moreover, at (1, 1), it's increasing for x < 1, and it's zeroes are at x = 0 and x = 2. But because we need to graph it, why not also find the y-intercept so it's easier? And we know that the y-intercept occurs when x = 0. So we plug in 0 for x in the function and get y = 0. So the y-intercept occurs at y = 0. This means that both our x and y intercepts occur at (0, 0). So that's everything. Now we just graph with the information we have. |dw:1363820753206:dw| @ephilo

    • one year ago
  22. Ephilo
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    @genius12 WOW! thanks thats perfect!

    • one year ago
  23. Ephilo
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    @ovenmitt12 what would that be?

    • one year ago
  24. genius12
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    @Ephilo No medal no fan? lol =[

    • one year ago
  25. Ephilo
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    @genius12 i did fan you ,lol

    • one year ago
  26. genius12
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    NOOO MEDALLLLLLLLL I WILLLL CRYYYYYYYYYYY

    • one year ago
  27. genius12
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    I JUST REMEMBERED U CANT GIVE ME A MEDAL LMFAO ITS OK

    • one year ago
  28. Ephilo
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    lmaoooo i cant give everyone a medal but if i could i swear i could that was a lot of help,btw i actually did read it @genius12

    • one year ago
  29. Ephilo
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    @genius12 ill give you medal if you help me on my other question

    • one year ago
  30. ovenmitt12
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    @Ephilo Zeros: x^3-4x+1 ?

    • one year ago
  31. SandeepReddy
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    Ephilo ask me

    • one year ago
  32. SandeepReddy
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    i mean ask us! :)

    • one year ago
  33. Ephilo
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    lol okay i have 5 more that i dont understand out of 15 questions i guess thats good :/

    • one year ago
  34. Ephilo
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    @ovenmitt12 whats that for? is that your question ?

    • one year ago
  35. Ephilo
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    find the absolute extrema and the x-value where they occur on the given interval f(x)x^3-12x [0,4]

    • one year ago
  36. Ephilo
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    @SandeepReddy

    • one year ago
  37. SandeepReddy
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    is that f(x) = x^3-12x

    • one year ago
  38. Ephilo
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    |dw:1363822851080:dw|

    • one year ago
  39. Ephilo
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    @SandeepReddy

    • one year ago
  40. SandeepReddy
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    absolute extrema is 16 and the x value is x =4

    • one year ago
  41. Ephilo
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    @SandeepReddy wow that was fast im still trying to understand the question :(

    • one year ago
  42. SandeepReddy
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    Ephilo do u want me to explain?

    • one year ago
  43. Ephilo
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    please.. @SandeepReddy

    • one year ago
  44. SandeepReddy
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    yeah wait

    • one year ago
  45. SandeepReddy
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    the one i wrote as answer was absolute maximum. absolute minimum also exists. at x = 0, the and the value is 0

    • one year ago
  46. SandeepReddy
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    two absolute extrema's one minimum and one maximum

    • one year ago
  47. SandeepReddy
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    Coming to the clarification part

    • one year ago
  48. ovenmitt12
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    yes that was my question... lol I know, delayed right

    • one year ago
  49. ovenmitt12
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    f ' (x) = 3x^2 - 12 critical numbers: 3(x^2-4) = 0 x = 2 x = -2 Plug those back into the original. Also plug the 0 and 4 from your interval into the original.

    • one year ago
  50. ovenmitt12
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    Which ever gives you the highest and lowest values are your extremas.

    • one year ago
  51. ovenmitt12
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    so f (2) = -16 and both f (-2) and f (4) give you 16 x value of 2 gives extrema -16 ; x values -2 and 4 give you 16

    • one year ago
  52. Ephilo
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    @ovenmitt12 and @SandeepReddy thanks guys ! :D

    • one year ago
  53. SandeepReddy
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    Example 1 Determine the absolute extrema for the function g(t) = 2t^3+3t^2-12t+4 and interval.[-4,2] Solution All we really need to do here is follow the procedure given below 1) Find all critical points of f(x) that are in the interval [a,b]. This makes sense if you think about it. Since we are only interested in what the function is doing in this interval we don’t care about critical points that fall outside the interval. Evaluate the function at the critical points found in step 1 and the end points. Identify the absolute extrema. Now, we need to get the derivative so that we can find the critical points of the function. g'(t) = 6t^2+6t-12 and make it equals to zero and find t's It looks like we’ll have two critical points, and . Note that we actually want something more than just the critical points. We only want the critical points of the function that lie in the interval in question. Both of these do fall in the interval as so we will use both of them. That may seem like a silly thing to mention at this point, but it is often forgotten, usually when it becomes important, and so we will mention it at every opportunity to make sure it’s not forgotten. Now we evaluate the function at the critical points and the end points of the interval. Absolute extrema are the largest and smallest the function will ever be and these four points represent the only places in the interval where the absolute extrema can occur. So, from this list we see that the absolute maximum of g(t) is 24 and it occurs at (a critical point) and the absolute minimum of g(t) is -28 which occurs at (an endpoint).

    • one year ago
  54. Ephilo
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    @SandeepReddy thanks that helped a lot, i have two more its okay if you cant help tho youve helped a lot

    • one year ago
  55. SandeepReddy
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    Its OKay

    • one year ago
  56. SandeepReddy
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    ask me!

    • one year ago
  57. SandeepReddy
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    ask me no problem!

    • one year ago
  58. Ephilo
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    okay if you say so @SandeepReddy find any critical numbers of f(x)=4x/x^2+1

    • one year ago
  59. Ephilo
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    |dw:1363825177046:dw|

    • one year ago
  60. SandeepReddy
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    A number a in the domain of a given function f is called a critical number of f if f '(a) = 0 or f ' is undefined at x = a

    • one year ago
  61. SandeepReddy
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    |dw:1363825539582:dw|

    • one year ago
  62. SandeepReddy
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    |dw:1363825649852:dw|

    • one year ago
  63. SandeepReddy
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    |dw:1363825701530:dw|

    • one year ago
  64. SandeepReddy
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    |dw:1363825784023:dw|

    • one year ago
  65. SandeepReddy
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    x = +1 or -1

    • one year ago
  66. SandeepReddy
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    @Ephilo Is it clear?

    • one year ago
  67. Ephilo
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    @SandeepReddy yes, i was just reading over it.

    • one year ago
  68. SandeepReddy
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    Okay!

    • one year ago
  69. Ephilo
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    lol @SandeepReddy you tired yet?

    • one year ago
  70. SandeepReddy
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    Nope !

    • one year ago
  71. SandeepReddy
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    Until I get the medal LOL

    • one year ago
  72. SandeepReddy
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    I wont get tired hahahahah

    • one year ago
  73. Ephilo
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    ohhh you want a medal lol okay !!@SandeepReddy

    • one year ago
  74. SandeepReddy
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    hey i m kidding~

    • one year ago
  75. SandeepReddy
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    u have any question?

    • one year ago
  76. Ephilo
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    lol yeah sadly, im so sleepy tho.sigh but this is due at 9 so id better get it done

    • one year ago
  77. Ephilo
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    Given the derivative, f ‘(x) = -6x2 + 12x a.State the critical values where f ‘(x) =0 b. Find the second derivative, f “(x) c. Use the second derivative to test the critical values found in part a. Analyze the results to determine whether these critical values produce a relative maximum or minimum.

    • one year ago
  78. SandeepReddy
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    a) 6x2 = 12x 6x2-12x = 6x(x-2) =0 => x=0 and x = 2 are critical points

    • one year ago
  79. SandeepReddy
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    b) its -12x+12

    • one year ago
  80. SandeepReddy
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    plug the values of crtical points we got in a) and check whether they produce minima or maxima

    • one year ago
  81. Ephilo
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    @SandeepReddy okay i understand that one

    • one year ago
  82. SandeepReddy
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    hmm good!

    • one year ago
  83. Ephilo
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    yes 2 freaking more !!

    • one year ago
  84. SandeepReddy
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    shoot

    • one year ago
  85. Ephilo
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    lol what ?

    • one year ago
  86. SandeepReddy
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    ask!

    • one year ago
  87. Ephilo
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    okay okay lol

    • one year ago
  88. Ephilo
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    apply the mean value theorem to f on [0,1] to find all values of c such that

    • one year ago
  89. Ephilo
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    |dw:1363826906752:dw|

    • one year ago
  90. SandeepReddy
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    Hey ask that last question. i have to go

    • one year ago
  91. SandeepReddy
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    @Ephilo

    • one year ago
  92. Ephilo
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    @SandeepReddy some ones already answering it , thanks for your help though i really appreciate it :)

    • one year ago
  93. SandeepReddy
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    Your Welcome and im leaving, have a good day, good luck with studies!!

    • one year ago
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