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Ephilo

  • one year ago

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

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  1. ovenmitt12
    • one year ago
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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)?

  2. Ephilo
    • one year ago
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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

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

  4. ovenmitt12
    • one year ago
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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....

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

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

  7. ovenmitt12
    • one year ago
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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.

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

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

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

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

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

  13. SandeepReddy
    • one year ago
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    lol!

  14. Ephilo
    • one year ago
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    lolololol!

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

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

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

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

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

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

  21. genius12
    • one year ago
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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

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

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

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

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

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

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

  28. Ephilo
    • one year ago
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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

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

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

  31. SandeepReddy
    • one year ago
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    Ephilo ask me

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

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

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

  35. Ephilo
    • one year ago
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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]

  36. Ephilo
    • one year ago
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    @SandeepReddy

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

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

  39. Ephilo
    • one year ago
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    @SandeepReddy

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

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

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

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

  44. SandeepReddy
    • one year ago
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    yeah wait

  45. SandeepReddy
    • one year ago
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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

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

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

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

  49. ovenmitt12
    • one year ago
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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.

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

  51. ovenmitt12
    • one year ago
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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

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

  53. SandeepReddy
    • one year ago
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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).

  54. Ephilo
    • one year ago
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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

  55. SandeepReddy
    • one year ago
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    Its OKay

  56. SandeepReddy
    • one year ago
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    ask me!

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

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

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

  60. SandeepReddy
    • one year ago
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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

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

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

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

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

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

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

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

  68. SandeepReddy
    • one year ago
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    Okay!

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

  70. SandeepReddy
    • one year ago
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    Nope !

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

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

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

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

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

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

  77. Ephilo
    • one year ago
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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.

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

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

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

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

  82. SandeepReddy
    • one year ago
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    hmm good!

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

  84. SandeepReddy
    • one year ago
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    shoot

  85. Ephilo
    • one year ago
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    lol what ?

  86. SandeepReddy
    • one year ago
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    ask!

  87. Ephilo
    • one year ago
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    okay okay lol

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

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

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

  91. SandeepReddy
    • one year ago
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    @Ephilo

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

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

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