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Math Help - Asymptotes and Limits of Infinity

  1. #1
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    Question Asymptotes and Limits of Infinity

    Directions: sketch the graph.

    f(t) = 3t^4 - 4t^2 + 3

    This is the first problem I've had like this. Since the polynominal equation isn't obvious, I think I take x to the limit of infinity:
    f(t) 3t^4 - 4t^2 +3 = 3t^2 -4 +3/t^2 = 3t^2 = 0
    lim approaches infinity +
    t=0
    f(0) = 3
    extrema point at (0,3)

    f'(t) = 12t^3 - 8t
    f'(t) = 4t(3t^2 - 2)
    f'(t) = 4t =0
    f'(t) = t=0
    f'(t) = 3t^2 -2 = 0
    f'(t) = 3t^2 = 2
    f'(t) = t^2 = 2/3
    f'(t) = t = 2/3
    f'(t) = 0.8164
    f'(.8164) = 12(.8164)^3 - 8(.8164)
    f'(.8164) = .6529 - 6.5312
    f'(.8164) = -5.8783

    critical points at (0,0) and (.8164, -5.8783)

    f"(t) = 36t^2 - 8
    f"(t) = 4(6t^2 - 2)
    f"(t) = 4
    f"(t) = 6t^2 - 2 = 0
    f"(t) = 6t^2 = 2
    f"(t) = t^2 = 1/3
    f"(t) = t = 1/3
    f"(t) = t =.577350269

    f"(4) = 36(16) - 8
    f"(4) = 568
    (4,568)
    f"(.577350268) = 36(.577350269)^2 - 8
    f"(.577350269) = 12 - 8
    f"(.577350269) = 4
    (.577350269,4)

    relative maximium (4,568)


    Anything I am forgetting (or just plain wrong)?
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  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by confusedagain View Post
    Directions: sketch the graph.

    f(t) = 3t^4 - 4t^2 + 3

    This is the first problem I've had like this. Since the polynominal equation isn't obvious, I think I take x to the limit of infinity:
    f(t) 3t^4 - 4t^2 +3 = 3t^2 -4 +3/t^2 = 3t^2 = 0
    lim approaches infinity +
    t=0
    f(0) = 3
    extrema point at (0,3)

    f'(t) = 12t^3 - 8t
    f'(t) = 4t(3t^2 - 2)
    f'(t) = 4t =0
    f'(t) = t=0
    f'(t) = 3t^2 -2 = 0
    f'(t) = 3t^2 = 2
    f'(t) = t^2 = 2/3
    f'(t) = t = 2/3
    f'(t) = 0.8164
    f'(.8164) = 12(.8164)^3 - 8(.8164)
    f'(.8164) = .6529 - 6.5312
    f'(.8164) = -5.8783

    critical points at (0,0) and (.8164, -5.8783)

    f"(t) = 36t^2 - 8
    f"(t) = 4(6t^2 - 2)
    f"(t) = 4
    f"(t) = 6t^2 - 2 = 0
    f"(t) = 6t^2 = 2
    f"(t) = t^2 = 1/3
    f"(t) = t = 1/3
    f"(t) = t =.577350269

    f"(4) = 36(16) - 8
    f"(4) = 568
    (4,568)
    f"(.577350268) = 36(.577350269)^2 - 8
    f"(.577350269) = 12 - 8
    f"(.577350269) = 4
    (.577350269,4)

    relative maximium (4,568)


    Anything I am forgetting (or just plain wrong)?
    Suppose I just show one way how to graph it. (I cannot follow very well your computions.]

    To graph is to find the critical points, the asymptotes, concavity, inflection points, some other points, etc.

    Critical points. Where f'(t) = 0:
    f(t) = 3t^4 - 4t^2 + 3
    f'(t) = 12t^3 -8t
    0 = 12t^3 -8t
    0 = 4t(3t^2 -2)
    t = 0 or sqrt(2/3) or -sqrt(2/3)
    t = 0, or 0.8165, or -0.8165 --------------at critical points.

    f(0) = 3. So point (0,3)
    f(0.8165) = 3(0.8165^4) -4(0.8165^2) +3 = 1.6667---so point (0.8165,1.6667)
    f(-0.8165) = 3(-0.8165)^4 -4(-0.8165)^2 +3 = 1.6667---so point (-0.8165,1.6667)

    Asymptotes: None. f(t) is polynomial in t, not in fraction form.

    Concavity at critical points, or f''(t) at critical points:
    f''(t) = 36t^2 -8

    f''(0) = -8 ----negative, graph concaves facing downward.
    f''(0.8165) = 36(0.8165^2) -8 = 16 ------positive, graph cocaves upward.
    f''(-0.8165) = 36(-0.8165)^2 -8 = 16 ------positive, graph cocaves upward.

    Plot those 3 critical points and sketch the concavities of the graph at those 3 points.
    There is a change in concavity between (-0.8165,1.6667) and (0,3). Likewise, there is also a change in concavity between (0,3) and (0.8165,1.6667). So there must be two inflection points.

    The inflection point:
    f''(t) = 0
    36t^2 -8 = 0
    t = +,-sqrt(8/36) = +,-0.4714 ----***
    f(+,-0.4714) = 3(+,-0.4714)^4 -4(+,-0.4714)^2 +3 = 2.2593
    Meaning, the inflection points are (-0.4714,2.2593) and (0.4714,2.2593).

    So now you have 5 points. Get at least two more. One to the left of (-0.8165,1.6667) and another to the right of (0.8165,1.6667).

    Then connect those 7 points with a smooth curve. There is your graph of the function. [White-Westinghouse would love that graph.]
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