Monday, September 14, 2009

Post #4

I had this typed up last night, and it wouldnt copy past..so yea, I gave up.

This week we also did differentiability, which was basically looking at the pictures/graphs and being able to recognize those. The one thing that I get, let's say, 98% of the time is the First and Second Derivative Tests. Now, Thursday/Friday, we added in the justifications. This didn't make it difficult necessarily, just tedius. So, let's do an example problem!!!:

this is the second derivative test.

x^3 - 6x^2 + 12x

take the first derivative:

3x^2 -12x + 12

take the second derivative:

6x - 12

Set equal to 0 and solve:

x=2

Set up intervals:

(-inf., 2) U (2, inf.)

Plug in values within each interval:

-ve and +ve

And if the question asked you about the concavity, it would be concave up because you are going from decreasing to increasing.

Now, for the justifications, I sometimes forget that you're not really supposed to write in 1st person, so:

Using the Second Derivative Test, set the 2nd derivative equal to zero to find possible points of inflection. In this case, x=2 is one. After setting up intervals, the graph is seen to be concave up
because it goes from decreasing (negative value) to increasing (positive value).

Now time for questions:

1. Does anyone want to go over the horizontal tangent thing, because I kind of forgot?
2. And, can anyone tell me when the limits do not exist (like removable, jump etc.)?

3 comments:

  1. Jumps only occur when you have a peicewise function. You find removables by factoring the top and bottom of your function..if something cancells out then its a removable at x=whatever that number is. Then after you cancel, if you have anything left at the bottom you have an infintite(asymptote) at that number. :)

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  2. Also to add to what stephanie said an example being

    lim->3 x^2+4
    -------
    x-3

    Plug in the 3 on the bottom and because you get 0 you know the function is not continuous then you do what stephanie said to get exactly what it is.

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  3. You asked about horizontal tangents to i think i can help explain how to do that. If you are given a problem and the directions say Find the points where y has horizontal tangents you simply just take ther derivative then set equal to zero, solving for x.

    An example of this would be Find the points where y has horizontal tangents:
    y=3x^3 + 4x^2 + 5 take the derivative
    y'=9x^2 + 8x
    then set it equal to zero
    9x^2 + 8x = 0
    factor out an x
    x(9x + 8) = 0
    you then get that your x's are 0 and -8/9.
    Plug your x's into the original function which gives you your y values.
    y=3(0)^3 + 4(0)^2 + 5 which gives you the point (0,5)
    then plug in the second x value
    y=3(-8/9)^3 + 4(-8/9)^2 + 5 which would give you the point (-8/9,6.053)

    The final answer would be (0,5) (-8/9,6.053)

    Hope that helped.

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