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How do I graph this?

  1. Nov 15, 2009 #1
    Given equation
    f(x) = x^4 + 4x^3 - 2

    so f ' (x) = 4x^3 + 12x^2
    which simplified to
    f ' (x) = 4x^2 (x+3), thus our critical points are 0 and -3

    I will use this to sketch the curve
    f ' (x) = 4x^2 (x+3)

    Results are below.
    According to my "sketch", the sequence is - , +, +, +
    B means "before"
    A means "after"
    I used a very close number to generate this list

    But I could not generate any sketch because (0,0) pair cannot have a +, +, + sequence. There must be a - sign, meaning the correct sequence should be -, +, -, +.
    I need to hite (0,0) but there is no way as (-3,0) is increasing.

    My calculator shows a different sequence. What is wrong with my sketch logic here?

    Attached Files:

    Last edited: Nov 15, 2009
  2. jcsd
  3. Nov 16, 2009 #2


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    Science Advisor

    Don't you mean the original curve, y= x^4- 4x^2- 2?

    If you actually are graphing y= 4x^2(x+3), then, yes, the graph is below the x-axis for x< -3 and above the x-axis for x> -3. The graph is then tangent to the x-axis at x= 0. It drops down from a maximum value (at (-2, 16)) to (0,0) but does NOT cross the axis there. It goes back up again.
  4. Nov 16, 2009 #3
    but as you see, in order to test the +, -, i did number line test which used the first derivative.
  5. Nov 16, 2009 #4
    i just found the problem.
    the test was corrected but the y value was wrong....
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