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Could really use some help.

  1. Jun 19, 2006 #1
    I have a test tomorrow and I'm not sure how to solve this.

    I need to find: inflection points, where f(x) is increasing, concave up, and local min/max points.

    [tex]f(x) = 3x^4 - 4x^3[/tex]

    [tex]f'(x) = 12x^3 - 12x^2[/tex]

    [tex]f'(x) = 12x(x-1)[/tex]

    So I get: x = 0 and x = 1 These are my critical points right? Are these the same as inflection points?

    anyway: [tex]x-1 > 0 [/tex] so, I get f(x) is increasing when x>1. Or (1, Infinity)

    Now I find concavity: [tex] f''(x) = 36x^2 - 24x [/tex]
    [tex] 6x(6x-4)[/tex]
    [tex] 6x = 0 , x = 0[/tex]and [tex] 6x-4 = 0 , x = 2/3[/tex]
    So I get: f(x) is concave up from (-infinity, 0)u(2/3, infinity)

    So I figured out about half of the problem. I guess what I really need help with is how to find the inflection points, and how those are different from the critical points. And also how to find the local min/max values.

    Any help is really appreciated.
    Last edited: Jun 19, 2006
  2. jcsd
  3. Jun 19, 2006 #2
    An "Inflection point" is a point in which the curve changes concavity. In other words, the second derivative = 0 and it changes concavity.

    For more on inflection points and an example:

    A critical point is a place in which the slope of the function is zero, or the first derivative is equal to zero at this point. You can use this to evaluate local min's and maximums. If the slope changes from negative to positive, then we know that the function was decreasing, hit a slope of zero, then increased. Therefore, that would be a maximum. If the function's derivative to the left was positive, then zero at the critical point, then to the right of the critical point negative, the function increased, zero slope'd, then decreased. Thus, that would be a maximum.

    Alternately, you could use the second derivative test (insert critical point's x value into the second derivative. If f ''(c) > 0, minimum. f ''(c) < 0, maximum. Concave down indicates maximum, concave up indicates minimum is one way of remembering.)

    Does that help?
  4. Jun 19, 2006 #3
    Ok, so my local min is -1. I don't think there is a max.

    Andy my inflection points are basically the points I had for change of concavity? So that's: 0, 2/3?

    Thanks for your help
  5. Jun 19, 2006 #4
    When doing these problems, it is useful to make a little chart.

    Here's an example

    To check for a min/maximum, we want their to be a "Change" from negative to zero to positive, or positive to zero to negative. That would indicate a min or a max (Look at y=x^3, x = 0. That does indeed have zero slope, but the slope goes positive, zero, positive. That means it is NOT a local min or max. That's why the "change" in slope is important.)

    Using your problem:
    We know that 0 and one are our critical points. Choose numbers to the left and right of 0. Easy numbers, that way we can see how the functions slope changes before and after the critical point x = 0.

    Let's try x = -1 (That's to the left of 0) and x = 1/2 (To the right of zero, and to the left of 1, our other critical point. We'll also use 2, as that is to the right of 1.)
    Put these test numbers into F '(x). We want to see if their is a sign change to the left and right of our critical points.

    _____________| 0|__________|1|___________________
    F'(1)=-24 < 0 F'(.5) < 0 F'(2) > 0

    So, we see that to the right of zero, slope is negative, to the left of zero, slope is negative. Thus, we can conclude that x = 0 is NOT a local min or max. However, notice the sign change of the slope from the left and right of 1. It goes from negative to positive. This indicates (as you said) a Local Minimum.

    Now, for inflection points, do the same thing! Make sure the concavity changes.

    Make sense a bit?
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