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First; Second Derivatives

  1. Nov 1, 2013 #1

    Qube

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    Gold Member

    1. The problem statement, all variables and given/known data

    http://i3.minus.com/j7uTkNLAl2aBy.png [Broken]

    2. Relevant equations

    Extrema occur at critical points; critical points are either where the first derivative fails to exist or equals 0.

    Horizontal tangent lines occur where the first derivative is 0.

    Points of inflections occur when the concavity changes across a point.

    3. The attempt at a solution

    14) Yes, the original function has a horizontal tangent line at x = 2. The graph of the derivative is 0 at the specified point. The original function also has an inflection point because the derivative of the derivative changes sign from positive to negative.

    15) No, the derivative is negative at x = -2 (the second tick mark on the x-axis to the left of the origin). It is concave up however since the derivative of the derivative is positive.

    16) x = 0 is a critical point since the derivative is going to the infinities. However, the derivative to the left is positive; the derivative to the left is negative. It's a local maximum.

    17) Yes; the second derivative is negative only from 2 to infinity.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 2, 2013 #2

    Mark44

    Staff: Mentor

    Corrected version:
    Extrema occur at critical points; critical points are points in the domain of the function either where the first derivative fails to exist or equals 0.
    I'm afraid you're missing some important ideas here. All of your answers are wrong.
    The function does have a horizontal tangent at x = 2, but there is NOT an inflection point here. An inflection point has to do with the second derivative changing sign, not the first derivative.
    No, the derivative is positive at x = -2. Notice that the derivative is positive all along the graph for x < 0. You should be able to look at the graph and see that the branch on the left is concave up and increasing.
    The function is not defined at x = 0. A critical point has to be in the domain of the function - 0 is not in the domain of this function, so isn't a critical point, and isn't a local maximum. There are no local or global minima. The local maximum is at x = 2. There is no global maximum.
    Not true, because that's not the only interval. By inspection, you should be able to notice that the graph is concave down for x > 0.
     
    Last edited by a moderator: May 6, 2017
  4. Nov 2, 2013 #3

    Qube

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    Wait. That is the graph of f'(x) not f(x). I know it's a bit small and I'm trying to upload another picture right now. I really appreciate your going through my answers and I apologize since the little prime symbol is difficult to see in the original picture.

    http://i3.minus.com/jbwsxB4NOJKivj.bmp [Broken]
     
    Last edited by a moderator: May 6, 2017
  5. Nov 2, 2013 #4

    Qube

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    Gold Member

    Correct. Critical points must exist in the domain of the original function. E.g. consider f(x) = 1/x. f'(x) = -x^-2. x = 0 appears to be a critical point but it does not exist in the domain of 1/x so it cannot be a critical point.
     
  6. Nov 2, 2013 #5

    Mark44

    Staff: Mentor

    OK, that changes things. I couldn't tell that what you first posted was the graph of f'. I'll take another look at your answers and get back to you.
     
    Last edited by a moderator: May 6, 2017
  7. Nov 2, 2013 #6

    Mark44

    Staff: Mentor

    In light of the fact that the graph was of f', your answers seem correct to me.
     
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