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Second derivatives to find max and min values then sketch graph

  1. Jul 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Sketch the graph of each function. List the coordinates of where extrema or points of inflection occur. State where the function is increasing or decreasing, as well as where it is concave up or concave down.

    9. f(x)=x3-12x

    2. Relevant equations



    3. The attempt at a solution

    f(x)=x3-12x

    f'(x)=3x2-12
    f''(x)=6x

    Critical Points: (2,-16) and (-2,16)

    f''(x)=6x=0 , which makes x = 0 which gives the inflection point (0,0)..

    I got all three of those points on the graph. I don't understand the concave up or concave down part and where the function is increasing or decreasing.
     
  2. jcsd
  3. Jul 13, 2010 #2
    Once you have the three points i dont understand why you have to put them into the second derivative to find the increasing or decreasing. You can see it by just looking at the graph and plugging in a few points.
     
  4. Jul 13, 2010 #3

    Office_Shredder

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    Staff Emeritus
    Science Advisor
    Gold Member

    You should be able to use the first derivative to tell whether a function is increasing or decreasing, and the second derivative to tell when it's concave up and concave down
     
  5. Jul 13, 2010 #4
    Points of inflection occur where the second derivative is equal to zero which occurs at zero.

    Take the first derivative and set it equal to zero to find the critical points. (only the x values) Once found, plug them into the second derivative function and only look at the sign of the answer. If it's positive then the function is concave up at that point ie a candidate for the min, if it's negative then it's concave down ie a candidate for the max.

    The function is increasing where the first derivative is positive and vice versa for the opposite.
     
  6. Jul 13, 2010 #5

    Mark44

    Staff: Mentor

    Calculus was developed centuries before graphing calculators or even ordinary scientific calculators. If you are taking a test in which a calculator is not permitted, you can still draw a very reasonable graph of a function using the first and second derivative.

    For an arbitrary function, f''(a) = 0 does not necessarily imply that there is an inflection point at (a, f(a)). For example, the graph of f(x) = x4 has no inflection points, even though f''(0) = 0.

    Similarly, g'(a) = 0 does not necessarily imply that there is a max or min at (a, g(a)). The graph of g(x) = x3 has no extreme points even though g'(0) = 0.

    A function f is increasing on any interval for which f'(x) > 0, and is decreasing on any interval for which f'(x) < 0.

    A function f is concave up on any interval for which f''(x) > 0, and is concave down on any interval for which f''(x) < 0. A function has an inflection point at a point where the concavity changes from concave down to concave up, or from concave up to concave down.
     
  7. Jul 13, 2010 #6

    Mark44

    Staff: Mentor

    This is an oversimplification that is not always true. For example, g(x) = x4 has no inflection points.
     
  8. Jul 13, 2010 #7
    Just find the gradient of point + or - 0.001 of the stationary point and the nature will seem obvious
     
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