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Inflection points using 2nd derivative

  1. Jul 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Find all inflection points,

    f(x) = -x^3/2


    2. Relevant equations
    n/a



    3. The attempt at a solution
    f'(x) = -3/2x^-1/2
    f''(x) = 3/4x^-3/2

    The f''(x) does not exist. I know that doesn't necessarily mean that there isn't any inflection points, but how do I go about finding them? Ususally I just set the 2nd derivate equal to 0 but it doesn't seem to work in this case. When I do set it equal to 0, I get x=0. And when I check points to the left and right of f''(0) the 2nd derivate doesn't change signs. Does that mean there is no inflection points??
     
  2. jcsd
  3. Jul 8, 2008 #2

    rock.freak667

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    Homework Helper

    I am assuming you meant that [itex]f(x) = \frac{-x^3}{2}[/itex]


    But you are correct in saying that f''(x)=0 => x=0 . f(0)=?

    Now check the sign of f'(x) at say x=0.1 and at x=-0.1. See if the sign changes.

    [You are to check if the sign of the gradient changes which is given by f'(x)]
     
  4. Jul 8, 2008 #3
    Judging by his derivatives, he actually means
    [tex]f(x) = -x^{3/2}[/tex]
    which should have been written as -x^(3/2).

    First of all, I don't think your derivatives are correct, check them again.
     
  5. Jul 8, 2008 #4
    Yes, I meant f(x) = -x^(3/2)

    wish I knew how to use w/e program that you guys are using so that there is less confusion..in looking at my derivative..they seem correct to me

    basically -(3/2)x^((3/2)-1) = -(3/2)x^(1/2)...no?

    ill check over them again and see
     
  6. Jul 8, 2008 #5
    [tex]\frac{d}{dx}f(x) = \frac{-3}{2}x^{1/2}[/tex] is correct, however that is not what you had in your first post.

    However there cannot be a point of inflection at x=0 because the domain is only defined from [0,inf) which automatically means that the second derivative cannot change signs at x=0 because it to is only defined on (0,inf)
     
  7. Jul 8, 2008 #6
    Ah, I see my mistake.So,

    f'(x) = (-3/2)x^(1/2)
    f''(x) = (-3/4)x^(-1/2)

    (-3/4)x^(-1/2)=0
    x=0

    Taking test points to the left and right of x=0 yields no sign change, thus there is no inflection point.

    Would what I said above be correct to say? We currently solve for inflection points setting the 2nd derivative equal to zero. We then use those zeros and divide the x-axis into intervals. Using a test point within those intervals we determine the sign of f''. If f'' changes sign then there is a inflection point.

    Just wanting a confirmation :smile:
     
  8. Jul 8, 2008 #7

    matt grime

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    Homework Helper

    You can't 'test' to the left of zero, i.e. negative numbers. As has been pointed out, this function isn't defined for negative x, and you just took a square root of a negative number in your last post as well in the higher derivatives.
     
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