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Curve sketching and concavity

  1. Jun 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Y=[itex]\frac{\sqrt{1-x^2}}{(2x+1)}[/itex]

    2. Relevant equations
    I took the first derivative which I believe to be:

    [itex]\frac{(-x-2)}{(2x+1)^2\sqrt{1-x^2}}[/itex]

    I also took the third which I believe to be:
    [itex]\frac{4x^3+12^2-7}{(2x+1)^3(1-x^2)^(3/2)}[/itex]

    I am trying to find zeros of the derivatives and they are all seemingly outside the domain of the original. I don't know what to do in a case like this. There appears to be no real solutions for the second derivative. What does this all mean?
     
  2. jcsd
  3. Jun 16, 2012 #2

    Mark44

    Staff: Mentor

    This is also what I get. You should identify this expression as y' = ...
    Third? Don't you mean second? I didn't check your work on this one, so I can't vouch for its correctness.
    Looking at y', its only zero is at x = -2, which is not in the domain of the original function. This means that the equation y' = 0 has no (real) solutions, so the original function has no points at which the tangent line is horizontal.

    If you graph the original function, you should see that it is strictly decreasing or strictly increasing on each of the two disjoint parts of its domain. Compare to y = 1/x, whose domain is x ≠ 0. On the left half of its domain, the graph is strictly increasing; on the right half of its domain, the graph is strictly increasing. Overall, the graph is neither increasing nor decreasing, because of the discontinuity at x = 0.
     
  4. Jun 16, 2012 #3

    rock.freak667

    User Avatar
    Homework Helper

    When you have a fractional function like that, what you should do also is find the asymptotes (horizontal and vertical).

    What value of x will make y tend to infinity? Also write down the domain of your function.

    And where does your function intersect the y-axis?
     
  5. Jun 16, 2012 #4
    The function's domain is [-1,-1/2)U(-1/2,1]

    VA @ x=-1/2
    HA @ y=+-1

    Correct?

    Also,

    Those two derivatives are y' and y''. My apologies, thanks for correcting.

    The function intersects the y axis at (0,1).
     
  6. Jun 16, 2012 #5

    How can I determine concavity if y'' has no real solutions?
     
  7. Jun 16, 2012 #6
    If the second derivitive has no real solutions, then there are no points of inflection. But there still is a concavity to the slope.
     
  8. Jun 16, 2012 #7

    Mark44

    Staff: Mentor

    To be more precise, what you should say is that the equation y'' = 0 has no real solutions. If so, that means that on each interval in the domain of y'', y'' is either positive at each point or else y'' is negative at each point. Where y'' > 0, the graph of the original function is concave up; where y'' < 0, the graph of the orig. function is concave down.
     
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