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Curve Sketching with Derivatives?

  1. Mar 24, 2008 #1
    1. The problem statement, all variables and given/known data
    I am given the following function:

    [tex]f(x)=\frac{2x^2}{x-1}[/tex]

    The question asks to find the critical points and classify each using the second derivative.

    2. Relevant equations



    3. The attempt at a solution
    I got the derivative of f(x) to be:
    [tex]f'(x)=\frac{2x^2-4x}{(x-1)^2}[/tex]
    So the critical points are when x=0, x=1, and x=2.

    Here's where I'm stuck. I don't know what it means when it asks to classify each using the second derivative.
    If it helps, the second derivative is:

    [tex] f''(x)=\frac{-8x^2+12x-4}{(x-1)^4}[/tex]
     
    Last edited: Mar 24, 2008
  2. jcsd
  3. Mar 24, 2008 #2
    What it means is that you can use the second derivative to determine if there is either a local minimum or local maximum or asymptote. For example, if f'(x)=0 and f''(x) is positive, you know that there is a local minimum.
     
  4. Mar 24, 2008 #3
    Second derivative determines concavity or convexity of a function. A function whose second derivative is everywhere positive is increasing at an increasing rate. This makes it convex downward, or just concave. On the other hand, a negative second derivative indicates a convex upward. When the 2nd derivative is zero, this is an inflection point.
     
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