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Finding Inflection Points (Applied Calc Question)

  1. Nov 6, 2003 #1


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    The problem: Find the inflection points, if any, for the following: f(x) = e^x + x^-1

    I know to find inflection points I have to:

    1. Compute f''(x)
    2. Determine the points in the domain of f for which f''(x) = 0 or f''(x)
    does not exist
    3. Determine the sign of f''(x) to the left and right of each point x = c
    found in step 2. If there is a change in the sign of f''(x) as we move
    across the point x = c, then (c, f(c)) is an inflection point of f.

    Well, this is what I came up with:

    f'(x) = e^x -x^-2
    f''(x)= e^x + 2x^-3

    Then, I don't know what to do from there because e^x can never be zero, right? but I don't know. My teacher is saying there are inflection points...

    Your help is much appreciated!

  2. jcsd
  3. Nov 6, 2003 #2


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    Staff Emeritus
    Science Advisor

    Yes, ex is never 0, but an inflection point is NOT where "ex= 0". It is where f"= ex+ 2/x3= 0.

    There is no "algebraic" way to solve that equation but it certainly has solutions: using Newton's method or a hand-dandy graphing calculator, we have a zero of f", and an inflection point, for x approximately 0.926.

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