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Finding a point of inflection

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Let f be the function f(x) = (x2 - 3)ex for all real #'s x.

    a. What values of x is f increasing?
    b. What are the x-coordinates of the inflection points for f?
    c. Where (by finding the x and y coordinates of a point) is f(x) at its absolute maximum.

    2. Relevant equations

    Product rule: uv' + vu'

    3. The attempt at a solution

    PART A

    f(x) = (x2 - 3)ex
    f '(x) = ex(x2 + 2x -3)

    x = 1 and x = -3

    If x < -3, f '(x) is positive which means that f is increasing.
    Same thing for x > 1

    :. f increases when x < -3 and x > 1

    PART B

    This is the part where I encounter trouble. Usually I have no problem with these types of problems ... but I can't find when the second derivative equals zero.

    f '(x) = ex(x2 + 2x -3)
    f ''(x) = ex(x2 +4x -1)

    This is a no-calculator-at-all problem so it has to be relatively easy to find when f ''(x) = 0. Here, it's not. Did I make a mistake somewhere?
     
  2. jcsd
  3. Oct 28, 2009 #2
    Use the quadratic formula

    [tex] x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} [/tex]

    No mistakes so far...
     
  4. Oct 28, 2009 #3
    Probably should have done that. I have a habit of viewing the quadratic formula as finding values with lots of decimals but don't round to a fraction. Definitely not the case.

    Anyway, here it is.

    x = -2 + (root20)/2
    x = -2 - (root20)/2

    Moving on to part c.

    *The absolute maximum point must either be at x=-3 or beyond the point where x=1.

    If x = -3, then ...

    f(-3) = 6e-3

    If x = a number above 1 (maybe 5)

    f(5) = 22e5

    f(5)>f(-3)

    Having greater values of x will only yield higher results; therefore there is no absolute maximum.

    Do you think this is ok as a final answer or should I add on to this?
     
  5. Oct 30, 2009 #4
    Anyone have their opinions on what I should add to the answer?
     
  6. Oct 30, 2009 #5

    LCKurtz

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    Homework Helper
    Gold Member

    Remember that a point where f''(x) = 0 is only a potential inflection point. To actually be one, you must show a change in concavity there. You can usually do this by observing that f''(x) changes sign at that point.

    Not every point where f''(x) = 0 is an inflection point. Consider f(x) = x4 at x = 0.
     
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