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I Critical points for difficult function

  1. Mar 30, 2017 #1
    Given that f'(x)=[(8cos(x)/(x2)-(⅛)], find the number of relative maxima and minima on the interval (1,10). Finding maxima and minima analytically wasn't fruitful for me, so instead I used a bit of handwaving. First I argued, using the intermediate value theorem, since f'(1)>f'(10), there exists at least one maximum or minimum. Then I said, well, the derivative is approximately equal to 8cos(x)/x2 and since cos(x) has 2 minima and one maximum on the interval (1,10), then f(x) must have 2 minima and one maximum on that same interval.

    Is there way to solve this without that handwaving?
     
  2. jcsd
  3. Mar 30, 2017 #2

    BvU

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    Take the complete derivative and work it out ?
     
  4. Mar 30, 2017 #3
    The derivative is the expression in the problem. Setting the derivative equal to zero results in 64cos(x)-x2=0. I don't see the way forward to solving that expression.
     
  5. Mar 30, 2017 #4

    BvU

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    $$ f' (x) = 0 \Leftrightarrow ( 8 \sqrt{\cos x} + x )( 8 \sqrt{\cos x} - x ) = 0 \quad ? $$
    [edit] Limits the search for extrema to ## [1, {5\over 2} \pi] = [1, u]## with ##u < 8##
     
    Last edited: Mar 30, 2017
  6. Mar 30, 2017 #5
    Yep, I got that. That's where I'm stuck, solving 8sqrt(cos(x)+/-x=0. I thought about expressing sqrt(cos(x)) as the first few terms of a Maclaurin series but that's less than exact or elegant.
     
  7. Mar 30, 2017 #6

    BvU

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    Exactness isn't required if you only need the number of extrema. A simultaneous plot of ##\cos x## and ##x^2/64## does the trick...
     
  8. Mar 30, 2017 #7
    I know. I was just unsatisfied with not being able to find an analytic solution.
     
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