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Quick Q

  1. Nov 7, 2006 #1
    I have a funtion which is ln(x)/x

    The derivative is 1-ln(x)\x^2

    I need to find the absolute max and absolute min. Now I know I need to find the critical numbers first then plug in to f(x) but I am having trouble finding the critical numbers..

    x^2 = 0 so x = 0

    1-ln(x) = 0 so x = e

    But now Iam confused on what to do. Thanks
     
  2. jcsd
  3. Nov 7, 2006 #2
    Also Iam not sure what to do or how to find the critical numbers of this:

    -[(cos(x))^2 + (sin x)(sin x) + 2]/(sin x + 2)^2
     
  4. Nov 7, 2006 #3
    anyone???????
     
  5. Nov 7, 2006 #4
    For ln(x)/x, you should try to sketch the function to see what it does. As far as I can tell, it starts at x=0, f(x) = -infinity, climbs sharply to a peak at x = e, f(x) = 1/e, and then slopes downwards towards x = infinity, f(x) tending to 0. So the only finite turning point is at x = e.
     
  6. Nov 7, 2006 #5
    So now would i just plug e in the function and see what number is produced to determine my max and min?
     
  7. Nov 8, 2006 #6

    HallsofIvy

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    Science Advisor

    For the first problem, x= 0 is NOT a critical number because the function itself, ln(x)/x is not defined at x= 0 but x= e is.

    Is that the original function or its derivative?
    I ask since the derivative of cos(x)/(sin(x)+ 2) is (-sin(x)(sin(x)+ 2)+ cos^2(x))/(sin(x)+ 2)^2, similar to what you have (but not exactly equal).
    If that is the derivative then you critical points are where the fraction is equal to 0 or does not exist. A fraction is 0 if and only if its numerator is 0, does not exist if and only if its denominator is 0 so you need to find where the numerator -sin(x)(sin(x)+ 2)+ cos^2(x)= 0 and where the denominator (sin(x)+ 2)^2= 0. Since sin(x) is never equal to -2, the denominator is never 0. Rewrite the first equation (and notice it is my version, not yours) as -sin^2(x)- 2sin(x)+ 1- sin^2(x)= 0 or 2sin^2(x)+ 2 sin(x)- 1= 0. I would write that as y^2+ 2y= 1 (with y= sin(x)) and complete the square: y^2+ 2y+ 1= 2 so (y+ 1)^2= 2. y= -1+ sqrt(2) and y= -1- sqrt(2). sin(x)= -1+ sqrt(2) gives x= arctan(sqrt(2)-1) but there is no x such that sin(x)= -1- sqrt(2) since that is less than -1.
     
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