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Find the intervals on which a function is increasing/decreasing?

  1. Sep 14, 2011 #1
    1. The problem statement, all variables and given/known data
    F(x)=6/x-(1/1-x)

    Find the intervals on which the function is increasing/decreasing?


    2. Relevant equations
    F(x)=6/x-(1/1-x)

    F'(x)= -6/x^2 -1/(1-x)^2




    3. The attempt at a solution

    Critical points are x=0 and x=1

    Function has a discontinuitiy at 0

    Checking to the right of the critical point:

    f'(2) <0

    f'(-1)<0

    Seems like for all values I put into the derivative I get a negative number in return. And I know from graphing the function it is increasing approxmiately on (2,infinity) and decreasing (-2,-infinity)
     
  2. jcsd
  3. Sep 14, 2011 #2

    Stephen Tashi

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    To me, it looks like the function is approaching 0 as x approaches infinity. How did you graph it?
     
  4. Sep 14, 2011 #3
    Using graph calc, I entered i as (6/x)/(1/1-x). Is the other work correct?
     
  5. Sep 14, 2011 #4

    Stephen Tashi

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    Shouldn't you do addition of 6/x and 1/(1-x) instead of division?
     
  6. Sep 14, 2011 #5

    uart

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    Why don't you solve the inequality F'(x) > 0

    To solve -6/x^2 -1/(1-x)^2 > 0 you can multiply both sides by x^2 (1-x)^2, and since this quantity is always positive it wont break the inequality. (Though obvious you still need to be careful of the points x=0 and x=1).

    BTW. If you do the above you will find that the LHS of the inequality is negative definite and hence there is nowhere in the domain of that function where the derivative is positive.
     
    Last edited: Sep 14, 2011
  7. Sep 14, 2011 #6

    Ray Vickson

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    For ANY x (except 0 and 1) both terms -6/x^2 and -1/(1-x)^2 are < 0, so F'(x) < 0 everywhere where it is defined (that is, everywhere except x=0 and x=1). When I graph it I do not get what you said. Are you sure you wrote the correct function here? (The function F does increase from negative values for x slightly below x = 1 to positive values just after x = 1, but it does so by "jumping", not by increasing in a smooth way.)

    RGV
     
  8. Sep 14, 2011 #7
    I graphed F(x) and it looked to be increasing on a certain interval, however when graphing F'(x) it is certainly negative everywhere it is defined.

    So it ok to say the function is decreasing everywhere it is defined?
     
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