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    if x>y, then x+1/x+2>y+1/y+2, if this is true, why is it?
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    If you say that x>y>2 then it is true.
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    To see why it might help if you see the graph of y=(x+1)/(x+2)
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    Let  f(x) = \frac{x+1}{x+2} .

    Then  f'(x) = \frac{1}{(x+2)^2} > 0 for all x. Ordinarily this would be sufficient to prove your result but there is a complication because of the discontinuity at x = -2. If we plot a graph of y = f(x) when we get:
    Attachment 473033

    Which is not of very high quality, but the vertical asymptote is at x = -2. The gradient is positive at all points of this graph though as we proved before, so for all x < -2 and x > -2, f(x) is strictly increasing. Hence f(q) > f(p) for q > p, provided that p,q are either both greater than -2 or are both less than -2.

    EDIT: My attachment does not seem to be working so just type "plot y = (x+1)/(x+2)" into wolframalpha.com to get a graph
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    thanks
    also is it correct to say n+1=<n+2 (where n is a natural number) is equivalent to saying (n+1)/(n+2)=<1
 
 
 
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Updated: October 29, 2015
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