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Showing a function is strictly decreasing on different intervals

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Define a function g on ℝ by f(x)=1/x if x < 0 and f(x)=-x if x≥0.
    Prove f is strictly decreasing on the intervals (-∞,0) and [0,∞), but that g is not decreasing on ℝ.


    2. Relevant equations



    3. The attempt at a solution
    I think I understand what I need to show, but I often have trouble properly showing it. I broke it up into the three cases that the directions seem to indicate.

    Show g(x)=1/x is strictly decreasing on (-∞,0)

    Let x and y be any elements in the interval (-∞,0) that satisfy the inequality x < y.
    Since the function on the interval is just the reciprocal, all of the image is between -1 and 0.

    For any x < y, it follows that f(x) > f(y) because the larger the magnitude of the element, the further away from 0 the element is and thus the closer its reciprocal is to 0.

    Show g(x)=-x is strictly decreasing on [0,∞)

    Since g is a linear function on the interval [0,∞) with a negative slope, this function is strictly decreasing on [0,∞).

    Show g is not strictly decreasing on ℝ

    Suppose g is strictly decreasing on ℝ
    Then g(.5) > g(0)→ -2 > 0
    This is not true, so it cannot be decreasing at this portion of the range.
    Hence, g is not strictly decreasing on ℝ.
     
  2. jcsd
  3. Mar 14, 2012 #2

    Mark44

    Staff: Mentor

    Is the function g or f?
    I think this is what you meant to say:
    What's this about the image being between -1 and 0? For example, let x = -1/2 and y = -1/3. Then 1/x = -2 and 1/y = -3.

    Since x < y, and both numbers are negative, then 1/x > 1/y, so g is decreasing. I think this is all you need to say.
     
  4. Mar 14, 2012 #3
    Yes, I meant how you corrected it. Sorry about that!

    Other than the first part, the rest looks good?
     
  5. Mar 14, 2012 #4

    Mark44

    Staff: Mentor

    I didn't look at anything very closely after the comments I inserted earlier. Here's the rest of your proof.
    Don't say "it follows that f(x) > f(y) because ..." Just show it like I did in my earlier post. What you have above, IMO, is more wordy than it should be, and less mathematical than it should be.

    Also, since you are presumably in a calculus class (you posted this in the Calculus & Beyond section), you can show that g'(x) < 0 for all x in (-∞, 0), and there's a connection between that fact and whether a function is increasing or decreasing.
    g(.5) = -.5, not -2. On the interval you chose -- [0, .5] -- g is in fact decreasing, so this does not further your argument. If you haven't sketched a graph of the function, you should do so, and this will help you understand why this function isn't decreasing on R even though it is decreasing on each half of its domain.
     
  6. Mar 15, 2012 #5
    g(.5) was a typo. I meant g(-.5) since when x < 0, g(x) = 1/x and that is why I said -2. That should show why it is not decreasing on all of R since -2 < 0.

    Thank you for the suggestions! I will revise my proof.
     
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