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Showing the inequality holds for an interval (?)

  1. Nov 28, 2011 #1
    Showing the inequality holds for an interval (???)

    1. The problem statement, all variables and given/known data
    Hi, my homework question is:

    Show that the inequality
    [itex]\sqrt{2+x}[/itex]<2+[itex]\frac{x}{4}[/itex] holds [itex]\forall[/itex]x[itex]\in[/itex][-2,0]

    2. Relevant equations



    3. The attempt at a solution
    I tried using IVT or bisection method, but they are just for existence of a root. How can I show it holds for all x in the interval [-2,0]? Would taking the derivative of the function
    [itex]\sqrt{2+x}[/itex]-2-[itex]\frac{x}{4}[/itex] lead me anywhere? Like finding maximum or minimum points? Thanks a lot for any help.
     
  2. jcsd
  3. Nov 28, 2011 #2

    Dick

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    Re: Showing the inequality holds for an interval (???)

    Find the roots of sqrt(2+x)-(2-x/4)=0. The use the roots to sketch a graph. Where your graph is negative is where sqrt(2+x)<(2-x/4), right?
     
  4. Nov 28, 2011 #3
    Re: Showing the inequality holds for an interval (???)

    Thanks, Dick, but when I try finding roots of [itex]\sqrt{2+x}[/itex]-(2+[itex]\frac{x}{4}[/itex])= 0, what I get is:

    2+x = [[itex]\frac{x}{4}[/itex]]2+4+x
    x2/16+2=0
    but it is not possible. How should I proceed from here? Is there a mistake?
     
  5. Nov 28, 2011 #4

    Dick

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    Re: Showing the inequality holds for an interval (???)

    No, no mistake. If there are no roots or singularities then that difference must always be positive or always be negative since it doesn't cross zero. Which is it?
     
  6. Nov 28, 2011 #5
    Re: Showing the inequality holds for an interval (???)

    x2/16+2 must always be positive for all intervals of x. If we put the inequality sign instead of 0, the statement is true. Should I express it like: 'Since x2/16+2>0 is true for all x, then it also holds for x[itex]\in[-2,0][/itex]' ? Is this a formal statement?
     
  7. Nov 28, 2011 #6

    Mark44

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    Re: Showing the inequality holds for an interval (???)

    That's formal enough for me.
     
  8. Nov 28, 2011 #7
    Re: Showing the inequality holds for an interval (???)

    Thanks a lot, I had thought I should solve for a specific case, but this is also true :)
     
  9. Nov 28, 2011 #8

    Mark44

    Staff: Mentor

    Re: Showing the inequality holds for an interval (???)

    If you want to get fancy, you can say this: Since the inequality is true for all real numbers, it's true a fortiori, for x [itex]\in[/itex] [-2, 0].

    Loosely translated, the italicized phrase means "even more strongly."
     
  10. Nov 28, 2011 #9

    Dick

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    Re: Showing the inequality holds for an interval (???)

    I think you should. So far you've only shown the curves sqrt(2+x) and 2+x/4 don't cross, since they are never equal. So one is always greater than the other one. To say WHICH one is greater, I would test them at, say, x=(-2).
     
  11. Nov 29, 2011 #10
    Re: Showing the inequality holds for an interval (???)

    Thanks, Dick. If I take x= -2, then I get [itex]\sqrt{-2+2}[/itex]< 2+ [itex]\frac{(-2)}{4}[/itex], which is 0 < [itex]\frac{3}{2}[/itex]. It also holds for 0, then [itex]\sqrt{2+x}[/itex] < 2+[itex]\frac{x}{4}[/itex] in the interval [-2,0]. Now how should I proceed? How can I show this holds for every number in this interval?
    I'm a bit confused :confused:
     
  12. Nov 29, 2011 #11

    Dick

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    Re: Showing the inequality holds for an interval (???)

    Because you showed the graphs don't cross because there are no values where they are equal. So sqrt(x+1) is ALWAYS greater than 2+x/4 for any x>=(-2). It's true on [-2.infinity). So it must also be true on [-2,0].
     
  13. Nov 29, 2011 #12
    Re: Showing the inequality holds for an interval (???)

    Wow, drawing their graphs shows everything clearly. sqrt(2+x) is always smaller than 2+x/4. Hence, this is true for every interval that they are defined (x>=-2).

    Thanks for your time and effort, Dick and Mark :) You were a great help.
     
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