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Showing the inequality holds for an interval (?)

  • #1
Showing the inequality holds for an interval (???)

Homework Statement


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]

Homework Equations





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.
 

Answers and Replies

  • #2
Dick
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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?
 
  • #3


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?
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?
 
  • #4
Dick
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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?
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?
 
  • #5


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?
 
  • #6
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That's formal enough for me.
 
  • #7


Thanks a lot, I had thought I should solve for a specific case, but this is also true :)
 
  • #8
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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."
 
  • #9
Dick
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Thanks a lot, I had thought I should solve for a specific case, but this is also true :)
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).
 
  • #10


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).
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:
 
  • #11
Dick
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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:
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].
 
  • #12


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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