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Find the domain for f

  1. Mar 20, 2006 #1
    Please Help!

    Let f be the function given by f(x)=2x/sqrt(x^2+x+1)

    a)Find the domain for f. Justify your answer.

    b)Write an equation for each horizontal asymptote of the graph f.

    c)Find the range of f. Use f'(x) to justify your answer.

    Note: f'(x)=(x+2)/(x^2 +x+1)^.5

    Please help me with this problem. I don't really know how to do much of it.
     
  2. jcsd
  3. Mar 21, 2006 #2

    HallsofIvy

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    Your are going to have to do some work yourself. Start by telling us what your text gives as the definition of "domain". If you are taking Calculus, you should have see that long ago.

    By the way, your "note" is wrong. f(x)= )=(x+2)/(x^2 +x+1)^.5, not f'.
     
  4. Mar 22, 2006 #3
    I know we have done domain before but it is a little fuzzy. I know domain is where the x values exist and the range is where the y values exist. These may not be the right defintions but this is how I was taught it. For b), I know that the horizontal asymptotes have the eqaution of y=+ or - 2. Back to domain and range- For a) , since there are no vertical asymptotes or holes , I would assume that the domain is all real numbers. For the range, since there are horizontal asymptotes at 2 and -2, I would say the range is -2<y<-2.
     
  5. Mar 22, 2006 #4
    You are right about the derivative being wrong. The correct derivative is (x+2)/(x^2+x+1)^3/2. It's wierd because the note about this wrong derivative was given on the worksheet.
     
  6. Mar 22, 2006 #5

    HallsofIvy

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    No, that is how you remember it. The domain is the set of values of x for which the function can be calculated. Are there any values of x for which you cannot calculate 2x? Are there any values of x for which you cannot calculate [itex]\sqrt{x^2+ x+ 1}[/itex]?
    Are there any values of x for which you cannot calculate [itex]\frac{2x}{\sqrt{x^2+ x+ 1}}[/itex]?
    Yes, the "range" is the set of all possible y values. Are there any values of y you cannot get for some x?

    Good, that's precisely what I was saying! (I hope you meant -2< y< 2 !!!)
     
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