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me123

  • 5 years ago

Find the domain of the function p(x)-x^-2x+1 What is the domain of p?

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  1. anonymous
    • 5 years ago
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    Can you specify the p function clearer please. The way that is now doesnt make any sense to me.

  2. anonymous
    • 5 years ago
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    Is this the equation my friend?

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  3. me123
    • 5 years ago
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    The only thing I can tell you is that the answer is either x is a real number and x= with a line through it or x is a real number or x is arel number and > 0 or x is a real number x= with a line through it 1

  4. me123
    • 5 years ago
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    sorry the first one should have a0 at the end

  5. me123
    • 5 years ago
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    yes it is one equation

  6. anonymous
    • 5 years ago
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    Take a look to the graph I attached. The domain goes from 0 to + infinite: Im trying to figure out how to mathematically get the answer. Please be pacient

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  7. me123
    • 5 years ago
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    I could not figure it I had no clue how to begin to find the answer to this question

  8. anonymous
    • 5 years ago
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    why you men by a "()" at the end?

  9. anonymous
    • 5 years ago
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    p(x)=x^-(2x+1) is this the function?

  10. anonymous
    • 5 years ago
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    please specify

  11. me123
    • 5 years ago
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    x is a real number x =(with a line through it) 0

  12. anonymous
    • 5 years ago
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    yeah I understand that. What it is not clear for me is if the function p. Is p(x)=x^-(2x+1) the function? Please write it down again

  13. me123
    • 5 years ago
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    p(x)-2^-2x+1

  14. me123
    • 5 years ago
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    sorry p(x)-x^-2x+1

  15. me123
    • 5 years ago
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    it is a - at 2x-1

  16. anonymous
    • 5 years ago
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    but were is the = sign? or this expression equated to zero? like this: p(x)-x^-2x+1=0

  17. me123
    • 5 years ago
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    px-x^-2x-1

  18. anonymous
    • 5 years ago
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    You need an = sign dude

  19. me123
    • 5 years ago
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    there is no equal sign

  20. anonymous
    • 5 years ago
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    ok. Im gonna suppose the = sign is at the end.Now I want you to tell me what of the two following equations is your equation. The way you write (without parenthesis) in not clear. a) p(x)-x^(-2x+1) b) p(x)-x^(-2x)+1 choose one. a or be

  21. me123
    • 5 years ago
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    b

  22. anonymous
    • 5 years ago
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    ok wait

  23. anonymous
    • 5 years ago
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    emun, sorry for just coming into the question like this... but i really think that the function would be p(x) = -x^(-2x) + 1 i have yet to see a question that gives the function as part of the equation as you suggested

  24. anonymous
    • 5 years ago
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    yeah that was what I thought. No problem

  25. me123
    • 5 years ago
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    thank you

  26. anonymous
    • 5 years ago
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    as for the domain, there's no mathematical operations to be done to put simply, you just look at the equation, pay special attention to the places where x is present. is there any values of x that would make the function undefined?

  27. anonymous
    • 5 years ago
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    The function have the shape that is shown in the graph attached. You were right. The domain for this function goes like this: (-infinity,0)U(0,+infinity) How do you find it? The most convenient way is to give x some values. The values I gave p are. p(3)=0.99 p(2)=0.937 p(1)=0 Until now we have found that x>0 is part of the domain => (0,+infinity) if you set x=0, we get p(0)=something divided by 0 = not a real number. So 0 is not part of the domain. Lets go a little bit to the left. P(-1)=something very small p(-2)=somithins smalle even. p for negative number exist anyway. So we can consider the negatives numbers as part of the domain of the function, that is (-infinity,0). Any question?

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  28. me123
    • 5 years ago
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    no and thank you

  29. anonymous
    • 5 years ago
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    No problem

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