me123
  • me123
Find the domain of the function p(x)-x^-2x+1 What is the domain of p?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Can you specify the p function clearer please. The way that is now doesnt make any sense to me.
anonymous
  • anonymous
Is this the equation my friend?
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me123
  • me123
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

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me123
  • me123
sorry the first one should have a0 at the end
me123
  • me123
yes it is one equation
anonymous
  • anonymous
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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me123
  • me123
I could not figure it I had no clue how to begin to find the answer to this question
anonymous
  • anonymous
why you men by a "()" at the end?
anonymous
  • anonymous
p(x)=x^-(2x+1) is this the function?
anonymous
  • anonymous
please specify
me123
  • me123
x is a real number x =(with a line through it) 0
anonymous
  • anonymous
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
me123
  • me123
p(x)-2^-2x+1
me123
  • me123
sorry p(x)-x^-2x+1
me123
  • me123
it is a - at 2x-1
anonymous
  • anonymous
but were is the = sign? or this expression equated to zero? like this: p(x)-x^-2x+1=0
me123
  • me123
px-x^-2x-1
anonymous
  • anonymous
You need an = sign dude
me123
  • me123
there is no equal sign
anonymous
  • anonymous
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
me123
  • me123
b
anonymous
  • anonymous
ok wait
anonymous
  • anonymous
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
anonymous
  • anonymous
yeah that was what I thought. No problem
me123
  • me123
thank you
anonymous
  • anonymous
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?
anonymous
  • anonymous
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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me123
  • me123
no and thank you
anonymous
  • anonymous
No problem

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