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