- me123

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

- katieb

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- anonymous

Can you specify the p function clearer please. The way that is now doesnt make any sense to me.

- anonymous

Is this the equation my friend?

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- 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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## More answers

- me123

sorry the first one should have a0 at the end

- me123

yes it is one equation

- 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

I could not figure it I had no clue how to begin to find the answer to this question

- anonymous

why you men by a "()" at the end?

- anonymous

p(x)=x^-(2x+1) is this the function?

- anonymous

please specify

- me123

x is a real number x =(with a line through it) 0

- 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

p(x)-2^-2x+1

- me123

sorry p(x)-x^-2x+1

- me123

it is a - at 2x-1

- anonymous

but were is the = sign? or this expression equated to zero? like this:
p(x)-x^-2x+1=0

- me123

px-x^-2x-1

- anonymous

You need an = sign dude

- me123

there is no equal sign

- 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

b

- anonymous

ok wait

- 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

yeah that was what I thought. No problem

- me123

thank you

- 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

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

no and thank you

- anonymous

No problem

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