Question

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

Answer 1

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

Answer 2

Is this the equation my friend?

Answer 3

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

Answer 4

sorry the first one should have a0 at the end

Answer 5

yes it is one equation

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

please specify

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

it is a - at 2x-1

Answer 16

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

Answer 17

px-x^-2x-1

Answer 18

You need an = sign dude

Answer 19

there is no equal sign

Answer 20

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

Answer 21

b

Answer 22

ok wait

Answer 23

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

Answer 24

yeah that was what I thought. No problem

Answer 25

thank you

Answer 26

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?

Answer 27

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?

Answer 28

no and thank you

Answer 29

No problem