If I have a function f(x) e.g. f(x)=x^2 and let the CP draw the graph I get the whole graph. Now i need only the part where (e.g.) x>0 (I think this is called "domain" in english). How do I add that? Is it possible?

~timppa

# Domains Of Functions

Started by
timppa1988
, Dec 06 2004 04:15 PM

7 replies to this topic

### #1

Posted 06 December 2004 - 04:15 PM

### #2

Posted 06 December 2004 - 04:30 PM

y=x^2|x>0

keyboard mth OPC | x > 0 enter

keyboard mth OPC | x > 0 enter

### #3

Posted 06 December 2004 - 04:34 PM

Thanks, that's what I was searching for

### #4

Posted 06 December 2004 - 04:54 PM

Ok it works with the graph but if I type solve((x^2=9)|x>0,x) it says {x=-3,x=3}. With solve((x^2=9)|x>3,x) it says "No Solution" which is correct. Why is that so?

### #5

Posted 06 December 2004 - 06:28 PM

Both are!

{x=-3,x=3} is the solution, by excluding it (x>3) there is "No Solution".

{x=-3,x=3} is the solution, by excluding it (x>3) there is "No Solution".

### #6

Posted 06 December 2004 - 09:10 PM

Yeah, of course both are solutions but i typed solve((x^2=9)|x>0,x) so that the only answer should be {x=3} shouldn't it?

### #7

Posted 07 December 2004 - 09:08 AM

Pardon, but I do not speak English

Graphic

y=x^2|x>0

y=9

G solve intersect

x=3

Graphic

y=x^2|x>0

y=9

G solve intersect

x=3

### #8 Guest_Salomon_*

Posted 13 February 2005 - 11:01 PM

Yeah, of course both are solutions but i typed solve((x^2=9)|x>0,x) so that the only answer should be {x=3} shouldn't it?

You should write solve(x^2=9)|x>0, no parenthesis outside the expresion!, because the operator "|" is a general operator, it is not a solve operator, therefore solve ignore it.

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