1. [SOLVED]Basic domain.

$y=$ $\sqrt{-x^3+x}$

$(-x^3+x)\geq0$
$\equiv$ $x(-x^2+1)\geq0$
$\equiv$ $x\geq0\cap(x\geq-1\cap~x\leq1)$

Domain = $0\leq~x\leq1$

Is that correct??

because in my calculator i have a domain = $]\infty, 1]$

2. Re: Basic domain.

No, it is not correct. You calculator is right (assuming you meant $]-\infty, 1]$ In order that ab> 0, a and b must have the same sign, NOT just "a> 0, b> 0". That is, either a> 0 and b> 0 or a< 0, and b< 0. x> 0 and 1- x^2> 0 gives 0< x< 1. x< 0 and 1-x^2< 1 gives x^2> 0 which is true for any x< 0. Of course, x= 0 and x= 1 are also included.

3. Re: Basic domain.

So you are telling me, that i need to assume that x<0 belongs to the function domain.

But, i solved the domain in incorrectly way.

I do not know how to solve the domain, too show x<0 in solution.

4. Re: Basic domain.

$y = \sqrt{x - x^3}$

$x - x^3 \ge 0$

$x(1 - x^2) \ge 0$

$x(1-x)(1+x) \ge 0$

note that the three values of x that make the expression equal to 0 are x = 0 , x = 1 , and x = -1

these three values break the possible values of x into four intervals ...

$(-\infty,-1]$ , $[-1, 0]$ , $[0, 1]$ , and $[1, \infty)$

pick any value in each interval and check the value of $x(1-x)(1+x)$.
if that value is positive, then that interval is in the domain.

5. Re: Basic domain.

Now i fully understood.
Thanks.

6. Re: Basic domain.

Originally Posted by Fabio010
So you are telling me, that i need to assume that x<0 belongs to the function domain.
No, I didn't say anything of the sort. I said that you cannot assume it does not!

But, i solved the domain in incorrectly way.

I do not know how to solve the domain, too show x<0 in solution.
I thought that was what I just showed you. You have, correctly, that $-x^3+ x\ge 0$. As I said before, if ab> 0, then either a> 0 and b> 0 or a<0 and b< 0. You must consider both possiblities.