Messing around with a function changes it's domain?

[Swallow]

messing around with a function changes it's domain??

consider a function f(x)= $$\sqrt[3]{2x^2-x^3}$$ if i take x^3 common from inside the cube root the function becomes f(x)= x*$$\sqrt[3]{2/x-1}$$

the domain of the orginal function includes all real numbers, but the domain of the "new" function (which should technically be the same as the origiinal function) becomes all real numbers except zero...
What's going on?

EDIT: does this mean that taking a factor common changes the nature of the function itself?

 

[disregardthat]

That factorization is only valid for non-zero x. If your manipulation had been valid for all x, e.g. factoring out the x^2, your resultant function would be well-defined on the whole domain. Technically the two functions you have are not equal.

 

[arildno]

It's all about that division by zero ain't allowed.

The two expressions, 2-x and x*(2/x-1) are not logically equivalent, because the latter contains the premise that x is non-zero,in order for the second factor to be a meaningful expression .
The first expression does not contain any such premise.

 

[Swallow]

Hi jarle, if it's not valid for x^3, why should it be valid for x^2, how i see it, taking anything common from under the root is the same as multilying and dividing by that factor, so if i divide and multilpy by x^2, even then it shouldn't be valid for zero x as arlidno mentions, you still can't divide by zero

 

[arildno]

Note, however, that your factorized expression will have the same limit as x goes to zero as your first one!
As x becomes tiny, the term 2x^2 in your first expression will be much larger than the x^3-term, so that the function behaves as 2^(1/3)*x^(2/3) as x goes to zero.

In your second expression, 2/x will swamp the 1 in your second factor, so that your function will go as x*2^(1/3)*x^(-1/3), i.e, as your first one when multiplying together the powers of x.

 

[arildno]

Well, you don't need to "divide" with x^2.

2*x^2-x^3=x^2*(2-x) by the distributive property of multiplication, and this holds irrespective of the existence of the multiplicative inverse (which is what division requires).

 

[disregardthat]

Hi jarle, if it's not valid for x^3, why should it be valid for x^2, how i see it, taking anything common from under the root is the same as multilying and dividing by that factor, so if i divide and multilpy by x^2, even then it shouldn't be valid for zero x as arlidno mentions, you still can't divide by zero

Your reasoning in this case is that 1/0 is a factor of 0 which is simply non-sensical. x^2 = 1 * x^2 for all x, but x^2 = x^3 * 1/x only for non-zero x. So you see that you cannot factor your expression this way simply because it's not defined when x is 0. The factorization assumes that 1/x exists; as a rule of thumb you can never divide by 0. You can define the function at 0 by taking the limit of your function as x approaches 0 and still have your original function. However as it stands it is not defined at x = 0.