Domain of a function.

[retsuL]

Hellow mates.

Please help me finding the correct domain of the following function:

~~~~~√ (x^3 -2x^2 - 5x +6)
f(x)= ------------------------
~~~~~√(x^2 -x -2)

~~~~√(x^2 -x -6)(x -1)
f(x)= --------------------- Factoring the 3rd degree polynomial expression, using Ruffini´s algorithm.
~~~~~√(x^2 -x -2)

~~~~√(x -3)(x+2)(x -1)
f(x)= ----------------------- Factoring all polynomials.
~~~~√(x -2)(x +1)

Now we have to find the real sets of values of x.

(x +3)(x+2)(x -1)
------------------- >= 0
(x -2)(x +1)

Finally I get that x is positive in this intervals :

-∞ -2 -1 1 2 3 +∞
<---------------------------------------------------->
- + - + - +

Dom f(x)= )-2 , -1) U )1 , 2) U )3 , +∞)
I do not have the answer to this, but when I graph this function in GeoGebra, the function does not exists in )1 , 2), but only in the other two sets I get in my answer.
Can some one explain my error?

 

[Dr.Peterson]

You found the domain of the function obtained by combining the two radicals into one radical of a fraction. For the function as given, you have to consider the numerator and denominator separately, requiring each to be positive.

Please ask the second question separately, as it is very different. Also, be sure to clarify whether you mean x - (9/x), which is what you said, or (x - 9)/x, which could be what you intended.

 

[retsuL]

You found the domain of the function obtained by combining the two radicals into one radical of a fraction. For the function as given, you have to consider the numerator and denominator separately, requiring each to be positive.

Please ask the second question separately, as it is very different. Also, be sure to clarify whether you mean x - (9/x), which is what you said, or (x - 9)/x, which could be what you intended.

Thank you very much sir. I think this is the answer I was looking for.
I will start a new topic for my second question and yes I meant x-(9/x), will use notation more carefully from now on.