you can see the Function in the picture.

what is the way to find the Domain definition ?

thanks.

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- Dec 8th 2010, 01:56 AMZOOZhow to find the Domain definition of Function
you can see the Function in the picture.

what is the way to find the Domain definition ?

thanks. - Dec 8th 2010, 02:04 AMProve It
Factorise the denominator. The function will be continuous everywhere except where the denominator is 0.

- Dec 12th 2010, 01:29 AMZOOZ
YES i know if there 0 in the Denominator so the Factorise are not continuous but Can have more points in the denominator to compare it to zero .

how i find Them?

thanks. - Dec 12th 2010, 02:07 AMHallsofIvy
So your question

**really**is "How do I factor $\displaystyle x^4- 3x^3+ 9x^2- 7x$?" It would have been better to say that in your first post.

"x" is an obvious factor: $\displaystyle x^4- 3x^3+ 9x^2- 7x= x(x^3- 3x^2+ 9x- 7)$. Now the only possible zeros of $\displaystyle x^3- 3x^2+ 9x- 7$ are 1, -1, 7, and -7 (because if it were to factor as (x-a)(x-b)(x-c) the last term would be abc=-7- and the only integer factors of 7 are 1 and 7). Trying x= 1 in the polynomial we get $\displaystyle 1^3- 3(1^2)+ 9(1)- 7= 1- 3+ 9- 7= 0$. That tells us that x- 1 is also a factor. Dividing $\displaystyle x^3- 3x^2+ 9x- 7$ by x- 1 leaves $\displaystyle x^2- 2x+ 7$ which has**discriminant**($\displaystyle b^2- 4ac$) equal to $\displaystyle (-2)^2- 4(1)(7)= 4- 28= -24$ so there are no more real number zeros of the polynomial. $\displaystyle x^4- 3x^3+ 9x^2- 7x= x(x-1)(x^2- 2x+ 7)$. - Dec 12th 2010, 05:56 AMZOOZ
thank you.

and yes i mean How do I factor.