you can see the Function in the picture.
what is the way to find the Domain definition ?
thanks.
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. $\displaystyle x^4- 3x^3+ 9x^2- 7x= x(x-1)(x^2- 2x+ 7)$.