# Am I on the right track with my homework?

I need a little help with my homework.

This is the assingment: If: $$U(x):= \frac{(x+2)^2}{x+1}$$ For what real numbers $x$ is defined: I tried to do that by solving the equation for $x$. I did that in following manner: $$\frac{(x+2)^2}{x+1} = x+3+\frac{1}{x+1}$$ I got this solution by dividing $(x+2)^2$ with $x+1$. At the end I got that $x$ is $-1$ and $-3$.

Am I on a good track?? Thanks.

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It's defined for every $x\ne -1$ – Riccardo.Alestra Oct 2 '13 at 16:11
Thanks. When I wrote question I figured that out. I fell a little bit stupid right now. – depecheSoul Oct 2 '13 at 16:14

What we know, however, is that $U(x)$ is defined for all $x\in \mathbb R, \;\;x \neq -1$. When $x = -1$, the denominator of $U(x)$ is zero, and division by zero is undefined.
A rain of $+$ for you. :-) – Babak S. Oct 21 '13 at 6:59