How do I work out the implied domains for $\displaystyle f(x)=\frac{x^2-1}{x+1}$ and also for $\displaystyle f(x)=\sqrt\frac{x-1}{x+2}$
Thanks heaps
Incorrect. If $\displaystyle x = -1$ then you have just committed the cardinal sin of dividing by zero when you do this. The domain is all real numbers EXCEPT $\displaystyle x = -1$. The graph of $\displaystyle y = \frac{x^2 - 1}{x + 1}$ has a 'hole' at $\displaystyle x = -1$.
Incorrect. See above.
Your bracket notation needs some minor corrections:
$\displaystyle {\color{red} (}-\infty,-2{\color{red})} \cup [1,+\infty{\color{red})}$
$\displaystyle \mathbb{R}/\left \{ -1 \right \}$ (sorry about later)
for the second one : study the sign of $\displaystyle x-1$ and $\displaystyle x+2 $ then study $\displaystyle \frac{x-1}{x+2}$
and see where $\displaystyle \frac{x-1}{x+2}$ is positive.
My judgement was that we don't actually divide by $\displaystyle x+1$ since the final expression is $\displaystyle x-1$, but I understand what you mean.
So, for example, $\displaystyle \frac{x^2}{x}$ will not be defined in $\displaystyle 0$ even though it's equal to $\displaystyle x$??