This is a really silly question, I know, but I just wanted to know if I'm too "strict" or if it's really an error.
An exercise our professor given us start with this: be $f$ a function defined as $$ f(x) = \begin{cases} \dfrac{x}{x-1} \sqrt{x^2-1} & x \leq -1\quad \text{or}\quad x \geq 1 \\\\ x^2 - 1 & -1 < x < +1 \end{cases}$$
- Question: is it legit to defined the function in the upper piece in the given range? I mean $x = 1$ is included, but I see $x-1$ in the denominator, and hence that range should have strict inequalities, that is: $x \leq -1$ or $x > 1$.
Am I right, or can this be defined in that way with no sin of error?
Thank you!
First of all i think you made a mistake when writing your function definition for the range of the upper piece, i guess you wanted to write $x \leq -1$. Second of all, yes you are right it should been a more strict range because when writing a function definition in such a manner we want to put the function where it makes sense. For example if i write
$ f(x) = \left\{ \begin{array}{ll} 1/x & \mbox{if } x \leq 0 \\ x & \mbox{if } x \geq 0 \end{array} \right. $
Then this function is well defined everywhere except around 0 where it doesnt make much more sense and it's a wrong way to write it. You can do it if your function is continuous around where your 2 ranges meet.