I need to find the domain of the function $\;f(x) =\dfrac{x+4}{x^2-9}.$
My answer was: $(-\infty, -3)\cup(3, \infty)$.
The book's answer was: $(-\infty, -3)\cup(-3,-3)\cup(3,+\infty)$
It's question 25 btw. Could the book have possibly made a mistake?
On
Your domain excludes all values between and including $-3$ and $3$. The only points to exclude are $x = -3, x=3$, and that's just what the book does although the middle interval should read $(-3, 3)$ (and not $(-3,-3)$, as you've typed it):
Domain: $$x \in (-\infty, -3)\cup (-3, 3) \cup (3, +\infty)$$
Test for yourself: pick a point in the interval $(-3, 3)$ and test it. You'll see the function is indeed defined there.
Another, perhaps simpler way to write the domain is as follows: $x\in \mathbb R - \{-3, 3\} = \mathbb R\setminus \{-3, 3\},\;$ essentially, "the domain consists of all real numbers except $-3$ and $3$.
On
Informally, the domain of the function is the set of all input values of the function. We can input any real number into this function, except $x=\pm3$ (in which case we're dividing by $0$, which is not allowed), so the domain is all real numbers that are not equal to $\pm 3$.
There are loads of different ways of writing $\text{dom}(f)$. e.g.
Your answer is wrong because you've forgotten about the numbers between $-3$ and $3$ (exclusive).
The book is right. It is not defined only for exactly two points, $\pm3$. I assume you meant $(-3,3)$ for the middle interval. It certainly is defined well for say $x=0, 1$.