algebra-precalculus
I'm asked here to find the domain and range of the graph in interval notation.
I know Domain: (-3,1], but the Range I'm not to sure of. Would the range be Range: [-4,0] or Range:[-4,0)? If it's included somewhere than it must be included everywhere right? I don't entirely agree with the explanation as to why the range is not including 0.
The domain of your function includes all values of $x$ for which the function is defined:
$$f(x)=\sqrt{x^2-1}$$
is defined if and only if $\quad x^2 - 1 = (x+1)(x-1) \geq 0$.
So we need to rule out all values for which $\;x^2 -1 < 0$, which happens when and only when $|x| < 1$, and that happens for any $x$ is in the interval $ (-1, 1)$. So we need to exclude the open interval $(-1, 1)$ from the domain of $f$, giving us a domain of $f$ all of $\mathbb R\setminus (-1,1)$, where $f$ is defined.
As you can see by your graph, the range of $f(x)$ includes all values $y = \sqrt{x^2 - 1} \geq 0$.
Your graph is "off" a bit, on the $x$-intercepts: the two curves you see should intersect the x-axis at the points $(-1, 0)$ and $(1, 0)$. Here's a zoomed-in graph of the function near the origin:
$f(x) = \sqrt{x^2 - 1}$
And here it the same function zoomed "out"
Firstly, I'd like to applaud you on your persistence in trying to get to a rigorous solution.
I'll give you a hint for a geometric solution: if you use the triangle definition of a tangent, can you think of a way of constructing a triangle giving any desired value of tangent?
Lets say I want to find a value of $x$ such that $\tan(x) = y$ for some $y$. Then I can construct a right angled triangle with sides of lengths $1, y$ and $\sqrt{1+y^2}$. Then if $x$ is the appropriate angle, then $\tan(x) = y$
Best Answer
It does look like f(0) = 0, which would make your range [-4,0].
If this is part of an assignment, I'd include a note explaining your reasoning for that.