Domain and range of a multiple non-connected lines from a function?

Question

How do you find the domain and range of a function that has multiple non-connected lines? Such as, $ f(x)=\sqrt{x^2-1}$. Its graph looks like this:

I'm wanting how you would write this with a set eg: $(-\infty, \infty)$.

^{P.S. help me out with the title. Not sure how to describe this.}

Answer 1

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.

• So that gives us that the domain of $f$ is equal to $(-\infty, -1] \cup [1, +\infty)$.

As you can see by your graph, the range of $f(x)$ includes all values $y = \sqrt{x^2 - 1} \geq 0$.

• Hence the range of the function lies withing the interval $[0, +\infty)$

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"

Answer 2

You can find domain of the function by simply analyzing the behavior of the function. For $$ f(x) = \sqrt{x^2-1} $$ you can conclude that the expression under the square root must be non-negative. So $$ x^2-1 \ge 0 \\ (x-1)(x+1) \ge 0 \\ x \in (-\infty, -1] \cup [1, +\infty) $$ Latter is your domain. $D[f] = (-\infty, -1] \cup [1, +\infty)$

Finding range of the function sometimes is more trickier, but for this particular function not so much. Again, you have to observe its behavior. It's square root, and square root in real analysis can take any non-negative value, so range is $E[f] = [0, +\infty)$.