$$f(x)=\sqrt{x^2−9}$$
I know that the domain of square root is greater than or equal to zero. I solve for when $x^2−9\ge 0$ and get $x^2\ge9$. Now I get $x\ge 3$ and x≤−3. So that the domain would be (−∞,−3]∪[3,∞).
But how can i find its range? Would anybody please show me the process of find the range of $f(x)=\sqrt{x^2−9}$.
Best Answer
Let's investigate what happens on the interval $[3,+\infty)$. We note that the function $f(x)=\sqrt{x^2-9}$ is symmetric around $x=0$, so we can justify not needing to look at the cases when $x\in(-\infty,-3]$ to find the range.
For $x\in[3,+\infty)$, we have the following:
Thus we can conclude that the range for $f(x)$ will be all nonnegative real numbers, namely the interval $[0,+\infty)$