Give the formula for a function with natural domain $(\infty ,0]\cup [10,\infty)$ and with range $(-\infty, -1]$

Question

As the title says, I need to create/generate a formula for a function with natural domain $(-\infty ,0] \cup [10, \infty )$ and with range $(-\infty , -1]$ where $-\infty $ means going off towards negative numbers' infinite, and $\infty$ means reaching towards positive infinite. $\cup$ means union, so all numbers from both sets $(-\infty ,0] \cup [10, \infty)$ combined. it means elements from $(-\infty ,0]$ "or" $[10, \infty)$

Also known as all numbers except for $(0,10)$ or $[1,9]$

how do you generate this type of function with a center section of the domain missing? I am able to create a function where there is only one type of restriction, but not a function where there's a set of numbers restricted.

Answer 1

$f(x)=-1-\sqrt{|x-5|-5}$ works for this.

Explanation: The term $|x-5|$ is centered at 5, the center of the interval $(0,10)$ which you wish to excluded from the domain, and $0 \le |x-5|<5$ on $(0,10)$. So next, we see that $|x-5|-5$ is negative on $(0,10)$ so that the squareroot will not be defined there.

So far, $\sqrt{|x-5|-5}$ is symmetric around the point $(5,0)$ and has range $[0,\infty).$ So if we change its sign, the range becomes $(-\infty,0]$, and since we desire a range of $(-\infty,-1]$ we finally subtract $1$ from the formula to shift it vertically down by $1$ unit.

The plot has two lower-half sideways parabolas, one with vertex at $(10,-1)$ going down and to the right, and the other with vertex at $(0,-1)$ going down and to the left.

Answer 2

We will change the problem a bit, to natural domain $(-\infty,-5]\cup [5,\infty)$. If we solve that problem we can shift to get what we want. We are doing this because symmetry is nice.

And although I have nothing really against $5$, but $1$ is a nicer number. So we will deal with natural domain $(-\infty,-1]\cup [1,\infty)$. You can do the suitable scaling.

So we want to run into trouble in $(-1,1)$. The following sounds good: $$\sqrt{1-\frac{1}{x^2}}.$$

But the range is $[0,\infty)$. That part is easy to fix: switch sign, and subtract $1$.