Answer: [tex]5x-5 \ge 0[/tex]
This is the same as [tex]x-1 \ge 0[/tex]
========================================================
Explanation:
We have [tex]f(x) = \sqrt{5x-5}+1[/tex]
The stuff under the square root (this stuff is called the radicand) is what we'll focus on (the +1 at the end does not affect the domain at all). We want this to be 0 or larger. This is to avoid applying the square root to negative values, which leads to complications.
So 5x-5 must be 0 or larger meaning we write [tex]5x-5 \ge 0[/tex]
Optionally we can divide all three terms (5x, -5 and 0) by 5 to go from [tex]5x-5 \ge 0[/tex] to [tex]x-1 \ge 0[/tex]
If you wanted to solve for x, you would get [tex]x \ge 1[/tex] to set up the domain. Meaning that x = 1 is the smallest x value you can plug into the function. The x value can be anything larger than 1 as well.
