# What is the domain of the function f(x)=sqrt(6 - 2x)?

Feb 5, 2015

In this case you do not want a negative argument for the square root (you cannot find the solution of a negative square root, at least as a real number).

What you do it is to "impose" that the argument is always positive or zero (you know the square root of a positive number or zero).

So you set the argument bigger or equal to zero and solve for $x$ to find the ALLOWED values of your variable:

$6 - 2 x \ge 0$
$2 x \le 6$ here I changed sign (and reversed the inequality).

And finally:
$x \le 3$

So the values of $x$ that you can accept (domain) for your function are all the values smaller than $3$ including $3$.
Check by yourself substituting for example $3$, $4$ and $2$ to confirm our deduction.