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

1 Answer
Gió
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:

#6-2x>=0#
#2x<=6# here I changed sign (and reversed the inequality).

And finally:
#x<=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.

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