Question 400822
"Domain" specifies the "range of values" x can take for the given function.
.
In your case, there are two conditions
- you don't want the stuff under the radical to be negative, because the square root of a negative produces an imaginary number
x-3 >= 0
x >= 3
that is, 'x' is greater than or equal to 3
.
- And, you don't want a zero in the denominator (because that produces an "undefined" number)
to find what value x can't take we set
x^2-8x+12
to zero and solve for x:
x^2-8x+12 = 0
(x+2)(x-6) = 0
x = {-2, 6}
.
Solution:
Domain x >= 3 AND "x does not = 6"
Or,
[3,6) U (6, +oo)
where
oo represents infinity