Thread: Domain of a expression involving a radical

This is to confirm that my knowledge is correct on this point
when the function has a radical like a sqrt and the interger is even (4) then the domain is equal to or greater then 0

Yes, an even root must be a non-negative value to be real.

So in terms of this specific question the Domain would be (0,infinity) in interval notation

No, you want to solve the inequality:



You will find two intervals in the solution.

OK I get 0 and 5 as my two interval this means then my domain is all real number except 0 and 5 so in interval notation
(-infinity,0) U (0,5) U (5, infinity)

You have found the correct critical numbers, but you need to test values within the sub-intervals (which are closed since we have a weak inequality):

We have:



a)

We can pick any value in the interval not at the ends, let's pick . For this value, we look at the resulting signs of the two factors on the left of our inequality. This will give us two negative factors, meaning the expression is positive, so this interval is part of the solution.

b)

Let's pick 1 as our test value. Now we have a positive factor and a negative factor, meaning the expression is negative, so this interval is not part of the solution.

c)

Let's pick 6 as our test value. Now we have two positive factors, meaning the expression is positive, so this interval is part of the solution.

Hence, the domain is:



This is a general method, but in this case we could simply consider the graph of the parabolic radicand, and note that is is only negative on the open interval between its roots.

Ohhh I see my mistake