How to find the domain of a rational function with square roots (we need three conditions)

This precalculus tutorial covers finding the domain of a rational function with square roots. The function is f(x)=sqrt(12-x)/(3-sqrt(4+x)). We need to set up three restrictions for this function. Be careful! Subscribe to @bprpmathbasics for more precalculus tutorials.

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#math #algebra #mathbasics

[Ninja]

We can also write the domain as [4, 12]\{5}, which is the way i prefer because its easier to think of and less to write. But you do you.

[epsi]

This one was easy, but fun!
So satisfying!

Also, you can combine the 4+x≥0 and 3-√(4+x)≠0 work since 4+x is a sub-expression of 3-√(4+x):

3 - √(4 + x) ≠ 0
-3- - √(4 + x) - -(-3)- ≠ 0 - (-3)
-√(4 + x)/(-1) ≠ -3/(-1)
√(4+x) ≠ 3
√(4+x)² ≠ 3²
4 + x ≠ 9

4 + x ≥ 0
0 ≤ 4 + x

0 ≤ 4 + x ≠ 9
0 - 4 ≤ -4- + x - -4- ≠ 9 - 4
-4 ≤ x ≠ 5

So the domain is [-4, 5) so far, but there's still the 12-x≥0 to deal with:

12 - x ≥ 0
12 - -x- + -x- ≥ 0 + x
12 ≥ x
x ≤ 12
5 ≠ x ≤ 12 # x ≠ 5, from above

This means we have the final result using interval notation:

[-4, 5) U (5, 12]

Note that this simply means -4 ≤ x ≤ 12 where x≠5; it does not mean x<5 and 5<x are both true at the same time, which would be a contradiction since x cannot be both less than 5 and greater than 5 simultaneously.

[JayTemple]

2:24 Me, teaching inequalities: The first classic blunder is failing to change the inequality when dividing by a negative. The second is only slightly less familiar. Never tangle with a Sicilian when death is on the line.

[GrimAxel]

What's interesting is, if you instead have the bottom as *4*-sqrt(4+x), the range becomes [4, 12)

[falkez]

but why cant we just assume this function also has a place in imaginary world?