How do you find the domain of #sqrt(x+4)/(x-3)#?

Question

Julia Adams · Accepted Answer

D: [-4, 3) #uu# (3, #oo#)

You would find the domain of the numerator and the domain of the denominator, see if there are any intersections, and then write the answer in interval notation. The numerator is a square root, so whatever is inside must be positive. So, you have #x + 4 >= 0#, which solves to see that #x# must be greater than or equal to -4. In the denominator, you can't get 0, as this will evaluate to undefined. So, if #x - 3 = 0#, you see that #x# can't equal 3. So, in interval notation, you show that #x# must be greater than or equal to -4, but not equal to 3.

Eva Abramson · Answer

#x inRR,x!=3# #"the denominator of the rational function cannot be"# #"zero as this would make it "color(blue)"undefined"# #"Equating the denominator to zero and solving gives the"# #"value that x cannot be"# #"solve "x-3=0rArrx=3larrcolor(red)"excluded value"# #"domain is "x inRR,x!=3#

Jace Anderson · Answer

undefined To find the domain of the function sqrt(x+4)/(x-3), we need to consider the values of x that make the function undefined. Since the square root function requires a non-negative argument, the expression inside the square root must be greater than or equal to zero. Additionally, the denominator cannot be equal to zero. Therefore, we set both conditions:

1. x + 4 ≥ 0
2. x - 3 ≠ 0

Solving these inequalities:
1. x + 4 ≥ 0 → x ≥ -4
2. x - 3 ≠ 0 → x ≠ 3

Therefore, the domain of the function is all real numbers greater than or equal to -4, excluding x = 3. In interval notation, the domain is (-∞, -4] ∪ (-4, 3) ∪ (3, ∞).