First of all the denominator cannot be equal to zero therefore you have to solve the irrational equation and exclude this solution(s). Otherwise is only defined if .
Okay so I am taking calculus 1 and thought I fully understood how to find the domain of functions until I saw one with a natural log on the top.
f(x)= [ln(12-2x)]/[sqrt(x+5)-3]
so far I have deduced that x can not equal -6 because a square root cant be a negative and that x can not equal 4 because the square root of 9 would be 3 and would give us a 0 on the bottom which we cant allow. However the entire top section has me puzzled. Can anyone please explain how to work with the top end of this function.
Thanks for your response I figured it out already. I just didn't know how to handle the natural log at first. If anyone else is looking at this thread with the same question I had... This is how it is solved.
So for the top portion we know that ln(x)>0 so if we solve for x we get 6 and know that we can not include six because the ln would be equal to 0 when it needs to be greater then. As for the bottom portion I described it in my initial question so the domain would be (in interval notation)
[-5,4)U(4,6)
What in the world do you mean by "ln(x)> 0"? There is no reason why the numerator cannot be zero or negative. What is true is that ln(x) is defined only for x> 0. Since you have ln(12- 2x) you must have 12- 2x> 0 which gives x< 6.