Thread: How to find the domain for this function?

the problem is p(x)= [³√(x-6)]/√(x²-x-30)
From what i know you have to make sure that the denominator isnt equal to zero that there is no negative in the square root sign. Those values that do make it equal to zero and a negative in the square root sign must be excluded from the domain. I dont have a clue on how to start? Im also wondering if the top cube root includes a value that must also be excluded from the domain?

Originally Posted by francisco1741

the problem is p(x)= [³√(x-6)]/√(x²-x-30)
From what i know you have to make sure that the denominator isnt equal to zero that there is no negative in the square root sign. Those values that do make it equal to zero and a negative in the square root sign must be excluded from the domain. I dont have a clue on how to start? Im also wondering if the top cube root includes a value that must also be excluded from the domain?

There is no restrictions on cube roots.
Note , solve that.

Originally Posted by Plato

There is no restrictions on cube roots.
Note , solve that.

So i did and got 6 and -5. Now that means that those two values cannot be the domain?
Would me answer be : All real numbers except where x=6 and x=-5

Originally Posted by francisco1741

So i did and got 6 and -5. Now that means that those two values cannot be the domain?
Would me answer be : All real numbers except where x=6 and x=-5

No that is not true. It is not for example.
We cannot have square roots of negative numbers.

So then we cannot have the denominator =0 and we cannot have square root of negative numbers right? Then how would my answer be like?

Originally Posted by francisco1741

So then we cannot have the denominator =0 and we cannot have square root of negative numbers right? Then how would my answer be like?

Can you do the elementary algebra to solve

I dont know why but i wrote = insteaad of >. I got x<-5 and x>6

Originally Posted by francisco1741

i dont know why but i wrote = insteaad of >. I got x<-5 and x>6

correct!