find the natural domain of the function:
f(x)= square root of (x^2-9)(x+5)/x-5
F(x) = sqrt(x^2-9)(x+5) / (x-5).
If x = 5, the denominator would be "0" and the function would be undefined.
So 5 is not in the domain. All real #s
>5 gives a positive # under the radical
and is, therefore, in the domain.
The # 4 gives a negative # under the radical and is, therefore, not in the
domain.
Conclusion: All real #s >5 and <4 are
in the domain.
Mathematically, 4 > X > 5.