Then you must mean this: √[(x + 1)/(x² + 3x - 10)]

The argument of the square root must be non-negative. It follows then that either the numerator and denominator have the same sign, or the numerator is zero while the denominator is non-zero.

numerator > 0 at x > -1

numerator = 0 at x = -1

numerator < 0 at x < -1

(x² + 3x - 10) = (x + 5)(x - 2)

denominator < 0 at -5 < x < 2

denominator > 0 at x < -5 or x > 2

numerator > 0 and denominator > 0:

x > -1 and (x < -5 or x > 2)

x > 2

numerator < 0 and denominator < 0:

x < -1 and -5 < x < 2

-5 < x < -1

numerator = 0 and denominator ≠ 0

x = -1

Domain: (-5, -1] U (2, ∞)

OK, so we know that the domain is all possible values for which our function is defined. So we know that our example allows all possible values for x, for which our function is defined.

OK, so our function is this........ √[(x+1)/(x^2+3x-10)].

OK, so our first group of possible values are for which (x^2+3x-10) ≠ 0. How do we know that? Well, inside the square_root is a denominator. We can't divide by zero! Our denominator can't be zero!

So, we have (x^2+3x-10) ≠ 0. OK, good. That's our first part.

OK, and we remember that we can never take the square root of a negative number. That would be undefined!

OK, good. So..... [(x+1)/(x^2+3x-10)] ≠ negative numbers.

So, our domain is any value of x such that (x^2+3x-10) ≠ 0 and [(x+1)/(x^2+3x-10)] ≠ negative numbers.

OK, we did it!!!!!!!!!!