What is the domain of g(x) = square root of x +1
f(x) is x +1 >= 0 {nl} {nl} x + 1 >= 0 {nl} x >= -1 {nl} {nl} for g(x), you cannot divide a number by zero. So, you set the denominator equal to zero and solve: {nl} x - 1 = 0 {nl} x != 1 (x is not equal to 1)
I think that's right? If not, I'm sorry!!
so would it be...
**all real numbers greater than or equal to 1
**all real numbers greater than or equal to 0
**all real numbers greater than 0
**all real numbers
It would be the First one, then.
The DOMAIN is the all of the values of x for which the function is valid
so as long as x+1 is positive (or zero) it is in the domain
x+1 >= 0
x>= -1 is the domain (from -1 to +infinity)
So it would be all real numbers?
NO.....real numbers include irrational numbers. x >= -1 is the domain
