h(x) = 1/x-3 + 1
I know the vertical and horizontal asymptotes but I just need to know the domain. Please help. I am very confused.
To find the domain of a function, you need to consider any restrictions on the variable, if any. In the case of the function h(x) = 1/(x - 3) + 1, there is one restriction that you need to be aware of.
The expression (x - 3) appears in the denominator of the fraction. A fraction is undefined when the denominator equals zero because division by zero is undefined. Therefore, the value of x that makes the denominator zero, which is 3 in this case, is not allowed in the domain.
To find the domain, you need to exclude 3 from the real numbers. Thus, the domain of h(x) = 1/(x - 3) + 1 is all real numbers except x = 3. This can be written as:
Domain: (-∞, 3) U (3, +∞)
In interval notation, this means that x can take any value less than 3 or any value greater than 3.