Suppose f(x)= 1/x and g(x)= x-2. Also, suppose h(x)=f(g(x)). Then h(x)= 1 / (x-2). Find the domain of h(x)= 1 / (x-2). Express the domain using inequality notation.
A-Rod, two posts below, had a similar question, take a look at my
reply to that one, and see if applies to yours as well.
I'm still a bit confused sorry.
The h(x)= is throwing me off. Then I also don't know how to express the domain using inequality notation.
x ≠ 2
x<2 OR x>2
x ε (-∞,2)U(2,∞)
even though f(x)= 1/x is imbedded in h(x)
we have to restrict f(x) and say x ≠ 0.
from h(x)=f(g(x)) = 1/(x-2), we must have x ≠ 2
Since I am old school when it comes to the new notation, somebody
might want to kick in here.
I would say:
x ∈ R, x ≠ 0,2
So would it be:
(-∞,2)U(2,∞)
OR
(-∞,0)U(0,2)U(2,∞)
my ad
(-∞,0)U(0,2)U(2,∞)
To find the domain of a function, we need to identify all the values of x for which the function is defined. In this case, we want to find the values of x that make h(x) = 1 / (x - 2) defined.
Since the function h(x) = 1 / (x - 2), we know that the denominator (x - 2) cannot be equal to zero, as division by zero is undefined. Therefore, we need to find the values of x that make the expression (x - 2) ≠ 0.
To solve the inequality (x - 2) ≠ 0, we can add 2 to both sides of the inequality:
x - 2 + 2 ≠ 0 + 2
x ≠ 2
So, the domain of h(x) = 1 / (x - 2) is all real numbers except x = 2. In inequality notation, we express this as:
x < 2 or x > 2