F(g(x)) problem, about the domain.

[AznBoi]

Ok, f(x)=x^2 g(x)=sq.rt.(2-x)

Problem: f(g(x))

You end up with the answer 2-x but how come you need a domain for the answer? How come the domain is (-infinity,2]?

 

[drpizza]

Because while your answer simplifies to 2-x, you have to remember that there's a sqrt(2-x) in there... you can't take the square root of a negative number.

Here's an example: $$f(x)=\frac{x^2+3x+2}{x+2} = \frac{(x+2)(x+1)}{x+2}$$

While it's obvious that the expression simplifies, you have to remember that you can't divide by zero. Thus, the value of -2 for x is not allowed in the original function. If you simplify the function, it becomes x+1, x doesn't equal 2.

 

[Architeuthis Dux]

I would like to clarify that the domain is a set of all possible input values for a function. In this case, the function f(g(x)) is composed of two functions, f(x) and g(x). The domain of f(x) is all real numbers, while the domain of g(x) is restricted to values where the expression inside the square root (2-x) is greater than or equal to 0. This is because the square root function is only defined for non-negative values. Therefore, the domain of g(x) is (-infinity, 2] as any value greater than 2 would result in a negative value inside the square root, which is not allowed.

When we substitute g(x) into f(x), we are essentially plugging in values from the domain of g(x) into the function f(x). Since the domain of g(x) is restricted to (-infinity, 2], the resulting domain for f(g(x)) would also be restricted to (-infinity, 2].

In summary, the domain of a composite function like f(g(x)) is determined by the individual domains of the component functions. In this case, the domain of f(g(x)) is restricted to (-infinity, 2] due to the domain restrictions of g(x).