f(x) = abs(sin(x)), -pi less than or equal to x less than or equal to pi
g(x) = x^2
let H(x) = g(f(x))
i got that H(x) would equal sin^2(x).
THIS IS WHERE I AM STUCK!
find the domain of H. do i still need to account for "-pi less than or equal to x less than or equal to pi" or is it all real numbers?
Since they put that restriction on the domain of f(x), the same restriction should apply to g(f(x)). Ignore my previous comment about the domain being all real numbers in this case.
To find the domain of the function H(x) = g(f(x)), we need to consider the domains of both the inner function f(x) and the outer function g(x).
Given that f(x) = |sin(x)|, we know that the absolute value of sin(x) will always be between 0 and 1, inclusive. Since the domain of sin(x) is all real numbers, the domain of f(x) will be the same, i.e., all real numbers.
On the other hand, g(x) = x^2 is a polynomial function, and its domain is also all real numbers. Any real number can be squared.
Therefore, since both f(x) and g(x) have a domain of all real numbers, the composition H(x) = g(f(x)) will also have a domain of all real numbers.
In conclusion, the domain of H(x) is all real numbers. You do not need to consider the restriction "-π ≤ x ≤ π" for the determination of the domain in this specific case.