To determine whether h(fg) is always an even function given that g is an even function, we first need to understand the definition of an even function. A function f(x) is called even if it satisfies the condition f(-x) = f(x) for all x in its domain.
Since g is an even function, we have:
- g(-x) = g(x)
Now, let’s examine the product fg, which involves two functions, f and g. The evenness of fg will depend on both functions. If f is also an even function, then:
- fg(-x) = f(-x)g(-x) = f(x)g(x) = fg(x)
This shows that if both f and g are even functions, then fg is also an even function. However, if f is not even, we cannot conclude that fg is even merely because g is even.
Next, we need to analyze the function h(fg). The evenness of h(fg) depends on the form of the function h and the properties of the function fg.
For h(fg) to be an even function, it must satisfy:
- h(fg(-x)) = h(fg(x))
So, without specific details about the functions h and f, we can conclude that h(fg) is not necessarily always an even function. It heavily relies on the nature of both h and f. If h is an even function and fg is even, then h(fg) will indeed be even.
In summary, the evenness of h(fg) cannot be established without additional information about f and h. Therefore, we cannot say that h(fg) is always an even function just because g is an even function.