Suppose g is an even function and f is any function such that the composition fog is defined. show that fog is an even function
To show that the composition function fog is even, we need to prove that for every value x in the domain of fog, the following holds:
fog(x) = fog(-x)
Given that g is an even function, we know that for every value x in its domain, the following holds:
g(x) = g(-x)
Now, we can use these properties to prove that fog is also an even function.
Let's start by recalling the definition of the composition of two functions:
fog(x) = f(g(x))
To prove that fog is even, we need to show that for every x in the domain of fog, fog(x) is equal to fog(-x):
fog(x) = fog(-x)
Now, let's substitute the definitions of fog(x) and fog(-x) using the composition definition:
f(g(x)) = f(g(-x))
Since for every value in the domain of g, g(x) = g(-x), we can replace g(x) with g(-x) in the equation:
f(g(-x)) = f(g(-x))
Therefore, we have shown that fog(x) = fog(-x), meaning that fog is an even function.
Note: It's important to mention that this proof assumes that f is defined for all values of g(x) and g(-x) in the domain of fog. If there are any restrictions or conditions on the composition of f and g, those would need to be considered separately.