Suppose f is an even function and g is an odd function such that the composition fog is defined.d show that fog is an even function
since f(x) is even, f(-z) = f(z)
g(-x) = -g(x) since g is odd
f(g(-x)) = f(-g(x)) = f(g(x)) so fog is even
To show that the composition fog is an even function, we need to demonstrate that fog satisfies the definition of an even function.
First, let's review the definitions of even and odd functions:
- An even function is a function where f(-x) = f(x) for all values of x. In other words, the function is symmetric about the y-axis.
- An odd function is a function where f(-x) = -f(x) for all values of x. In other words, the function is symmetric about the origin.
Now, let's prove that the composition fog is an even function using the given information that f is even and g is odd.
Since g is an odd function, we have g(-x) = -g(x) for all values of x.
Now, consider the composition of fog for some value of x:
(fog)(x) = f(g(x))
To show that fog is even, we need to prove that (fog)(-x) = (fog)(x).
Using the definition of the composition, we have:
(fog)(-x) = f(g(-x))
Now, since g is an odd function, we can substitute -g(x) for g(-x):
(fog)(-x) = f(-g(x))
Since f is an even function, we know that f(-x) = f(x) for all values of x.
Therefore, we can write the above expression as:
(fog)(-x) = f(g(x))
Now, comparing this to our expression for (fog)(x), we have shown that:
(fog)(-x) = (fog)(x)
This demonstrates that the composition fog is an even function as the output for -x is the same as the output for x.
In conclusion, if f is an even function and g is an odd function, then the composition fog is an even function.