Suppose that f(x) is an even function, and g(x) is an odd function.?

Question

Suppose that f(x) is an even function, and g(x) is an odd function. Consider the function h(x) = (f ◦ f)(x) + (f ◦ g)(x) + (g ◦ f)(x). Is h(x) odd, even or neither?

Answer 1

Let's find h(-x)

h(-x) = (f ◦ f)(-x) + (f ◦ g)(-x) + (g ◦ f)(-x) = (f ◦ f)(x) + f(-g(x)) +
+ (g(f(-x)) = (f ◦ f)(x) + f(g(x)) + g(f(x)) =
= (f ◦ f)(x) + (f ◦ g)(x) + (g ◦ f)(x) = h(x)

Thus, h is even.

Answer 2

whether an incredible variety of the above solutions are sturdy, i'm uncertain in the event that they actually answer the question coping with products of even and mind-blowing functionality. So, i'm going to offer it a whirl. the applicable definition of a good functionality (a minimum of in my view) is that that's represented as a series (even no count if that's infinite) if words containing x to even exponents. for this reason, it takes on the variety: f(x)=a(x^0)+b(x^2)+c(x^4) ... or a+b(x^2)+c(x^4) ... for some constants a,b,c. an identical case would properly be made for mind-blowing purposes. Now analyzing the manufactured from 2 even purposes: f(x)*g(x)= [a+b(x^2)+c(x^4) ... ]*[d+e(x^2)+f(x^4) ... ] =a*d+a*e(x^2)+a*f(x^4) ... +b(x^2)*d+b(x^2)*e(x^4).... = advert+ae(x^2)+af(x^4) ... + bd(x^2) +be(x^6) ... and properly, it may get gruesome, yet you will observe the exponents of x continuously stay even, the manufactured from 2 even purposes are even. As for mind-blowing purposes... h(x)*j(x)= [a(x^a million)+b(x^3)+c(x^5) ... ]*[d(x^a million)+e(x^3)+f(x^5) ... ] = a(x)*d(x)+a(x)*e(x^3) ... +b(x^3)*d(x^a million) +b(x^3)*e(x^3) ... = advert(x^2)+ae(x^4) ... +bd(x^4) +be(x^6) ... those exponents shrink to even exponents, so returned, the product is often even.

Answer 3

ODD