Question

THIS QUESTION WILL SHOCK YOU?!?! 100% MEDAL!!! INSANE GLITCHED POST!!!

Given that function f(x) is even, prove that g(f(x)) is even regardless of whether g(x) is even, odd, or neither.

Answer 1

Even and odd have to do with the behavior of the function when you replace x with negative x (what happens to y when x changes sign)

if (x , y) becomes (-x, y), then it is even, of f(-x) = f(x)
if (x , y) becomes (-x, -y), then it is odd, or g(-x) = -g(x)
if (x , y) becomes (-x , z), then it is neither h(-x) <> h(x) nor -h(x)

1) f(x) / g(x)
f(x) doesn't change signs, but g(x) does, so f(-x) / g(-x) = f(x) / -g(x)
this is odd (the quotient changes signs when x changes sign)

2) g(-x) = -g(x) (odd function)
f(g(-x)) = f(-g(x)) = f(g(x)), since f is even; therefore, the composite is even

3) f(-x) = f(x) (since f is even)
g(f(-x)) = g(f(x)) , therefore this is even (changing sign of x doesn't change the sign of the composite

4) f(-x) = f(x)
g(-x) = -g(x)
f(-x) - g(-x) = f(x) + g(x); therefore neither even nor odd

an example of an even function: f(x) = x^2
f(2) = f(-2) = 4

an example of an odd function: g(x) = x^3
g(2) = 8; g(-2) = -8

an example of neither: h(x) = x^2 - x
h(1) = 0
h(-1) = 2

Answer 2

Wow :0