a
f(-x)g(-x)=-f(x)*-g(x)=f(x)g(x)
b
f(-x)g(-x)=f(x)*-g(x)=-f(x)g(x) lets call f(x)g(x)=h(x) -h(x)=h(-x)
c
f(-x)g(-x)=f(x)g(x)
A function f is even if: f(-x) = f(x)
A function f is off if: f(-x) = -f(x)
a) Suppose that f and g are both odd. Prove that y = f(x)g(x) is even.
b) suppose that f is even and g is odd. Prove that y = f(x)g(x) is odd.
c) Suppose that f and g are botheven. Prove that y = f(x)g(x) is odd.
My attempt for a), but I'm not sure if I did it correctly.
f(-x)g(-x) = -f(x) (-g(x))
Then I brought the (-)'s out on the left side
-f(x) (-g(x))= -f(x) (-g(x))
How would I do b and c) if it's not correct?
Okay.
So I always start the left side with f(-x) or g(-x)?
b) f(-x) * g(-x) = f(x) * (-g(x))
Then I take out the -1's like you said, for left side:
(-1) (-1) f(x) g(x) = - f(x)g(x)
f(x) g(x) = -f(x)g(x)
...? lol.
c) f(-x) g(-x) = f(x) g(x)
(-1)(-1) f(x) g(x) = f(x) g(x)
f(x) g(x) = f(x) g(x)
No, no, no....there's no "taking out -1's on the left." This doesn't make sense. You simply apply the def of even and odd.
f(-x)g(-x) = f(x)[-g(x)]=f(x)(-1)g(x)=(-1)f(x)g(x) = -f(x)g(x).
I'm simply working from left to right here.
Across the first equal sign I'm applying the definitions of even and odd.
Across the second equal sign I simply revealed the "hidden" 1.
Across the third equal sign I just used the fact that multiplication is commutative to move the -1 to the front.
And finally I hid the 1 again.
(In general hiding and unhiding the 1 is unnecessary. I just did it here because I think that this may be what is confusing you.)
To summarize, to check algebraically if a function is even or odd you replace x by -x, and see what happens,
The function we are checking here is f(x)g(x).
So I am replacing x in this function by -x. Thus I'm "checking" what f(-x)g(-x) is equal to.
Note: I am abusing notation a bit here to make things less confusing. This particular abuse of notation is pretty standard however, so I won't confuse you further by writing the technical details.