Expert: Alon Mandes - 10/25/2009

Question
prove that f(x)-(f-x) is an odd function but f(x)+f(-x) is a an even function of X .

Show with detail working do not write only even function f(x)= f(-x) even function and f(-x)= -f(x)

Answer
Hello Gaurav
1st of all we must remember the definition of even %26 odd, %26 here they are :
Function h(x) is even if %26 only if : h(x)=h(-x)
Function h(x) is even if %26 only if : h(-x)=-h(x)
This is in general.
Now, we have a function f(x)that could be either 1 of two or none .

Let's set g(x) to be : g(x)=f(x)-f(-x) . We have to prove now that g(x) is odd. To prove it
we have to show that g(x) obay the rule : g(-x)=-g(x).Ok, here we go :
# Let's write an expression for g(-x) :
 If g(x)=f(x)-f(-x) then g(-x)=f(-x)-f(-(-x))=f(-x)-f(x) .
# Let's write an expression for -g(x) :
 If g(x)=f(x)-f(-x) then -g(x)=-f(x)-(-f((-x))=-f(x)+f(-x) .
It's clear that g(-x)=-g(x) . Q.E.D

Let's set h(x) to be : h(x)=f(x)+f(-x) . We have to prove now that g(x) is odd. To prove it
we have to show that h(x) obey the rule : h(-x)=h(x).Ok, here we go :
# Let's write an expression for h(-x) :
 If h(x)=f(x)+f(-x) then g(-x)=f(-x)+f(-(-x))=f(-x)+f(x) .
It's clear that g(-x)=g(x) . Q.E.D

Alon.