Let f be any function with the property that x is in the dom

Let f be any function with the property that -x is in the domain of f whenever x is in the domain of f, and let E and O be the functions defined by E(x) = 1\\2[f(x) + f(-x)] and O(x) = 1/2[f(x) -f(-x)] (A) Show that E is always even. (B) Show that O is always odd. (C) Show thatf(x) = E(X) + O(x). what is your conclusion?

Solution

a) E(x) = (1/2)[ f(x) +f(-x) ]

To prove E(x) is even , E(-x) = E(x); plug x= -x

So, E(-x) = (1/2[ f(-x) +f(x) ]

So, (1/2)[ f(-x) +f(x) ] = E(x)

Proved that E(x) is always even

b) O(x) = (1/2)[ f(x) - f(-x) ]

To prove odd funtion we have to show f(-x) = -f(x)

plug x= -x

O(-x) = (1/2)[ f(-x) - f(x) ]

= -(1/2)[ f(x) - f(-x) ]

= -O(x)

Hence O(x) is odd function

c) Add E(x) +O(x) = (1/2)[ f(-x) +f(x) ] + (1/2)[ f(x) - f(-x) ]

= 1/2 [ f(-x) +f(x) +f(x) - f(-x) ]

=1/2[2*f(x)]

= f(x)

Hence Proved