Proof Involving Functions g & h

Section 1.8
Question 73

See attachment.

Can you set up part (a) for me?

I need your set up for part (a) to answer part (b).

What exactly is part (c) asking to do?

 

a) given function f, prove that g is even and h is odd, where

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

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

to prove g(x) is even show that g(-x)=g(x)

g(-x)=(1/2)[f(-x)+f(-(-x))]
g(-x)=(1/2)[f(-x)+f(x)]
g(-x)=(1/2)[f(x)+f(-x)]
g(-x)=g(x) -> proven that g is even

to prove h(x) is odd, show that h(-x)=-h(x) => prove it

 

I will do this later on my break at the job. Now going to sleep for a few hours.

 

Let me see.

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

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

How do I make the right side -h(x)?

Stuck....

 

For part (b), I must add g(x) + h(x).

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

g(x) + h(x) = (1/2)f(x) + (1/2)f(-x) + (1/2)f(x) - (1/2)f(-x)

g(x) + h(x) = (1/2)f(x) + (1/2)f(x)

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

If this is right, I have no idea what I just did. It is just mechanical after a certain step for me.

 

If my work for part (b) is right, for part (c), I simply
add f(x) + k(x).

Yes?

 

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


h(-x)=-h(x)

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

h(-x)=(1/2)[f(-x)-f(x)]...factor out -1

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

h(-x)=-h(x)

 

correct

 

Very good. Moving on to Section 1.9 or Inverse Functions.