Question (1 Viewer)

[withoutaface]

I have a question which I can't get, its probably something simple, here it is:

Suppose f(x) can be expressed as f(x)=h(x)+g(x)

g(x) is even and h(x) is odd

express g(x) in terms of f(x)

 

[mojako]

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

doesn't that equal zero?

Is that the only possible solution? can you give more details on the process?
thanks.

 

[Xayma]

No. It doesnt.

If g(x) the part of f(x) that is even. g(-x)=g(x).

h(-x)=-h(x)

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

∴ f(-x)=-h(x)+g(x)
f(x)=g(x)+h(x) from above
∴ f(-x)+f(x)=2g(x)
∴ g(x)=[f(-x)+f(x)]/2

 

[mojako]

oops... I was thinking of f(x) being h(x) and hence an odd function ^^
i need more sleep haha