Let f(x) and g(x) be two odd functions defined for all real values of x. Given that s(x)=f(x)+g(x), prove that s(x) is also an odd function.

We recall that a function f(x) is said to be an odd function when f(-x)=-f(x).

We are told that f(x) and g(x) are odd functions, so we know from the above definition that:

1. f(-x)=-f(x)

2. g(-x)=-g(x)

Solution

We want to show that s(x) is an odd function. In other words, we want to show that s(-x)=-s(x) (that it satisfies the above definition).

We are told that s(x)=f(x)+g(x), so substituting x for -x, we get that

s(-x)=f(-x)+g(-x)

=-f(x)-g(x) (using 1 and 2)

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

=-s(x) as required!

We have now shown that s(-x)=-s(x) and thus we have proven that s(x) is indeed an odd function.