Prove that (f+g)(x) is an odd function, if f and g are odd functions (Stewart, Calculus)

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Suppose f(x) and g(x) are odd functions. Prove that (f+g)(x) is also an odd function.  Answer:  1. Strategy By definition, f is an odd function if and only if f(-x) = - f(x) To show (f+g) is an odd function, we need to show (f+g)(-x) = - (f+g)(x) 2. Explanation Since $f(x)$ and $g(x)$ are odd functions $\Rightarrow f(-x) =-f(x)$ and $g(-x) =-g(x)$ By definition of sum of functions. $(f+g)(-x) =f(-x)+g(-x)$ $=-f(x)-g(x)$ $=-(f(x)+g(x))$ $=-(f+g)(x)$ (by definition of sum of functions) $\Rightarrow(f+g)(-x) =-(f+g)(x)$ $\Rightarrow f+g$ is an odd function. Q.E.D. 
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Proof: f + g is an even function, if f and g are even functions. (Stewart, Calculus)

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Question: Suppose f(x) and g(x) are even functions. Prove that (f+g)(x) is also an even function. 


Answer:

1. Strategy

Definition: a function h(x) is an even function if h(x) = h(-x)

To prove (f+g)(x) is an even function, we need to show (f+g)(x) = (f+g)(-x) 


2. Explanation 

Recall $(f+g)(x)=f(x)+g(x)$ (by definition of sum of functions)

Since $f(x)$ and $g(x)$ are even functions

$\Rightarrow f(x)=f(-x)$ and $g(x) = g(-x)$

$\Rightarrow(f+g)(x)=f(x)+g(x)=f(-x)+g(-x)=(f+g)(-x)$

$\Rightarrow(f+g)(x)=(f+g)(-x)$

$\Rightarrow(f+g)$ is an even function. Q.E.D. 


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