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 ⇒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) ⇒(f+g)(−x)=−(f+g)(x) ⇒f+g is an odd function. Q.E.D. 
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[Math] A proof of Triangle Inequality (1)

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Prove: |x+y|≤|x|+|y| (triangle inequality) 


xy≤|xy| 

⇔xy≤|x||y| 

⇔2xy≤2|x||y|

⇔x2+y2+2xy≤x2+y2+2|x||y| 

⇔x2+y2+2xy≤|x|2+|y|2+2|x||y| 

⇔(x+y)2≤(|x|+|y|)2 [equation 1]


case 1) if x+y>0

then |x+y|=x+y 

[equation1] ⇒(|x+y|)2≤(|x|+|y|)2 [equation 2]

⇔|x+y|≤|x|+|y| 

case 2) if x+y<0

then |x+y|=−(x+y) 

note that (−(x+y))2=(x+y)2=|x+y|2

⇒ [equation 2] ⇒ |x+y|≤|x|+|y| ◼


Published on: January 10, 2023 

By: ComputeFinance

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

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