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Let e(x) be an even function, and let o(x) be an odd function, such that \(e(x) + o(x) = \frac{6}{x + 2} + x^2 + 2^x\)
for all real numbers x. Find o(1).

 Oct 2, 2019

Best Answer 

 #1
avatar+6046 
+1

\(\text{any function $f(x)$ can be decomposed into its even and odd parts}\\ f(x) = e(x)+o(x)=\dfrac{6}{x+2}+x^2+2^x\\ e(x) = \dfrac 1 2 \left(f(x)+f(-x)\right)\\ o(x) = \dfrac 1 2 \left(f(x) - f(-x)\right)\\~\\ o(1) = \dfrac 1 2(f(1) -f(-1))\\~\\ \text{I leave you to plug and chug}\)

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 Oct 2, 2019
 #1
avatar+6046 
+1
Best Answer

\(\text{any function $f(x)$ can be decomposed into its even and odd parts}\\ f(x) = e(x)+o(x)=\dfrac{6}{x+2}+x^2+2^x\\ e(x) = \dfrac 1 2 \left(f(x)+f(-x)\right)\\ o(x) = \dfrac 1 2 \left(f(x) - f(-x)\right)\\~\\ o(1) = \dfrac 1 2(f(1) -f(-1))\\~\\ \text{I leave you to plug and chug}\)

Rom Oct 2, 2019
 #2
avatar+106539 
+1

Thanks, Rom......I learned something new today from this !!!

 

To see how Rom arrived at his answer, look at this easy-to-understand proof  :

 

http://mrhonner.com/archives/8957

 

 

cool cool cool

 Oct 3, 2019
edited by CPhill  Oct 3, 2019
edited by CPhill  Oct 3, 2019
 #3
avatar+6046 
+1

I should have been more careful.

 

My statement is true if for all x such that f(x) is defined, then f(-x) is defined as well.

Rom  Oct 3, 2019

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