Question 980559
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*[tex \LARGE f(x)\ =\ f(-x)\ \ \ \ \ \ ]Definition of even function


*[tex \LARGE g(x)\ =\ g(-x)\ \ \ \ \ \ ]Definition of even function


*[tex \LARGE f(x)\ +\ g(x)\ =\ f(-x)\ +\ g(-x)\ \ \ \ \ \ ]Addition property of Equality


*[tex \LARGE h(-x)\ =\ f(-x)\ +\ g(-x)\ \ \ \ \ \ ]Definition *[tex \LARGE h(x)]


*[tex \LARGE h(x)\ =\ h(-x)\ \ \ \ \ \]Substitution


*[tex \LARGE h(x)] is even by definition of even function.  Q.E.D.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \