MATH116 Lecture Notes - Lecture 16: Opata Language, Hypotenuse
9 Dec 2015
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Tutorial solutions 2: recall from the practice problems that the even and odd parts of any function f are de ned, respectively, as fe(x) = fo(x) = Let f (x) = 6x5 4x2 + 2x + 1. (a) find the even and odd parts of f (x). (f (x) + f ( x)), (f (x) f ( x)). 2 (6x5 4x2 + 2x + 1 ( 6x5 4x2 2x + 1)) = 2 (6x5 4x2 + 2x + 1 (6( x)5 4( x)2 + 2( x) + 1)) 2 (12x5 + 4x) = 6x5 + 2x (b) we realize that neither the even part nor the odd part is f itself, which suggests that f is neither even nor odd. An alternate approach is to go straight to the de nition: f ( x) = 6( x)5 4( x)2 + 2( x) + 1 = 6x5 4x2 2x + 1.
