precalculus solutions 1.2 #3

C x 19 f x x 5 2 2 x x 2 10 x 25 2 x

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Unformatted text preview: appears even Prove: c( − x) = c( x) (− x) 2 x2 c( − x) = = = c( x) (− x) 2 + 4 x 2 + 4 ∴ c is even 18. c( x) = 19. f ( x) = ( x − 5) 2 − 2 x = x 2 − 10 x + 25 − 2 x = x 2 − 12 x + 25 appears neither Prove: f ( − x) ≠ f ( x) f ( − x) = (− x ) 2 − 12(− x) + 25 = x 2 + 12 x + 25 ≠ f ( x) ∴ f is not even Prove: f ( − x) ≠ − f ( x) from above we know that f ( − x) = x 2 + 12 x + 25 − f ( x) = −[ x 2 − 12 x + 25] = − x 2 + 12 x − 25 f ( − x) ≠ − f ( x) ∴ f is not odd x3 2x2 − 9 appears odd Prove: g (− x ) = − g ( x) ( − x)3 − x3 g (− x) = = 2(− x) 2 − 9 2 x 2 − 9 20. g ( x) = [ x3 − x3 − g ( x) = − =2 2x2 − 9 2x − 9 g (− x ) = − g ( x) ∴ g is odd...
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