5 but 2 45 should get some sort of consolation prize

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Unformatted text preview: (−x), so i is even. The calculator supports our conclusion. 5. x −1 100 −x j (−x) = (−x)2 − −1 100 x −1 j (−x) = x2 + 100 j (x) = x2 − The expression for j (−x) doesn’t seem to be equivalent to j (x), so we check using x = 1 to 1 1 get j (1) = − 100 and j (−1) = 100 . This rules out j being even. However, it doesn’t rule out j being odd. Examining −j (x) gives x −1 100 x −j (x) = − x2 − −1 100 x −j (x) = −x2 + +1 100 j (x) = x2 − The expression −j (x) doesn’t seem to match j (−x) either. Testing x = 2 gives j (2) = and j (−2) = 151 , so j is not odd, either. The calculator gives: 50 149 50 70 Relations and Functions The calculator suggests that the graph of j is symmetric about the y -axis which would imply that j is even. However, we have proven that is not the case. There are two lessons to be learned from the last example. The first is that sampling function values at particular x values is not enough to prove that a function is even or odd...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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