Unformatted text preview: candidates for points of in±ection. (a) When f ′ ( x ) = 0, we have x 21 = 0, giving that x = ± 1. (b) When f ′ ( x ) is unde²ned, we need x 2 + 3 = 0, which is not possible. Therefore, the only candidates are x = ± 1. Step 3 . We construct the following table: intervals (∞ ,1) (1 , 1) (1 , ∞ ) test values10 10 sign of f ′′ ++ concavity up down up Conclusion . The graph is concave up on (∞ ,1) and (1 , ∞ ) , and it is concave down on (1 , 1) . There is a change of concavity at x ± 1. Since f (1) = f (1) = 1 4 (are de²ned), we see that the points of in±ection are (1 , 1 4 ) and (1 , 1 4 ). The graph of the function is given on the right....
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 Fall '08
 DANIELYAN
 Calculus, Chain Rule, Derivative, Jinghan Meng, Matthew Fleeman, Arbin Rai, Vindya Kumari

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