# 3.3.notes - candidates for points of in±ection(a When f...

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MAC 2233/001-006 Business Calculus, Fall 2011 CRN: 81700–81702, 81705, 81716, 81715 Assistants : Matthew Fleeman, Arbin Rai, Tadesse Zerihun, Vindya Kumari, Junyi Tu, Jing- han Meng An example on concavity Example . (Similar to #5, #6 in § 3.3) Determine the intervals on which the graph of the function given below is concave up, and intervals on which the graph is concave down. Find also all point of inflection. f ( x ) = 1 x 2 + 3 . Solution . Step 1 . We need to find the first and second derivatives. The following is one of many ways. We first re-write the function as f ( x ) = ( x 2 + 3) 1 , and then use the chain rule to find f ( x ) = - 2 x ( x 2 + 3) 2 . For the second derivative, we use the product rule, and then the chain rule to get f ( x ) = ( - 2)( x 2 + 3) 2 + ( - 2 x )( - 2)( x 2 + 3) 3 (2 x ) = 2 x 3 - 18 x ( x 3 + 3) 3 = 6( x 2 - 1) ( x 2 + 3) 2 .
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Unformatted text preview: candidates for points of in±ection. (a) When f ′ ( x ) = 0, we have x 2-1 = 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 values-10 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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