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# 3.1.notes - 5 5 1 5 2 sign of 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 finding intervals on which f is increasing or f is decreasing Example . (Similar to #31 in Section 3.1) Find open intervals on which the function below is increasing or decreasing: f ( x ) = x 3 x 2 1 . Solution . Step 1 . The derivative is f ( x ) = (3 x 2 )( x 2 1) ( x 3 )(2 x ) ( x 2 1) 2 = x 4 3 x 2 ( x 2 1) 2 = x 2 ( x 2 3) ( x 2 1) 2 . Step 2 . (a) When f ( x ) = 0, we have x 2 ( x 2 3) = 0, giving that x = 0 , ± 3. (b) When f ( x ) is undefined, we have x 2 1 = 0, giving that x = ± 1. (NOTE: Only x = 0 , ± 3 are critical numbers because f ( x ) is not defined at x = ± 1.) Step 3 . intervals ( −∞ , 3) ( 3 , 1) ( 1 , 0) (0 , 1) (1 , 3) (

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Unformatted text preview: . 5 . 5 1 . 5 2 sign of f ′ (+)(+) (+) = + (+)( − ) (+) = − (+)( − ) (+) = − (+)( − ) (+) = − (+)( − ) (+) = − (+)(+) (+) = + conclusion ր ց ց ց ց ր Step 4 . The function f is increasing on ( −∞ , − √ 3), ( √ 3 , ∞ ), and it is decreasing on ( − √ 3 , − 1), ( − 1 , 1), (1 , √ 3). (NOTE: We combine the intervals ( − 1 , 0), (0 , 1) because f (0) is de±ned.) The solution of the problem is now complete. We get the following if we graph the function: x K 4 K 3 K 2 K 1 1 2 3 4 y K 4 K 3 K 2 K 1 1 2 3 4 The vertical blue lines are asymptotes. The green line is an inclined asymptote....
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