MATH 12002 - CALCULUS I 3.3: Concavity & Inflection Points - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I § 3.3: Concavity & Inflection Points - Example Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 11

Example Example Let f ( x ) = 3 x 4 − 4 x 3 . Determine intervals where f is increasing, intervals where f is decreasing, the location of all local maxima and minima, intervals where f is concave up, intervals where f is concave down, and the location of all inflection points. D.L. White (Kent State University) 2 / 11

Example We need to determine the signs of f ′ and f ′′ for f ( x ) = 3 x 4 − 4 x 3 . We have 12 x 3 − 12 x 2 = 12 x 2 ( x − 1) , f ′ ( x ) = 36 x 2 − 24 x = 36 x ( x − 2 f ′′ ( x ) = 3 ) . Hence f ′ and f ′′ are never undefined and f ′ ( x ) = 0 when x = 0 or x = 1; f ′′ ( x ) = 0 when x = 0 or x = 2 3 . In order to determine the signs of both f ′ and f ′′ on each interval, we put all of these numbers in our sign chart: D.L. White (Kent State University) 3 / 11

Example f ( x ) = 3 x 4 − 4 x 3 , f ′ ( x ) = 12 x 2 ( x − 1), f ′′ ( x ) = 36 x ( x − 2 3 ) 2 0 1 3 12 x 2 0 + + + + 0 x − 1 − − − + ✲ f ′ ( x ) 0 0 + − − − 0 + + + 36 x − x − 2 0 + + − − 3 ✛ f ′′ ( x ) 0 0 + − + + ✲ D D D I Inc-Dec MIN ✛ U D U U Concave INF INF ☞ ✍ ✍ ✌ Shape D.L. White (Kent State University) 4 / 11

Example f ( x ) = 3 x 4 − 4 x 3 , f ′ ( x ) = 12 x 2 ( x − 1), f ′′ ( x ) = 36 x ( x − 2 3 ) 2 0 1 3 D D D I Inc-Dec MIN U D U U Concave INF INF ✟ ✡ ✡ ✠ Shape f is increasing on (1 , ∞ ), f is decreasing on ( −∞ , 1); f has a local minimum at x = 1. f is concave up on ( −∞ , 0) ∪ ( 2 3 , ∞ ), f is concave down on (0 , 2 3 ); f has an inflection point at x = 0 and at x = 2 3 . D.L. White (Kent State University) 5 / 11

Example In order to sketch the graph of f , we will need to plot the points whose x coordinates are in the sign chart. We need to evaluate f ( x ) = 3 x 4 − 4 x 3 at these points: 3(0 4 ) − 4(0 3 ) = 0 f (0) = 3 ) 4 − 4( 2 f ( 2 3( 2 3 ) 3 3 ) = ( 2 3 ) 3 � 3( 2 � = 3 ) − 4 ( 8 = 27 )(2 − 4) ( 8 27 )( − 2) = − 16 = 27 ≈ − 0 . 59 3(1 4 ) − 4(1 3 ) = 3 − 4 = − 1 f (1) = Hence the points (0 , 0), ( 2 3 , − 16 27 ), and (1 , − 1) are on the graph. D.L. White (Kent State University) 6 / 11

Example 2 0 1 3 D D D I Inc-Dec MIN U D U U Concave INF INF ✟ ✡ ✡ ✠ Shape ✻ 1 ✛ ✲ q − 1 1 2 (0 , 0), INF q ( 2 3 , − 16 27 ), INF q − 1 (1 , − 1), MIN ❄ D.L. White (Kent State University) 7 / 11

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