MATH 12002 - CALCULUS I 3.3: Increasing & Decreasing Functions - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I § 3.3: Increasing & Decreasing Functions Professor Donald L. White Department of Mathematical Sciences Kent State University D.L. White (Kent State University) 1 / 11

First Derivative Tests Increasing/Decreasing Test Let y = f ( x ) be a function. If f ′ ( x ) > 0 on an interval I, then f is increasing on I. If f ′ ( x ) < 0 on an interval I, then f is decreasing on I. First Derivative Test for Local Max/Min Let c be a critical number of a continuous function f . If f ′ changes sign from positive to negative at x = c, then f has a local maximum at c. If f ′ changes sign from negative to positive at x = c, then f has a local minimum at c. If f ′ does not change sign at x = c, then f has neither a local maximum nor a local minimum at c. D.L. White (Kent State University) 2 / 11

Example Example Let f ( x ) = 3 x 4 − 20 x 3 + 36 x 2 − 17 . Determine intervals where f is increasing, intervals where f is decreasing, and the location of all local maxima and minima. We need to determine the signs of the derivative of f . We have f ′ ( x ) = 12 x 3 − 60 x 2 + 72 x = 12 x ( x 2 − 5 x + 6) = 12 x ( x − 2)( x − 3) . The critical numbers of f are therefore x = 0, x = 2, and x = 3. Since f ′ is a polynomial, it is continuous everywhere, and so can change sign only where it is 0; that is, at x = 0, 2, or 3. We will use a sign chart to determine the signs of f ′ . D.L. White (Kent State University) 3 / 11

Example f ( x ) = 3 x 4 − 20 x 3 + 36 x 2 − 17 f ′ ( x ) = 12 x ( x − 2)( x − 3) 0 2 3 0 − + + + 12 x 0 x − 2 − − + + 0 x − 3 + − − − f ′ ( x ) 0 0 0 + + − − D I D I Inc-Dec MIN MAX MIN ❅ � ❅ � ❅ � ❅ � Shape f is increasing on (0 , 2) ∪ (3 , ∞ ); f is decreasing on ( −∞ , 0) ∪ (2 , 3). Local maximum at x = 2; local minimum at x = 0 and at x = 3. D.L. White (Kent State University) 4 / 11

Examples 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 − 20 x 3 + 36 x 2 − 17 at these points: 3(0 4 ) − 20(0 3 ) + 36(0 2 ) − 17 = − 17 f (0) = 3(2 4 ) − 20(2 3 ) + 36(2 2 ) − 17 = 48 − 160 + 144 − 17 = 15 f (2) = 3(3 4 ) − 20(3 3 ) + 36(3 2 ) − 17 = 243 − 540 + 324 − 17 = 10 f (3) = Hence the points (0 , − 17), (2 , 15), and (3 , 10) are on the graph. D.L. White (Kent State University) 5 / 11

Examples 0 2 3 D I D I Inc-Dec MIN MAX MIN ❅ ❅ � � ❅ ❅ � � Shape ✻ 20 (2 , 15), MAX 15 q 10 q (3 , 10), MIN 5 ✛ ✲ − 1 1 2 3 4 − 5 − 10 − 15 q (0 , − 17), MIN − 20 ❄ D.L. White (Kent State University) 6 / 11

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