2.10 Concavity and the Second Derivative Test 0 +1 -1+2 -2 +3 Concave Down -3 a b y=f(x) *** If the first derivative is decreasing on an interval [ ],a bthen the original function f(x) is concave down on [ ],a b.Concave Up-3 +3 -2 +2 -1 +1 a 0 by=f(x)*** If the first derivative is increasing on an interval [ ],a bthen f(x) is concave up on[ ],a b.Example 1: Let ( )3 23 1f x x x= - +Find intervals on which f(x) is concave up and concave down.( )2' 3 6f x x x= -( )'' 6 6f x x= -1x = (Possible point of Inflection (PPOI); might be a place where the original function’s concavity changes.)( ) ( ),1 1,- � �Test # 0 2 f''sign -- +ConclusionsAbout F Concave Concave Down UpPOI at ( )1, 1-*(Remember that the Y value for the POI comes from( )( )1, 1f)We did this problem in 4.3 using the first derivative to find where we had a relative max or min. 4.3:( )2' 3 6f x x x= -( )3 2x x= -( ) ( ) ( ), , ,0 0 2 2- � �Test # -1 1 3Sign f’ + - +Conclusions Increasing Decreasing Increasing About f Relative Max at Relative min at ( )0,1( )2, 3- Dec Concave Down Inc Concave Down (POI) Inc Concave UpDec Concave Up************************************************************************************************************Example 2:Find the X value of the POI: ( )xf x xe-=( ) ( )2, ,2- � �( ) ( )' 1 1x xf x x e e- -= � �- + �Test # 0 3( )'x xf x e xe- -= -Sign f” -- +( ) ( )'' 1x x xf x e e xe- - -� �= �- - -� �POI at (2, f(2)) = ( )2222, 2 2,ee-� �=� �� �( )''x x xf x e e xe- - -=- - +( )'' 2x xf x xe e- -= -( ) ( )'' 2xf x e x-= -( )2x PPOI=Example 3:( )4f x x=( ) ( )0, ,0- � �( )3' 4f x x=Test # -1 1( )2'' 12f x x=Sign f” + + at x=0PPOI Conclusions: NO POI!!! No sign change!Example 4:( )sing x x= On the interval 52 2xp p� �- � �� �� �( )' cosg x x=( )'' sing x x=-0, , 2PPOI at x p p=( ) ( )5, , ,0 ,20 2 22p p pp pp� � � �-� � � �� � � �Test # 4p- 2p 32p 94pSign g’’ + -- + --POI at x=0, π, 2πReview: 2.9 I. ( )'f xtells us when ( )f xis increasing and decreasing.II. First derivative test; used to find relative mins and relative maxes.( ) ( ) ( )Test#Sign of f’ + - + rel max, rel min2.10I. ( )''f xtells us when ( )f xis concave up and concave down. (find PPOI) ( ) ( ) ( )Test #Sign of f”’ + - - POI, No POIII. The second derivative test is used to find relative mins and relative maxes.Find the critical numbers from f’( ) ( ) ( )1 2 3, ,c c cEvaluate ( ) ( ) ( )( )1 1 1'' rel min at ,f c c f c= +( ) ( ) ( )( )2 2 2'' rel max at ,f c c f c= -( )3'' 0 = test failed!! f c undefined=Example:1. Locate the relative mins and maxes for 4 2( ) 2f x x x= -using the second derivative test.( )3' 4 4f x x x= -( )24 1x x= -( ) ( )4 1 1x x x= - - 1, 0, 1x =-( )2'' 12 4f x x= -Sign of ( ) ( )'' 0 rel max at 0,0f =- *remember that the “y” value comes from ( , f(0))( ) ( )'' 1 rel min at 1, 1f - =+ - -( ) ( )'' 1 rel min at 1, 1f =+ -Redo example one using first derivative test to compare and contrast the differences:( )4 22f x x x= -( ) ( ) ( ) ( ), 1 1, 0 0,1 1,- �- - �( )3' 4 4f x x x= -test # -2 -0.5 0.5 2( )24 1x x= -sign f’ - + - +( ) ( )4 1 1x x x= - - rel min rel max rel min1, 0, 1x =-2. ( )3 25 3f x x x x k= - + +Find the value of kfor which fhas 11as its relative minimum.Work:( )2' 3 10 3f x x x= - +( ) ( )0 3 1 3x x= - -133x x= =( )'' 6 10f x x= -1''3f� �=-� �� �( ) ( )'' 3 rel minf =+( )3 11f =( ) ( ) ( )3 23 5 3 3 3 11k- + + =20k