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 b y=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 -- + Conclusions About F Concave Concave Down Up POI 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 3 Sign 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 Up Dec 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 2xπ π − ≤ ≤   ()' cosg x x= ()'' sing x x= − 0, , 2PPOI at xπ π= ( ) ( )5, , ,0 ,20 2 22π π ππ ππ   −       Test # 4π− 2π 32π 94π Sign g’’ + -- + -- POI at x=0, π, 2π Review: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 min 2.10 I. ()''f xtells us when ()f xis concave up and concave down. (find PPOI) ( ) ( ) ( ) Test # Sign of f”’ + - - POI, No POI II. The second derivative test is used to find relative mins and relative maxes. Find the critical numbers from f’()()()1 2 3, ,c c c Evaluate ()()()()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 min 1, 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