2.10 Concavity and the Second Derivative Test
0
+1
+2
-1
-2
+3
Concave Down
a
-3
b
y=f(x)
*** If the first derivative is decreasing on an interval
a, b then the original function f(x) is concave down on a, b .
Concave Up
-3
+3
-2
+2
-1
a
+1
0
y=f(x)
b
*** If the first derivative is increasing on an interval
Example 1:
a, b then f(x) is concave up on a, b .
f  x   x3  3x2  1
Let
Find intervals on which f(x) is concave up and concave down.
f '  x   3x 2  6 x
f ''  x   6 x  6
x  1 (Possible point of Inflection (PPOI); might be a place where the original function’s concavity changes.)
 ,1
1, 
Test #
0
2
sign f''
--
+
Conclusions
About F Concave
Down
POI at
Concave
Up
1, 1 *(Remember that the Y value for the POI comes from 1, f 1  )
We did this problem in 4.3 using the first derivative to find where we had a relative max or min.
4.3:
f '  x   3x 2  6 x
 3x  x  2 
 , 0
 0, 2
 2, 
Test #
-1
1
3
Sign f’
+
-
+
Conclusions
About f
Increasing
Decreasing Increasing
Relative Max at
Relative min at
 2, 3
 0,1
Dec Concave Down
Inc Concave
Down
(POI)
Inc Concave Up
Dec Concave Up
************************************************************************************************************
Example 2:
Find the X value of the POI:
f  x   xe x
 , 2
 2, 
f '  x   x  e x   1  e x 1
Test #
0
3
f '  x   e x  xe x
Sign f”
--
+
f ''  x   e  x   1  e  x  xe  x 
POI at (2, f(2)) =
 2, 2e    2, e2 
2
2
f ''  x   e x  e x  xe x
f ''  x   xe x  2e x
f ''  x   e x  x  2
x2
 PPOI 
Example 3:
f  x   x4
f '  x   4 x3
f ''  x   12x2
PPOI at x=0
 , 0
 0, 
Test #
-1
1
Sign f”
+
+
Conclusions: NO POI!!! No sign change!
Example 4:
5 
 
g  x   sin x On the interval    x 
2 
 2
g '  x   cos x
g ''  x    sin x
PPOI at x  0,  , 2
Test #
Sign g’’
  
  ,0
 2 


4
+
 0,  
 , 2 

2
3
2
5 

 2 , 
2 

9
4
--
+
--
POI at x=0,
Review:
π,
2π
2.9
I.
f '  x  tells us when f  x  is 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 ''  x  tells us when f  x  is 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’
Evaluate
 c1  ,  c2  ,  c3 
f ''  c1      rel min at  c1 , f  c1  
f ''  c2      rel max at  c2 , f  c2  
f ''  c3   0 undefined = test failed!!
Example:
1. Locate the relative mins and maxes for
f ( x)  x 4  2 x 2 using the second derivative test.
f '  x   4 x3  4 x
 4 x  x 2  1
 4 x  x 1 x 1
x  1, 0, 1
f ''  x   12x2  4
Sign of
f ''  0   rel max at  0,0 *remember that the “y” value comes from (
f ''  1   rel min at  1, 1
f '' 1   rel min at 1, 1
Redo example one using first derivative test to compare and contrast the differences:
f  x   x4  2x2
f '  x   4x  4x
3
 4 x  x 2  1
 , 1  1,0  0,1 1, 
test #
-2
-0.5
0.5
sign f’
-
+
-
 4 x  x 1 x 1
x  1, 0, 1
2.
rel min
rel max
2
+
rel min
f  x   x3  5x2  3x  k Find the value of k for which f has 11 as its relative minimum.
Work:
f '  x   3x2  10 x  3
0   3x 1 x  3
1
x
x3
3
f ''  x   6x 10
1
f ''    
3
f ''  3    rel min 
f  3  11
 3
3
 5  3  3  3  k  11
2
k  20
, f(0))