3.4
Concavity and the
2nd Derivative Test
concave up:
f’ is increasing.
tangent lines are
below the graph.
concave down:
f’ is decreasing.
tangent lines are
above the graph.
Test for Concavity
1. If f”(x) > 0 for all x in I, then f is concave up.
2. If f”(x) < 0 for all x in I, then f is concave down.
2nd Derivative Test (Use C.N.’s of 1st Derivative)
1. If f”(c) > 0, then f(c) is a relative min.
2. If f”(c) < 0, then f(c) is a relative max.
3. If f”(c) = 0, then the test fails. No min. or max.
Ex. 1 Determine where f(x) = 6(x2 + 3)-1 is
increasing, decreasing, has max’s or min’s, is
concave up or down, has inflection points.
f’(x) =
-6(x2
+
3)-2(2x)
=
- 12 x
(x
2
+
(- •,0)
inc.
1st der.
test
)
+3
0
(0,2)
max.
2
-
(0, • )
dec.
C.N. = 0
f '(x) =
-12x
(x
x
(
f "(x) =
2
2
f "(x) =
=
+ 3)
+ 3) (-12) - (-12x )(2)( x 2 + 3)(2x )
2
[
(x
(x
2
+ 3)
4
2
2
2
12
x
+
3
x
+
3
+
4
x
( )(
) (
)
36( x 2 -1)
2
2
+ 3)
3
(x
2
+ 3)
4
C.N.’s -1, 1
+
]=
2
2
12
x
+
3
3x
( )(
)[ - 3]
(x
+ 3)
+
-
down
concave up -1
(- •,-1)
(- 1,1)
3 )
(-1,
Inflection points
2
2
1
up
(1, • )
(1, 3 2 )
4
2nd Derivative Test
C.N. from
1st
Plug C.N.’s of 1st der. into 2nd
derivative.
der. was 0.
f " (0) < 0 \
f " ( x) =
0 is a maximum
(
)
36 x 2 - 1
(x
2
)
+3
3
Remember, a neg. in the 2nd der. means concave down.
Therefore,
the point is a maximum.
There are no x-intercepts
(-1, 3/2) Inf. pt.
(0,2) max
(1, 3/2) Inf. pt.
Ex. 2 Determine where f(x) = x4 – 4x3 is
increasing, decreasing, has max’s or min’s, is
concave up or down, has inflection points.
f’(x) = 4x3 – 12x2 = 4x2(x – 3)
-
0
dec.
1st der. test
-
dec.
C.N.’s 0, 3
+
3
inc.
(3, -27 ) min.
f”(x) = 12x2 – 24x = 12x(x – 2) = 0
+
-
+
up
0
down
2
up
Inf. pt.’s (0, 0 )
(2, -16 )
(- •,3)(3, • )
dec.
C.N.’s
inc.
0, 2
2nd der. test
f”(0) = 0
f”(3) > 0
(3,-27) is a min.
9
-9
-16
-27
Given the graph of f, trace the graph of f’, and f”,
on the same axes.
f’
f
f”