12.2 Second Derivative and
Graphs
“If we think the derivative as a rate of
change, then the second derivative is the
rate of change of the rate of change”
* The second derivative is the derivative of
the derivative
Compare f(x) and g(x)
Both are increasing functions but they don’t look
quite the same.
Compare f’(x) and g’(x)
Both f’(x) and g’(x) are positive, however, f’(x) – the slope
of the tangent line - is increasing but g’(x) is decreasing
Concavity Tests
Theorem. The graph of a function f is concave upward on
the interval (a,b) if f ’(x) is increasing on (a,b), and is
concave downward on the interval (a,b) if f ’(x) is
decreasing on (a,b).
For y = f (x), the second derivative of f, provided it exists, is
the derivative of the first derivative:
d2 f
y ' '  f ' ' ( x)  2 ( x)
dx
Theorem. The graph of a function f is concave upward on
the interval (a,b) if f ’’(x) is positive on (a,b), and is concave
downward on the interval (a,b) if f ’’(x) is negative on (a,b).
Relationship between F and F’’
F’’(x)
F’(x)
F(x)
F: increasing on (-∞,-1) and (1,∞)
decreasing on (-1,1)
So F’ > 0 on (-∞,-1) and (1,∞)
F’ < 0 on (-1,1)
F’: decreasing on (- ∞, 0)
increasing on (0, ∞)
So F’’ < 0 on (- ∞, 0)
F’’ > 0 on (0, ∞)
F: concave down (- ∞, 0)
concave up (0, ∞)
So F’’ < 0 on (- ∞, 0)
F’’ > 0 on (0, ∞)
Concavity
Concave down
Concave up
Concavity
up
down
up
Example 1
Determine the intervals on which the graph is concave
upward and the intervals on which it’s concave downward.
A) f(x) = -e-x
Domain (-∞∞), no critical point
f’(x) = -e-x (-1) = e-x
f’’(x) =
e-x (-1) = -e-x
Test some numbers in the domain (review section 12.1
if you forgot), we will see that f’’ is always negative.
Therefore, the graph of f(x) is concave downward on (∞∞)
Example 1 (continue)
Determine the intervals on which the graph is concave
upward and the intervals on which it’s concave downward.
B) f(x) = ln (1/x) (that should equal ln 1 – lnx)
Domain (0,∞), no critical point
f’(x) = -1/x = -x-1
f’’(x) = x-2 = 1/x2
Test some numbers in the domain, we will see that f’’ is
always positive. Therefore, the graph of f(x) is concave
upward on (0,∞)
Example 1 (continue)
Determine the intervals on which the graph is concave
upward and the intervals on which it’s concave downward.
C) f(x) = x1/3
Domain (-∞,∞), critical value is x = 0
1
f ' ( x)  x
3

2
3

5
1
3x
2
3
2 3
2
f ' ' ( x) 
x  5
9
9x 3
Since there is a critical point, we want to test some points on the left of
0 and some on the right of 0. We will see that f’’ is always positive on
the left of 0 and always negative on the right of 0. Therefore, the graph
of f(x) is concave upward on (-∞, 0) and concave downward on (0, ∞).
Note that this graph changes from concave upward to concave downward at
(0,0). This point is called an inflection point.
Inflection Points
An inflection point is a point on the graph where the concavity
changes from upward to downward or downward to upward.
This means that if f ’’(x) exists in a neighborhood of an
inflection point, then it must change sign at that point.
Theorem 1. If y = f (x) is continuous on (a,b) and has an
inflection point at x = c, then either f ’’(c) = 0 or f ’’(c) does not
exist.
Example 2
Find the inflection point(s) of f(x) = x3 – 9x2 +24x -10
F’(x) = 3x2 – 18x + 24
F’’(x) = 6x – 18 = 0
6(x-3) = 0
x=3
x
2
3
4
F’’
--
0
+
Concave down
Note: It’s important to do
this test because the
second derivative must
change sign in order for
the graph to have an
inflection point.
Concave up
Therefore 3 is the infection point of f(x)
Example 3: A special case
Find the inflection point(s) of f(x) = x4
F’(x) = 4x3
F’’(x) = 12x2 = 0
x =0
x
-1
0
1
F’’
+
0
+
Concave up
Concave up
Therefore 0 is not the inflection point of f(x)
There is no inflection point for this graph
Example 4
Find the inflection point(s) of f(x) = ln(x2 - 2x + 5)
2x  2
f ' ( x)  2
x  2x  5
2( x 2  2 x  5)  (2 x  2)(2 x  2)
f ' ' ( x) 
( x 2  2 x  5) 2
 2x2  4x  6
f ' ' ( x)  2
0
2
x
-2 -1 0
( x  2 x  5)
f ' ' ( x)  2( x 2  2x  3)  0
 2( x  1)(x  3)  0
x = -1 and x =3
F’’
-
0
+
3
4
0
-
Concave up
Concave down
Concave down
Therefore there are two inflection points at x= -1 and x=3
Example 5
The given graph shows the graph of the derivative function f’(x).
Discuss the graph of f and sketch a possible graph of f.
x
F’(x)
F(x)
(-∞,-1)
Positive
Decreasing
Increasing
Concave down
X= -1
Local minimum
Inflection point
(-1,1)
positive
increasing
Increasing
Concave up
X=1
Local maximum
Inflection point
(1,2)
Positive
decreasing
Increasing
Concave down
X=2
X-intercept
Local maximum
(2, ∞)
Negative
decreasing
Decreasing
Concave down
F’(x)
• With today technology, graphing calculator
and computer can produce graphs.
However, important points on a plot may
be difficult to identify.
• Therefore, it’s useful to learn how to
sketch a graph by hand.
Curve Stretching
• Analyze f(x). Find the domain and intercepts. (Set x=0,
solve for f(x); set f(x)=0, solve for x).
• Analyze f’(x): Find critical values. Determine increasing
and decreasing intervals as well as local maximum
and/or minimum. (set f’(x)=0).
• Analyze f’’(x): Find inflection point. Determine the
intervals on which the graph is concave upward and
concave downward. (set f’’(x)=0).
• Plot additional points as needed and sketch the graph.
Example 6
Sketch f(x) = x4 + 4x3 by hand
Step 1: Domain: (-∞,∞)
X: intercept: x4 + 4 x3 = 0
x3 (x+4) = 0, so x=0 or x = -4
Y: intercept: f(0) = 0
Step 2: f’(x) = 4x3 + 12x2 = 0
4x2 (x+3) = 0 so x= 0 or x=-3 both critical v.
Test numbers on the left and on the right of 0 and -3, we see that -3 is a
local minimum. Also, f(x) is decreasing on (- ∞, -3) and increasing on
(-3, ∞).
Step 3: f’’(x) = 12x2 + 24x = 0
12x(x+2) = 0 so x = 0 or x = -2
Test numbers on the left and on the right of -2 and 0, we see that both of
them are inflection points. Also, the graph is concave upward on (- ∞, -2),
concave downward on (-2,0), and concave upward on (0, ∞)
Continue: Sketch f(x) = x4 + 4x3
X F(x) Note
-4 0
x-int
-3 -27
-2 -16
0 0
min
Inflection point
x-int, y-int
Inflection point
(- ∞, -3)
(-3, ∞)
(- ∞, -2)
(-2,0)
(0, ∞)
decreasing
increasing
concave up
concave down
concave up
Example 7
Sketch f(x) = 3x2/3 - x by hand
Step 1: Domain: (-∞,∞)
X: intercept: 3x2/3 - x = 0
x (3x-1/3 - 1) = 0, so x=0 or 3x-1/3 – 1 = 0
x-1/3
= 1/3, (x-1/3)-3 = (1/3)-3 , so x = 27
Y: intercept: f(0) = 0
Step 2: f’(x) = 2x-1/3 -1 = 0
x-1/3
= 1/2, (x-1/3)-3 = (1/2)-3 , so x= 8
Also, f’(x) is discontinuous at 0.
Test numbers on the left and on the right of 0 and 8, we see that 0 is a min
and 8 is a max. Also, f(x) is decreasing on (- ∞, 0) and (8, ∞) and increasing
on (0,8)
Step 3: f’’(x) = (-2/3) x-4/3 = 0
x-4/3 = 0; so x = 0
F’’ is also discontinuous at 0. Test numbers on the left and on the right of 0,
we see that there is no inflection point. The graph is concave downward on
(- ∞, 0) and on (0, ∞)
Continue: Sketch f(x) = 3x2/3 - x
X
0
F(x) Note
0
X-int, y-int,
min
8 4
max
27 0
X-int
(- ∞, 0)
(0,8)
(8, ∞)
(0, ∞)
Decreasing
Concave down
increasing
decreasing
Concave down
May need to add more points on the
left of 0 to have a better graph
• The value of x where rate of change of
sales changes from increasing to
decreasing is called the point of
diminishing returns. This is also the
point where the rate of change has a
maximum value.
Example 8
• A discount appliance store is selling 200 television sets monthly. If the
store invests $x thousand in an advertising campaign, the ad company
estimates that sales will increase to N(x) = 4x3 -.25x4 + 500, with 0≤x≤12.
When is rate of change of sales increasing and when is it decreasing?
What is the point of diminishing returns and the maximum rate of change
of sales.
• The rate of change of sales with respect to advertising expenditures is the
derivative N’(x). To determine when N’(x) is increasing or decreasing, we
find N’’(x)
• N’(x) = 12x2 – x3
N’’(x) = 24x – 3x2 = 0
X -1 0 1 8 9
3x (8 – x) = 0 so x = 0 or x = 8
N’’ 0 + 0 Therefore, the rate of change of sales
is increasing on (0, 8) and decreasing
on (8,12). The point of diminishing returns
Is x = 8 and the maximum rate of change
We do not need to know
Is N’(8) = 256.
the outcome on the left of
0 because 0≤x≤12