Objectives for Section 12.2
Second Derivatives and Graphs
■ The student will be able to use
concavity as a graphing tool.
■ The student will be able to find
inflection points.
■ The student will be able to analyze
graphs and do curve sketching.
■ The student will be able to find the
point of diminishing returns.
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Concavity
The term concave upward (or simply concave up) is used to
describe a portion of a graph that opens upward. Concave
down(ward) is used to describe a portion of a graph that
opens downward.
Concave down
Concave up
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Definition of Concavity
A graph is concave up on the interval (a,b) if any secant connecting
two points on the graph in that interval lies above the graph.
It is concave down on (a,b) if all secants lie below the graph.
down
up
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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:
2
d 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).
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Example 1
Find the intervals where the graph of
f (x) = x3 + 24x2 + 15x - 12.
is concave up or concave down.
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Example 1
Find the intervals where the graph of
f (x) = x3 + 24x2 + 15x - 12.
is concave up or concave down.
f ’(x) = 3x2 + 48x + 15
f ”(x) = 6x + 48
f ”(x) is positive when 6x + 48 > 0 or x > -8, so it is concave
up on the region (-8, ).
f ”(x) is negative when 6x + 48 < 0 or x < -8, so it is
concave down on the region (-, -8).
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Example 1
(continued)
f (x)
f ’’ (x)
- 8
-25 < x < 20,
– 400 < y <14,000
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-10 < x < 1
–2 < y < 6
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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.
continued
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Inflection Points
(continued)
The theorem means that an inflection point can occur only at
critical value of f ’’. However, not every critical value
produces an inflection point.
A critical value c for f ’’ produces an inflection point for the
graph of f only if f ’’ changes sign at c, and c is in the
domain of f.
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Summary
Assume that f satisfies one of the conditions in the table
below, for all x in some interval (a,b). Then the other
condition(s) to the right of it also hold.
f ’(x) > 0
f is increasing
f ’(x) < 0
f is decreasing
f ’(x) = 0
f is constant
f ’’(x) > 0
f ’(x) increasing
f is concave up
f ’’(x) < 0
f ’(x) decreasing
f concave down
f ’’(x) = 0
f ’(x) is constant
f is linear
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Example 2
Find the inflection points of f (x) = x3 + 24x2 + 15x - 12.
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Example 2
Find the inflection points of f (x) = x3 + 24x2 + 15x - 12.
Solution:
In example 1, we saw that f ’’(x) was negative to the left of
–8 and positive to the right of –8. At x = – 8, f ’’ (x) = 0.
This is an inflection point because f changes from concave
down to concave up at this point.
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Example 2
(continued)
Find the inflection point using a graphing calculator.
Inflection points can be difficult to recognize on a graphing
calculator, but they are easily located using root approximation
routines. For instance, the above example when f is graphed
shows an inflection point somewhere between -6 and -10.
f (x)
-25 < x < 20,
– 400 < y <14,000
continued
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Example 2
(continued)
Graphing the second derivative and using the zeros
command on the calc menu shows the inflection point at –8
quite easily, because inflection points occur where the second
derivative is zero.
f ’’(x) = 6x + 48
- 10 < x < 1
– 2<y<6
-8
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Second Derivative Test
Let c be a critical value for f (x), then
f ’(c)
f ’’(c)
graph of f is
f (c) is
0
+
concave up
local minimum
0
–
concave down
local maximum
0
0
?
test fails
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Curve Sketching
Graphing calculators and computers produce the graph
of a function by plotting many points. Although quite
accurate, important points on a plot may be difficult to
identify. Using information gained from the function
and its dervative, we can sketch by hand a very good
representation of the graph of f (x). This process is
called curve sketching and is summarized on the
following slides.
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Graphing Strategy
 Step 1. Analyze f (x).
Find the domain and the intercepts. The x intercepts are the
solutions to f (x) = 0, and the y intercept is f (0).
 Step 2. Analyze f ’(x).
Find the partition points and critical values of f ’(x).
Construct a sign chart for f ’(x), determine the intervals
where f is increasing and decreasing, and find local
maxima and minima.
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Graphing Strategy
(continued)
 Step 3. Analyze f ”(x).
Find the partition numbers of f ”(x). Construct a sign chart for
f ”(x), determine the intervals where the graph of f is concave
upward and concave downward, and find inflection points.
 Step 4. Sketch the graph of f.
Locate intercepts, local maxima and minima, and inflection
points. Sketch in what you know from steps 1-3. Plot
additional points as needed and complete the sketch.
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Graphing Strategy
Example
Sketch the graph of y = x3/3 – x2 – 3x
■ Step 1. Analyze f (x).
This is a polynomial function, so the domain is all reals. The
y intercept is 0, and the x intercepts are 0 and (3 + 45)/2.
■ Step 2. Analyze f ’(x).
f ’(x) = x2 – 2x – 3 = (x+1)(x–3), so f has critical values
at –1 and 3.
■ Step 3. Analyze f ”(x).
f ”(x) = 2x – 2, so f ’’ has a critical value at x = 1.
A combined (steps 2 and 3) sign chart for this function is shown
on the next slide.
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Sign chart for f ’ and f ”
(- , -1)
(- 1, 3)
(3, )
f ’’ (x) - - - - - - - 0 + + + + + + + +
f ’ (x) + + + 0 - - - - - - 0 + + + + +
-1
f (x) increasing
f (x)
1
decreasing
maximum
3
increasing
minimum
f (x) concave down - inflection - concave up
point
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Analyzing Graphs - Applications
A company estimates that it will sell N(x) units of a product after
spending $x thousand on advertising, as given by
N(x) = -2x3 + 90x2 – 750x + 2000 for 5 ≤ x ≤ 25
(a) When is the rate of change
of sales, N ’(x), increasing?
Decreasing?
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Analyzing Graphs - Applications
A company estimates that it will sell N(x) units of a product after
spending $x thousand on advertising, as given by
N(x) = -2x3 + 90x2 – 750x + 2000 for 5 ≤ x ≤ 25
(a) When is the rate of change
-5 < x < 50 and –1000 < y < 1000
of sales, N ’(x), increasing?
Decreasing?
N ’(x) = -6x2 + 180x –750.
N ’(x) is increasing on (5, 15),
then decreases for (15, 25).
15
Note: This is the graph of the
derivative of N ’(x)
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Application
(continued)
(b) Find the inflection points for the graph of N.
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Application
(continued)
(b) Find the inflection points for the graph of N.
N ’(x) = -6x2 + 180x –750.
N ’’(x) = -12x + 180
0 < x < 70 and –0.03 < y < 0.015
Inflection point at x = 15.
15
15
Note: This is N (x).
Note: This is N ’’(x).
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Application
(continued)
(c) What is the maximum rate of change of sales?
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Application
(continued)
(c) What is the maximum rate of change of sales?
We want the maximum of the derivative.
N ’(x) = -6x2 + 180x –750.
- 5 < x < 50 and – 1000 < y < 1000
Maximum at x = 15.
N ’(15) = 600.
15
Note: This is the graph of N ’(x).
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Point of Diminishing Returns
If a company decides to increase spending on advertising, they
would expect sales to increase. At first, sales will increase at an
increasing rate and then increase at a decreasing rate. The
value of x where the 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. Money spent after this point may increase
sales, but at a lower rate. The next example illustrates this
concept.
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Maximum Rate of Change Example
Currently, a discount appliance store is selling 200 large-screen
television sets monthly. If the store invests $x thousand in an
advertising campaign, the ad company estimates that sales will
increase to
N (x) = 3x3 – 0.25x4 + 200
0<x<9
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?
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Example
(continued)
Solution:
The rate of change of sales with respect to advertising
expenditures is
N ’(x) = 9x2 – x3 = x2(9-x)
To determine when N ’(x) is increasing and decreasing, we
find N ”(x), the derivative of N ’(x):
N ”(x) = 18x – 3x2 = 3x(6-x)
The information obtained by analyzing the signs of N ’(x)
and N ”(x) is summarized in the following table (sign charts
are omitted).
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Example
(continued)
x
N ”(x)
N ’(x)
N ’(x)
N (x)
0<x<6
+
+
Increasing
Increasing,
concave up
x=6
0
+
Local Max
Inflection
Point
6<x<9
-
+
Decreasing
Increasing,
concave down
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Example
(continued)
Examining the table, we see that N ’(x) is increasing
on (0, 6) and decreasing on (6, 9). The point of
diminishing returns is x = 6, and the maximum rate of
change is N ’(6) = 108. Note that N ’(x) has a local
maximum and N (x) has an inflection point at x = 6.
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Summary
■ We can use the second derivative to determine when a
function is concave up or concave down.
■ When the second derivative is zero, we may get an
inflection point in f (x) (a change in concavity).
■ The second derivative test may be used to determine if a
point is a local maximum or minimum.
■ The value of x where the rate of change changes from
increasing to decreasing is called the point of
diminishing returns.
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