Download - 12.2 Second Derivative and Graphs
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12.2 Second Derivative and Graphs
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“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
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Compare f(x) and g(x)
Both are increasing functions but they don’t look quite the same.
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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
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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:
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).
)()(''''2
2
xdx
fdxfy
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Relationship between F and F’’
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, ∞)
F’’(x)F(x)
F’(x)
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Concavity
Concave down
Concave up
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Concavity
up
down
up
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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 11.1 if you forgot), we will see that f’’ is always negative. Therefore, the graph of f(x) is concave downward on (-∞∞)
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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,∞)
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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
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.
3
2
3
1
3
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3
2
x
xxf
3
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5
9
2
9
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xxf
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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.
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Example 2Find 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
+0--F’’
432x
Concave down Concave up
Therefore 3 is the infection point of f(x)
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.
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Example 3: A special caseFind the inflection point(s) of f(x) = x4
F’(x) = 4x3
F’’(x) = 12x2 = 0
x = 0
+0+F’’
10-1x
Concave up Concave up
Therefore 0 is not the inflection point of f(x)
There is no inflection point for this graph
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Example 4Find the inflection point(s) of f(x) = ln(x2 - 2x + 5)
x = -1 and x =3
+
0
0
3
-0-F’’
4-1-2x
Concave downConcave up
Therefore there are two inflection points at x= -1 and x=3
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0)62(2)('' 2 xxxf0)3)(1(2 xx
Concave down
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Example 5The given graph shows the graph of the derivative function f’(x). Discuss the graph of f and sketch a possible graph of f.
F’(x)
Local maximumX-interceptX=2
Increasing
Concave up
positive
increasing
(-1,1)
Inflection pointLocal maximumX=1
Increasing
Concave down
Positive
decreasing
(1,2)
Decreasing
Concave down
Negative
decreasing
(2, ∞)
Inflection pointLocal minimumX= -1
Increasing
Concave down
Positive
Decreasing
(-∞,-1)
F(x)F’(x)x
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• 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.
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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.
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Example 6Sketch f(x) = x4 + 4x3 by hand
Step 1: Domain: (-∞,∞)X: intercept: x4 + 4 x3 = 0 x3 (x+4) = 0, so x=0 or x = -4Y: 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 = -2Test 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, ∞)
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Continue: Sketch f(x) = x4 + 4x3
X F(x) Note
-4 0 x-int
-3 -27 min
-2 -16 Inflection point
0 0 x-int, y-int
Inflection point
(- ∞, -3) decreasing
(-3, ∞) increasing
(- ∞, -2) concave up
(-2,0) concave down
(0, ∞) concave up
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Example 7Sketch f(x) = 3x2/3 - x by handStep 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 = 27Y: 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= 8Also, 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 = 0F’’ 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, ∞)
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Continue: Sketch f(x) = 3x2/3 - x
X F(x) Note
0 0 X-int, y-int, min
8 4 max
27 0 X-int
(- ∞, 0) Decreasing
Concave down
(0,8) increasing
(8, ∞) decreasing
(0, ∞) Concave down
May need to add more points on the left of 0 to have a better graph
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• 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.
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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 3x (8 – x) = 0 so x = 0 or x = 8Therefore, 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 Is N’(8) = 256.
X -1 0 1 8 9
N’’ - 0 + 0 -
We do not need to know the outcome on the left of 0 because 0≤x≤12