CHAPTER 3 SECTION 3.4 CONCAVITY AND THE SECOND DERIVATIVE TEST
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Transcript of CHAPTER 3 SECTION 3.4 CONCAVITY AND THE SECOND DERIVATIVE TEST
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CHAPTER 3SECTION 3.4
CONCAVITY AND THE SECOND DERIVATIVE
TEST
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Definition of Concavity and Figure 3.24
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Sketch 4 graphs a)1 decreasing and concave upb)1 increasing and concave up, c)1 decreasing and concave down,
d)1 increasing and concave down
x
y
x
y
x
y
x
y
a b
cd
x x
x x
y y
y y
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Concave upward
• Look at these two graphs. Each is concave upward, but one is decreasing and the other is increasing. We need to be able to determine concavity from the function and not just from the graph. For each of the graphs above sketch the tangent lines to the graph at a number of different points.
x
y
x
y
x x
y y
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Concave upward
• As we move from left to right the slopes of the tangent lines are getting less negative. That is they are increasing.x
y
x
y
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Concave upward
• As we move from left to right the slopes of the tangent lines are getting larger. That is they are increasing.x
y
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When a graph is concave upward
The slope of the tangent lines are increasing.
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Concave downward
• Look at these two graphs. Each is concave downward, but one is decreasing and the other is increasing. We need to be able to determine concavity from the function and not just from the graph. For each of the graphs above sketch the tangent lines to the graph at a number of different points.
x
y
x
y
x x
y y
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Concave downward
• As we move from left to right, the slopes of the tangent lines are getting more negative.
• They are decreasing.x
y
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Concave downward
• As we move from left to right the slopes of the tangent lines are getting smaller. That is they are decreasing.
x
y
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When a graph is concave downward
The slopes of the tangent lines are decreasing.
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Putting it all together
• For a function f that is differentiable on an interval I, the graph of f is
• (i) Concave up on I, if the slope of the tangent line is increasing on I or
• (ii) Concave down on I, if the slope of the tangent line is decreasing on I
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Linking knowledge
• (i) Concave up on I, if the slope of the tangent line is increasing on I.
• If the slope of the tangent line is increasing and the slope of the tangent line is represented by the first derivative and to determine when something is increasing we had to take the derivative, then to find where the slope of the tangent line (f ‘(x)) is increasing we will need to take the derivative of f ‘(x) or find the second derivative f “(x)
I know, this is a very large run on sentence.
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Linking knowledge
• (ii) Concave down on I, if the slope of the tangent line is decreasing on I
• If the slope of the tangent line is decreasing and the slope of the tangent line is represented by the first derivative and to determine when something is decreasing we had to take the derivative, then to find where the slope of the tangent line (f ‘(x)) is decreasing we will need to take the derivative of
• f ‘(x) or find the second derivative f “(x)
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Definition of concavity
• For a function f that is differentiable on an interval I, the graph of f is
• (i) Concave up on I, if f’ is increasing on I or• (ii) Concave down on I, if f’ is decreasing on I
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Theorem 3.7Test for concavity
I.in downward concave is ofgraph e th then I,in allfor 0)( If 2.
I.in upward concave is ofgraph e th then I,in allfor 0)( If 1.
I. intervalopen an on exists derivative second hosefunction w a be fLet
fxxf
fxxf
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Putting it all together
• Given the function f(x)• f(x) = 0 x-intercepts• f(x) undefined vertical asymptote• f(x)>0 Q-1 or Q-2• f(x)<0 Q-3 or Q-4
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Putting it all together
decreasing is 0)(increasing is 0)(
asymptote) tical ver orpt ner number(cor critical undefined )(
number critical 0)()( Find
fxffxf
xfxf
xf
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Putting it all together
downward concave is decreasing is 0)(
upward concave is ;increasing is 0)(
??????? undefined )(number cal"hypercriti" 0)(
)( Find
ffxfffxf
xfxf
xf
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Determining concavity
• Determine the open intervals on which the graph is concave upward or concave downward.
• Concavity find second derivative.
• Find hypercritical numbers.
• Set up a chart• Find concavity
36)( 2
x
xf
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32
2
42
222
22
22
12
3136)(
3)12)(2)(3(2312)(
312)(
)2()3(6)(
36)(
xxxf
xxxxxxf
xxxf
xxxf
xxf
c = 1; c = -1 and f” is defined on the entire # line
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32
2
3136)(
xxxf Setting up the chart
interval Test points
Sign of f” f ‘ concave
(-∞, -1) -2 + inc upward
(-1,1) 0 - dec downward
(1,∞) 2 + inc upward
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Points of inflection
• A point of inflection for the graph of f is that point where the concavity changes.
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Theorem 3.7 Test for Concavity
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Definition of Point of Inflection and Figure 3.28
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Theorem 3.8 Points of Inflection
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Theorem 3.9 Second Derivative Test and Figure 3.31
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• Example 1: Graph the function f given by
• and find the relative extrema.• 1st find graph the function.
f (x) x3 3x2 9x 13,
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• Example 1 (continued): • 2nd solve f (x) = 0.
• Thus, x = –3 and x = 1 are critical values.
3x2 6x 9 0x2 2x 3 0
(x 3)(x 1) 0
x 3 0x 3
orx 1 0
x 1
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• Example 1 (continued): • 3rd use the Second Derivative Test with –3 and 1.
• Lastly, find the values of f (x) at –3 and 1.
• So, (–3, 14) is a relative maximum and (1, –18) is a • relative minimum.f ( 3) ( 3)3 3( 3)2 9( 3) 13 14
f (1) (1)3 3(1)2 9(1) 13 18
f ( 3) 6( 3) 6 18 6 12 0 : Relative maximum
f (1) 6(1) 6 6 6 12 0 : Relative minimum
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Second Derivative Testa. If f’’(c) > 0 then ________________________
If c is a critical number of f’(x) and…
b. If f’’(c) < 0 then ________________________
c. If f’’(c) = 0 or undefined then __________________________________
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Second Derivative Testa. If f’’(c) > 0 then ________________________ (c, f(c)) is a relative min
If c is a critical number of f’(x) and…
b. If f’’(c) < 0 then ________________________
c. If f’’(c) = 0 or undefined then __________________________________
(c, f(c)) is a relative max
the test fails (use 1st Derivative test)
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The second derivative gives the same information about the first derivative that the first derivative gives about the original function.
If f’’(x) > 0 ______________ If f’’(x) < 0 ______________ If f’’(x) = 0 ______________
Concave upward Concave downward
Inflection Points
If f’(x) > 0 ______________ If f’(x) < 0 ______________ If f’(x) = 0 ______________
______________ ______________ ________________
For f(x) to increase, _____________
For f’(x) to increase, _____________
For f(x) to decrease, _____________
For f’(x) to decrease, _____________
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The second derivative gives the same information about the first derivative that the first derivative gives about the original function.
If f’’(x) > 0 ______________ If f’’(x) < 0 ______________ If f’’(x) = 0 ______________
Concave upwardSlopes increase
Concave downwardSlopes decrease
f(x) increases f(x) decreases f(x) is constant
Inflection Points Where concavity changes
Occur at critical numbers of f”(x)
If f’(x) > 0 ______________ If f’(x) < 0 ______________ If f’(x) = 0 ______________
f’(x) decreasesf’(x) increases
f(x) is conc up f(x) is conc down
f’(x) is constant
f(x) is a straight line______________ ______________ ________________
For f(x) to increase, _____________
For f’(x) to increase, _____________
For f(x) to decrease, _____________
For f’(x) to decrease, _____________
f’(x) > 0
f’’(x) > 0
f’(x) < 0
f’’(x) < 0
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Sketch 126 3f x x
Include extrema, inflection points, and intervals of concavity.
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Sketch 126 3f x x
22' 6 3 2f x x x
22
12
3
x
x
Critical numbers: 12 0x 2 3 0x 0x
No VA’s
smooth
None
Include extrema, inflection points, and intervals of concavity.
22 2
42
3 12 12 2 3 2''
3
x x x xf x
x
2 2 2
42
12 3 3 4
3
x x x
x
2
32
12 3 3
3
x
x
2
32
36 1
3
x
x
Critical numbers: 2 1 0x 2 3 0x 1x None
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1. Find the extrema of 2sin cos2 on 0,2f x x x
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1. Find the extrema of 2sin cos2 on 0,2f x x x
' 2cos 2sin2f x x x Crit numbers: 0 2cos 2sin2x x
'' 2sin 4cos2f x x x
0 2cos 2 2sin cosx x x 0 2cos 1 2sinx x
0 2cos 0 1 2sinx x 120 cos sinx x
3 52 2 6 6, ,x x
2'' 2f
32'' 6f
56'' 3f
6'' 3f 326f
2 1f
32 3f
25 36f
rel min at
rel max at
rel min at
rel max at
36 2,
2 ,1
5 36 2,
32 , 3
2nd Derivative Test
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2. Sketch 22sin on 0,4xf x
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2. Sketch 22sin on 0,4xf x
2' cos xf x Crit numbers:
,3 ,5 ,7x
12'' 3f
12''f 2f
23f rel min at
rel max at ,2
3 , 2
2nd Derivative Test
20 cos x3
2 2 2,x 5 72 2,
12 2'' sin xf x Crit numbers: 20 sin x
2 ,2 ,3 ,4x 2 ,4 ,6 ,8x Intervals: 0, 2 ,4
Test values: 3
Inf pt
2x
2 0f
2 ,0
f ’’(test pt) f(x) down up
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2sin xf x
2
-2
24
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Find a Function
Describe the function at the point x=3 based on the following:
4)3( f
0)3( f
6)3( f
3
(3, 4)
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Find a Function
Describe the function at the point x=5 based on the following:
0)5( f
0)5( f
0)5( f 5
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Find a Function
Given the function is continuous at the point x=2, sketch a graph based on the following:
3)2( fDNEf )2(
20)(20)(20)(
xallforxfxforxfxforxf
2
(2,3)
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WHY? BECAUSE f’(x) is POSITVE!!!!!!!!!!!!!!!
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