Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c)...
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Transcript of Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c)...
![Page 1: Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:](https://reader035.fdocuments.net/reader035/viewer/2022071807/56649e9e5503460f94b9ffc7/html5/thumbnails/1.jpg)
Second Derivative Test,
Graphical Connections
Section 4.3b
![Page 2: Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:](https://reader035.fdocuments.net/reader035/viewer/2022071807/56649e9e5503460f94b9ffc7/html5/thumbnails/2.jpg)
Do Now: #30 on p.204 (solve graphically)
21 2 4y x x x
1 2 4
(a) Local Maximum at
2x
(b) Local Minimum at
4x
(c) Points of Inflection:
1,x 1.634,3.366x
y
![Page 3: Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:](https://reader035.fdocuments.net/reader035/viewer/2022071807/56649e9e5503460f94b9ffc7/html5/thumbnails/3.jpg)
Second Derivative Test for Local Extrema
Let’s see these with the graphs of the squaringfunction and the negative squaring function…
1. If and , then has a local maximum at x = c.
0f c 0f c f
2. If and , then has a local minimum at x = c.
0f c 0f c f
![Page 4: Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:](https://reader035.fdocuments.net/reader035/viewer/2022071807/56649e9e5503460f94b9ffc7/html5/thumbnails/4.jpg)
Another Quick ExampleFind the extreme values of 3 12 5f x x x
2 23 12 3 4f x x x
2 12 0f
6f x x
2 12 0f
Critical Points???
x = 2, –2
Support our answers graphically???
has a local max. at x = –2
has a local min. at x = 2f
f
![Page 5: Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:](https://reader035.fdocuments.net/reader035/viewer/2022071807/56649e9e5503460f94b9ffc7/html5/thumbnails/5.jpg)
Let 3 24 12f x x x 1. Identify where the extrema of occur.
2. Find the intervals on which is increasing and the intervals on which is decreasing.
3. Find where the graph of is concave up and where it is concave down.
4. Sketch a possible graph for .
f
ff
f
f
![Page 6: Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:](https://reader035.fdocuments.net/reader035/viewer/2022071807/56649e9e5503460f94b9ffc7/html5/thumbnails/6.jpg)
Let 3 24 12f x x x is continuous since exists.
The domain of is all reals, so the domain of is the same.
Thus, the critical points of occur only at the zeros of .
24 3f x x x CP at x = 0, 3
Intervals
Sign of
Behavior of
x < 0
–Dec.
0 < x < 3
–Dec.
x > 3
+Inc.
Use first derivative test: Local Min. at x = 3
Decreasing on , Increasing on
f f f f
f f
f f
,0 , 0,3 3,
![Page 7: Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:](https://reader035.fdocuments.net/reader035/viewer/2022071807/56649e9e5503460f94b9ffc7/html5/thumbnails/7.jpg)
Let 3 24 12f x x x
The zeros of (IP) are at x = 0, 2
Intervals
Sign of
Behavior of
x < 0
+Conc. up
0 < x < 2
–Conc. down
x > 2
+Conc. up
Concave up on ,
212 24f x x x
Concave down on
12 2x x f
f
f
,0 , 2,
0,2
![Page 8: Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:](https://reader035.fdocuments.net/reader035/viewer/2022071807/56649e9e5503460f94b9ffc7/html5/thumbnails/8.jpg)
Let 3 24 12f x x x Summarizing info. from both tables:
x < 0
Decreasing
Concave up
0 < x < 2
Decreasing
Concave down
2 < x < 3
Decreasing
Concave up
x > 3
Increasing
Concave up
One possibility for the graph of :
x = 0
x = 2
x = 3
f
![Page 9: Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:](https://reader035.fdocuments.net/reader035/viewer/2022071807/56649e9e5503460f94b9ffc7/html5/thumbnails/9.jpg)
Let 3 24 12f x x x
One possibility for the graph of :
x = 0
x = 2
x = 3
Note: We are able to recover almost everything about adifferentiable function by examining its first derivative…
We cannot determine how to place the graph in the x-y plane(vertically) to position the graph we would need only thevalue of at one point!!!f
f
![Page 10: Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:](https://reader035.fdocuments.net/reader035/viewer/2022071807/56649e9e5503460f94b9ffc7/html5/thumbnails/10.jpg)
A function is continuous on , , , and derivatives have the following properties:
x –2 < x < 0
+
+
x = 0
does not exist
does not exist
0 < x < 2
–
+
x = 2
0
0
2 < x < 4
–
–
1. Find where all absolute extrema of occur.
2. Find where the points of inflection of occur.
3. Sketch a possible graph of .
f : 2, 4D 2 5f 4 1f
f f
f
f
f
![Page 11: Section 4.3b. Do Now: #30 on p.204 (solve graphically) (a) Local Maximum at (b) Local Minimum at (c) Points of Inflection:](https://reader035.fdocuments.net/reader035/viewer/2022071807/56649e9e5503460f94b9ffc7/html5/thumbnails/11.jpg)
A function is continuous on , , , and derivatives have the following properties:
x –2 < x < 0
+
+
x = 0
does not exist
does not exist
0 < x < 2
–
+
x = 2
0
0
2 < x < 4
–
–
f : 2, 4D 2 5f 4 1f
f f
Absolute Maximum occurs at x = 0(cannot determine its value)
Point of Inflection at x = 2
Absolute Minimum of 1 at x = 4