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..
Sec on 4.3Deriva ves and the Shapes of Curves
V63.0121.001: Calculus IProfessor Ma hew Leingang
New York University
April 11, 2011
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Announcements
I Quiz 4 on Sec ons 3.3,3.4, 3.5, and 3.7 thisweek (April 14/15)
I Quiz 5 on Sec ons4.1–4.4 April 28/29
I Final Exam Thursday May12, 2:00–3:50pm
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ObjectivesI Use the deriva ve of a func onto determine the intervals alongwhich the func on is increasingor decreasing (TheIncreasing/Decreasing Test)
I Use the First Deriva ve Test toclassify cri cal points of afunc on as local maxima, localminima, or neither.
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ObjectivesI Use the second deriva ve of afunc on to determine theintervals along which the graphof the func on is concave up orconcave down (The ConcavityTest)
I Use the first and secondderiva ve of a func on toclassify cri cal points as localmaxima or local minima, whenapplicable (The SecondDeriva ve Test)
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OutlineRecall: The Mean Value Theorem
MonotonicityThe Increasing/Decreasing TestFinding intervals of monotonicityThe First Deriva ve Test
ConcavityDefini onsTes ng for ConcavityThe Second Deriva ve Test
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Recall: The Mean Value TheoremTheorem (The Mean Value Theorem)
Let f be con nuous on [a, b]and differen able on (a, b).Then there exists a point c in(a, b) such that
f(b)− f(a)b− a
= f′(c)....
a..
b
..
c
Another way to put this is that there exists a point c such that
f(b) = f(a) + f′(c)(b− a)
![Page 7: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/7.jpg)
Recall: The Mean Value TheoremTheorem (The Mean Value Theorem)
Let f be con nuous on [a, b]and differen able on (a, b).Then there exists a point c in(a, b) such that
f(b)− f(a)b− a
= f′(c)....
a..
b
..
c
Another way to put this is that there exists a point c such that
f(b) = f(a) + f′(c)(b− a)
![Page 8: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/8.jpg)
Recall: The Mean Value TheoremTheorem (The Mean Value Theorem)
Let f be con nuous on [a, b]and differen able on (a, b).Then there exists a point c in(a, b) such that
f(b)− f(a)b− a
= f′(c)....
a..
b..
c
Another way to put this is that there exists a point c such that
f(b) = f(a) + f′(c)(b− a)
![Page 9: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/9.jpg)
Recall: The Mean Value TheoremTheorem (The Mean Value Theorem)
Let f be con nuous on [a, b]and differen able on (a, b).Then there exists a point c in(a, b) such that
f(b)− f(a)b− a
= f′(c)....
a..
b..
c
Another way to put this is that there exists a point c such that
f(b) = f(a) + f′(c)(b− a)
![Page 10: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/10.jpg)
Why the MVT is the MITCMost Important Theorem In Calculus!
TheoremLet f′ = 0 on an interval (a, b). Then f is constant on (a, b).
Proof.Pick any points x and y in (a, b) with x < y. Then f is con nuous on[x, y] and differen able on (x, y). By MVT there exists a point z in(x, y) such that
f(y) = f(x) + f′(z)(y− x)
So f(y) = f(x). Since this is true for all x and y in (a, b), then f isconstant.
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OutlineRecall: The Mean Value Theorem
MonotonicityThe Increasing/Decreasing TestFinding intervals of monotonicityThe First Deriva ve Test
ConcavityDefini onsTes ng for ConcavityThe Second Deriva ve Test
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Increasing FunctionsDefini onA func on f is increasing on the interval I if
f(x) < f(y)
whenever x and y are two points in I with x < y.
I An increasing func on “preserves order.”I I could be bounded or infinite, open, closed, orhalf-open/half-closed.
I Write your own defini on (muta s mutandis) of decreasing,nonincreasing, nondecreasing
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Increasing FunctionsDefini onA func on f is increasing on the interval I if
f(x) < f(y)
whenever x and y are two points in I with x < y.
I An increasing func on “preserves order.”I I could be bounded or infinite, open, closed, orhalf-open/half-closed.
I Write your own defini on (muta s mutandis) of decreasing,nonincreasing, nondecreasing
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The Increasing/Decreasing TestTheorem (The Increasing/Decreasing Test)
If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0on an interval, then f is decreasing on that interval.
Proof.It works the same as the last theorem. Assume f′(x) > 0 on aninterval I. Pick two points x and y in I with x < y. We must showf(x) < f(y). By MVT there exists a point c in (x, y) such that
f(y)− f(x) = f′(c)(y− x) > 0.
So f(y) > f(x).
![Page 15: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/15.jpg)
The Increasing/Decreasing TestTheorem (The Increasing/Decreasing Test)
If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0on an interval, then f is decreasing on that interval.
Proof.It works the same as the last theorem. Assume f′(x) > 0 on aninterval I. Pick two points x and y in I with x < y. We must showf(x) < f(y).
By MVT there exists a point c in (x, y) such that
f(y)− f(x) = f′(c)(y− x) > 0.
So f(y) > f(x).
![Page 16: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/16.jpg)
The Increasing/Decreasing TestTheorem (The Increasing/Decreasing Test)
If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0on an interval, then f is decreasing on that interval.
Proof.It works the same as the last theorem. Assume f′(x) > 0 on aninterval I. Pick two points x and y in I with x < y. We must showf(x) < f(y). By MVT there exists a point c in (x, y) such that
f(y)− f(x) = f′(c)(y− x) > 0.
So f(y) > f(x).
![Page 17: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/17.jpg)
The Increasing/Decreasing TestTheorem (The Increasing/Decreasing Test)
If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0on an interval, then f is decreasing on that interval.
Proof.It works the same as the last theorem. Assume f′(x) > 0 on aninterval I. Pick two points x and y in I with x < y. We must showf(x) < f(y). By MVT there exists a point c in (x, y) such that
f(y)− f(x) = f′(c)(y− x) > 0.
So f(y) > f(x).
![Page 18: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/18.jpg)
Finding intervals of monotonicity IExample
Find the intervals of monotonicity of f(x) = 2x− 5.
Solu onf′(x) = 2 is always posi ve, so f is increasing on (−∞,∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solu on
Since f′(x) =1
1+ x2is always posi ve, f(x) is always increasing.
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Finding intervals of monotonicity IExample
Find the intervals of monotonicity of f(x) = 2x− 5.
Solu onf′(x) = 2 is always posi ve, so f is increasing on (−∞,∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solu on
Since f′(x) =1
1+ x2is always posi ve, f(x) is always increasing.
![Page 20: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/20.jpg)
Finding intervals of monotonicity IExample
Find the intervals of monotonicity of f(x) = 2x− 5.
Solu onf′(x) = 2 is always posi ve, so f is increasing on (−∞,∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solu on
Since f′(x) =1
1+ x2is always posi ve, f(x) is always increasing.
![Page 21: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/21.jpg)
Finding intervals of monotonicity IExample
Find the intervals of monotonicity of f(x) = 2x− 5.
Solu onf′(x) = 2 is always posi ve, so f is increasing on (−∞,∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solu on
Since f′(x) =1
1+ x2is always posi ve, f(x) is always increasing.
![Page 22: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/22.jpg)
Finding intervals of monotonicity IIExample
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
.
.
min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).I In fact we can say f is decreasing on (−∞, 0] and increasing on[0,∞)
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Finding intervals of monotonicity IIExample
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
I f′(x) = 2x, which is posi ve when x > 0 and nega ve when x is.
.
.
min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).I In fact we can say f is decreasing on (−∞, 0] and increasing on[0,∞)
![Page 24: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/24.jpg)
Finding intervals of monotonicity IIExample
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
I f′(x) = 2x, which is posi ve when x > 0 and nega ve when x is.I We can draw a number line:
.. f′.− ..0.0. +
.
min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).
I In fact we can say f is decreasing on (−∞, 0] and increasing on[0,∞)
![Page 25: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/25.jpg)
Finding intervals of monotonicity IIExample
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
I f′(x) = 2x, which is posi ve when x > 0 and nega ve when x is.I We can draw a number line:
.. f′.f
.− .↘
..0.0. +.
↗
.
min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).I In fact we can say f is decreasing on (−∞, 0] and increasing on[0,∞)
![Page 26: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/26.jpg)
Finding intervals of monotonicity IIExample
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
.. f′.f
.− .↘
..0.0. +.
↗
.
min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).
I In fact we can say f is decreasing on (−∞, 0] and increasing on[0,∞)
![Page 27: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/27.jpg)
Finding intervals of monotonicity IIExample
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
.. f′.f
.− .↘
..0.0. +.
↗
.
min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).I In fact we can say f is decreasing on (−∞, 0] and increasing on[0,∞)
![Page 28: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/28.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.
![Page 29: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/29.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.
![Page 30: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/30.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.
![Page 31: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/31.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3
![Page 32: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/32.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×
![Page 33: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/33.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.−
![Page 34: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/34.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +
![Page 35: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/35.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
![Page 36: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/36.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
![Page 37: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/37.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
![Page 38: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/38.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
.+
![Page 39: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/39.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
.+
.
f′(x)
.
f(x)
![Page 40: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/40.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
.+
.
f′(x)
.
f(x)
..
−4/5
.
0
![Page 41: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/41.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
.+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
![Page 42: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/42.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
.+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
![Page 43: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/43.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
.+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
..
−
![Page 44: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/44.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
.+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
..
−
..
+
![Page 45: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/45.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
.+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
..
−
..
+
.
↗
![Page 46: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/46.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
.+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
..
−
..
+
.
↗
.
↘
![Page 47: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/47.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
.+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
..
−
..
+
.
↗
.
↘
.
↗
![Page 48: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/48.jpg)
The First Derivative Test
Theorem (The First Deriva ve Test)
Let f be con nuous on [a, b] and c a cri cal point of f in (a, b).I If f′ changes from posi ve to nega ve at c, then c is a local
maximum.I If f′ changes from nega ve to posi ve at c, then c is a local
minimum.I If f′(x) has the same sign on either side of c, then c is not a local
extremum.
![Page 49: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/49.jpg)
Finding intervals of monotonicity IIExample
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
.. f′.f
.− .↘
..0.0. +.
↗
.
min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).I In fact we can say f is decreasing on (−∞, 0] and increasing on[0,∞)
![Page 50: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/50.jpg)
Finding intervals of monotonicity IIExample
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
.. f′.f
.− .↘
..0.0. +.
↗.
min
I So f is decreasing on (−∞, 0) and increasing on (0,∞).I In fact we can say f is decreasing on (−∞, 0] and increasing on[0,∞)
![Page 51: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/51.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
.+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
..
−
..
+
.
↗
.
↘
.
↗
![Page 52: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/52.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
.+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
..
−
..
+
.
↗
.
↘
.
↗
.
max
![Page 53: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/53.jpg)
Finding intervals of monotonicity IIIExample
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solu on
f′(x) = 23x
−1/3(x+ 2) + x2/3
= 13x
−1/3 (5x+ 4)
The cri cal points are 0 andand−4/5.
.. x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.−
.+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
..
−
..
+
.
↗
.
↘
.
↗
.
max
.
min
![Page 54: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/54.jpg)
OutlineRecall: The Mean Value Theorem
MonotonicityThe Increasing/Decreasing TestFinding intervals of monotonicityThe First Deriva ve Test
ConcavityDefini onsTes ng for ConcavityThe Second Deriva ve Test
![Page 55: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/55.jpg)
ConcavityDefini onThe graph of f is called concave upwards on an interval if it liesabove all its tangents on that interval. The graph of f is calledconcave downwards on an interval if it lies below all its tangents onthat interval.
.
concave up
.
concave down
![Page 56: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/56.jpg)
ConcavityDefini onThe graph of f is called concave upwards on an interval if it liesabove all its tangents on that interval. The graph of f is calledconcave downwards on an interval if it lies below all its tangents onthat interval.
.
concave up
.
concave down
![Page 57: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/57.jpg)
ConcavityDefini on
.
concave up
.
concave downWe some mes say a concave up graph “holds water” and a concavedown graph “spills water”.
![Page 58: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/58.jpg)
Synonyms for concavity
Remark
I “concave up” = “concave upwards” = “convex”I “concave down” = “concave downwards” = “concave”
![Page 59: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/59.jpg)
Inflection points mean change in concavityDefini onA point P on a curve y = f(x) is called an inflec on point if f iscon nuous at P and the curve changes from concave upward toconcave downward at P (or vice versa).
..concavedown
.
concaveup
..inflec on point
![Page 60: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/60.jpg)
Testing for Concavity
Theorem (Concavity Test)
I If f′′(x) > 0 for all x in an interval, then the graph of f is concaveupward on that interval.
I If f′′(x) < 0 for all x in an interval, then the graph of f is concavedownward on that interval.
![Page 61: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/61.jpg)
Testing for ConcavityProof.Suppose f′′(x) > 0 on the interval I (which could be infinite). Thismeans f′ is increasing on I.
Let a and x be in I. The tangent linethrough (a, f(a)) is the graph of
L(x) = f(a) + f′(a)(x− a)
By MVT, there exists a c between a and x with
f(x) = f(a) + f′(c)(x− a)
Since f′ is increasing, f(x) > L(x).
![Page 62: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/62.jpg)
Testing for ConcavityProof.Suppose f′′(x) > 0 on the interval I (which could be infinite). Thismeans f′ is increasing on I. Let a and x be in I. The tangent linethrough (a, f(a)) is the graph of
L(x) = f(a) + f′(a)(x− a)
By MVT, there exists a c between a and x with
f(x) = f(a) + f′(c)(x− a)
Since f′ is increasing, f(x) > L(x).
![Page 63: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/63.jpg)
Testing for ConcavityProof.Suppose f′′(x) > 0 on the interval I (which could be infinite). Thismeans f′ is increasing on I. Let a and x be in I. The tangent linethrough (a, f(a)) is the graph of
L(x) = f(a) + f′(a)(x− a)
By MVT, there exists a c between a and x with
f(x) = f(a) + f′(c)(x− a)
Since f′ is increasing, f(x) > L(x).
![Page 64: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/64.jpg)
Testing for ConcavityProof.Suppose f′′(x) > 0 on the interval I (which could be infinite). Thismeans f′ is increasing on I. Let a and x be in I. The tangent linethrough (a, f(a)) is the graph of
L(x) = f(a) + f′(a)(x− a)
By MVT, there exists a c between a and x with
f(x) = f(a) + f′(c)(x− a)
Since f′ is increasing, f(x) > L(x).
![Page 65: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/65.jpg)
Finding Intervals of Concavity IExample
Find the intervals of concavity for the graph of f(x) = x3 + x2.
Solu on
I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x+ 2.I This is nega ve when x < −1/3, posi ve when x > −1/3, and 0
when x = −1/3
I So f is concave down on the open interval (−∞,−1/3), concaveup on the open interval (−1/3,∞), and has an inflec on pointat the point (−1/3, 2/27)
![Page 66: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/66.jpg)
Finding Intervals of Concavity IExample
Find the intervals of concavity for the graph of f(x) = x3 + x2.
Solu on
I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x+ 2.
I This is nega ve when x < −1/3, posi ve when x > −1/3, and 0when x = −1/3
I So f is concave down on the open interval (−∞,−1/3), concaveup on the open interval (−1/3,∞), and has an inflec on pointat the point (−1/3, 2/27)
![Page 67: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/67.jpg)
Finding Intervals of Concavity IExample
Find the intervals of concavity for the graph of f(x) = x3 + x2.
Solu on
I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x+ 2.I This is nega ve when x < −1/3, posi ve when x > −1/3, and 0
when x = −1/3
I So f is concave down on the open interval (−∞,−1/3), concaveup on the open interval (−1/3,∞), and has an inflec on pointat the point (−1/3, 2/27)
![Page 68: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/68.jpg)
Finding Intervals of Concavity IExample
Find the intervals of concavity for the graph of f(x) = x3 + x2.
Solu on
I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x+ 2.I This is nega ve when x < −1/3, posi ve when x > −1/3, and 0
when x = −1/3
I So f is concave down on the open interval (−∞,−1/3), concaveup on the open interval (−1/3,∞), and has an inflec on pointat the point (−1/3, 2/27)
![Page 69: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/69.jpg)
Finding Intervals of Concavity IIExample
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solu on
We have
f′′(x) =109x−1/3 − 4
9x−4/3
=29x−4/3(5x− 2)
.. x−4/3..0.×.+ . +.
5x− 2
..
2/5
.
0
.−
.+
.
f′′(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢
.
⌢
.
⌣
.
IP
![Page 70: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/70.jpg)
Finding Intervals of Concavity IIExample
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solu on
We have
f′′(x) =109x−1/3 − 4
9x−4/3
=29x−4/3(5x− 2)
.. x−4/3..0.×.+ . +.
5x− 2
..
2/5
.
0
.−
.+
.
f′′(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢
.
⌢
.
⌣
.
IP
![Page 71: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/71.jpg)
Finding Intervals of Concavity IIExample
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solu on
We have
f′′(x) =109x−1/3 − 4
9x−4/3
=29x−4/3(5x− 2)
.. x−4/3..0.×.+ . +.
5x− 2
..
2/5
.
0
.−
.+
.
f′′(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢
.
⌢
.
⌣
.
IP
![Page 72: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/72.jpg)
Finding Intervals of Concavity IIExample
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solu on
We have
f′′(x) =109x−1/3 − 4
9x−4/3
=29x−4/3(5x− 2)
.. x−4/3..0.×.+ . +.
5x− 2
..
2/5
.
0
.−
.+
.
f′′(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢
.
⌢
.
⌣
.
IP
![Page 73: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/73.jpg)
Finding Intervals of Concavity IIExample
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solu on
We have
f′′(x) =109x−1/3 − 4
9x−4/3
=29x−4/3(5x− 2)
.. x−4/3..0.×.+ . +.
5x− 2
..
2/5
.
0
.−
.+
.
f′′(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢
.
⌢
.
⌣
.
IP
![Page 74: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/74.jpg)
Finding Intervals of Concavity IIExample
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solu on
We have
f′′(x) =109x−1/3 − 4
9x−4/3
=29x−4/3(5x− 2)
.. x−4/3..0.×.+ . +.
5x− 2
..
2/5
.
0
.−
.+
.
f′′(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢
.
⌢
.
⌣
.
IP
![Page 75: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/75.jpg)
Finding Intervals of Concavity IIExample
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solu on
We have
f′′(x) =109x−1/3 − 4
9x−4/3
=29x−4/3(5x− 2)
.. x−4/3..0.×.+ . +.
5x− 2
..
2/5
.
0
.−
.+
.
f′′(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢
.
⌢
.
⌣
.
IP
![Page 76: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/76.jpg)
Finding Intervals of Concavity IIExample
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solu on
We have
f′′(x) =109x−1/3 − 4
9x−4/3
=29x−4/3(5x− 2)
.. x−4/3..0.×.+ . +.
5x− 2
..
2/5
.
0
.−
.+
.
f′′(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢
.
⌢
.
⌣
.
IP
![Page 77: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/77.jpg)
Finding Intervals of Concavity IIExample
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solu on
We have
f′′(x) =109x−1/3 − 4
9x−4/3
=29x−4/3(5x− 2)
.. x−4/3..0.×.+ . +.
5x− 2
..
2/5
.
0
.−
.+
.
f′′(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢
.
⌢
.
⌣
.
IP
![Page 78: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/78.jpg)
Finding Intervals of Concavity IIExample
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solu on
We have
f′′(x) =109x−1/3 − 4
9x−4/3
=29x−4/3(5x− 2)
.. x−4/3..0.×.+ . +.
5x− 2
..
2/5
.
0
.−
.+
.
f′′(x)
.
f(x)
..
2/5
.
0
..
0
.
×
..
−−
..
−−
..
++
.
⌢
.
⌢
.
⌣
.
IP
![Page 79: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/79.jpg)
The Second Derivative TestTheorem (The Second Deriva ve Test)
Let f, f′, and f′′ be con nuous on [a, b]. Let c be be a point in (a, b)with f′(c) = 0.
I If f′′(c) < 0, then c is a local maximum of f.I If f′′(c) > 0, then c is a local minimum of f.
Remarks
I If f′′(c) = 0, the second deriva ve test is inconclusiveI We look for zeroes of f′ and plug them into f′′ to determine iftheir f values are local extreme values.
![Page 80: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/80.jpg)
The Second Derivative TestTheorem (The Second Deriva ve Test)
Let f, f′, and f′′ be con nuous on [a, b]. Let c be be a point in (a, b)with f′(c) = 0.
I If f′′(c) < 0, then c is a local maximum of f.I If f′′(c) > 0, then c is a local minimum of f.
Remarks
I If f′′(c) = 0, the second deriva ve test is inconclusiveI We look for zeroes of f′ and plug them into f′′ to determine iftheir f values are local extreme values.
![Page 81: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/81.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
.. f′′ = (f′)′.f′
...c.+
..+ .. +.↗
.↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 82: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/82.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is con nuous,f′′(x) > 0 for all xsufficiently close to c. .. f′′ = (f′)′.
f′...c.+
..+ .. +.↗
.↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 83: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/83.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is con nuous,f′′(x) > 0 for all xsufficiently close to c. .. f′′ = (f′)′.
f′...c.+..+
.. +.↗
.↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 84: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/84.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is con nuous,f′′(x) > 0 for all xsufficiently close to c. .. f′′ = (f′)′.
f′...c.+..+ .. +
.↗
.↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 85: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/85.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is con nuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we knowf′ is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +
.↗
.↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 86: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/86.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is con nuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we knowf′ is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗
.↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 87: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/87.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is con nuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we knowf′ is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 88: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/88.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′(c) = 0 and f′ isincreasing, f′(x) < 0 for xclose to c and less than c,and f′(x) > 0 for x closeto c and more than c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 89: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/89.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′(c) = 0 and f′ isincreasing, f′(x) < 0 for xclose to c and less than c,and f′(x) > 0 for x closeto c and more than c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 90: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/90.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′(c) = 0 and f′ isincreasing, f′(x) < 0 for xclose to c and less than c,and f′(x) > 0 for x closeto c and more than c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 91: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/91.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
I This means f′ changessign from nega ve toposi ve at c, whichmeans (by the FirstDeriva ve Test) that f hasa local minimum at c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 92: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/92.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
I This means f′ changessign from nega ve toposi ve at c, whichmeans (by the FirstDeriva ve Test) that f hasa local minimum at c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 93: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/93.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
I This means f′ changessign from nega ve toposi ve at c, whichmeans (by the FirstDeriva ve Test) that f hasa local minimum at c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 94: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/94.jpg)
Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.
I This means f′ changessign from nega ve toposi ve at c, whichmeans (by the FirstDeriva ve Test) that f hasa local minimum at c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
![Page 95: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/95.jpg)
Using the Second Derivative Test IExample
Find the local extrema of f(x) = x3 + x2.
Solu on
I f′(x) = 3x2 + 2x = x(3x+ 2) is 0 when x = 0 or x = −2/3.I Remember f′′(x) = 6x+ 2I Since f′′(−2/3) = −2 < 0,−2/3 is a local maximum.I Since f′′(0) = 2 > 0, 0 is a local minimum.
![Page 96: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/96.jpg)
Using the Second Derivative Test IExample
Find the local extrema of f(x) = x3 + x2.
Solu on
I f′(x) = 3x2 + 2x = x(3x+ 2) is 0 when x = 0 or x = −2/3.
I Remember f′′(x) = 6x+ 2I Since f′′(−2/3) = −2 < 0,−2/3 is a local maximum.I Since f′′(0) = 2 > 0, 0 is a local minimum.
![Page 97: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/97.jpg)
Using the Second Derivative Test IExample
Find the local extrema of f(x) = x3 + x2.
Solu on
I f′(x) = 3x2 + 2x = x(3x+ 2) is 0 when x = 0 or x = −2/3.I Remember f′′(x) = 6x+ 2
I Since f′′(−2/3) = −2 < 0,−2/3 is a local maximum.I Since f′′(0) = 2 > 0, 0 is a local minimum.
![Page 98: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/98.jpg)
Using the Second Derivative Test IExample
Find the local extrema of f(x) = x3 + x2.
Solu on
I f′(x) = 3x2 + 2x = x(3x+ 2) is 0 when x = 0 or x = −2/3.I Remember f′′(x) = 6x+ 2I Since f′′(−2/3) = −2 < 0,−2/3 is a local maximum.
I Since f′′(0) = 2 > 0, 0 is a local minimum.
![Page 99: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/99.jpg)
Using the Second Derivative Test IExample
Find the local extrema of f(x) = x3 + x2.
Solu on
I f′(x) = 3x2 + 2x = x(3x+ 2) is 0 when x = 0 or x = −2/3.I Remember f′′(x) = 6x+ 2I Since f′′(−2/3) = −2 < 0,−2/3 is a local maximum.I Since f′′(0) = 2 > 0, 0 is a local minimum.
![Page 100: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/100.jpg)
Using the Second Derivative Test IIExample
Find the local extrema of f(x) = x2/3(x+ 2)
Solu on
I Remember f′(x) =13x−1/3(5x+ 4) which is zero when x = −4/5
I Remember f′′(x) =109x−4/3(5x− 2), which is nega ve when
x = −4/5
I So x = −4/5 is a local maximum.I No ce the Second Deriva ve Test doesn’t catch the local
minimum x = 0 since f is not differen able there.
![Page 101: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/101.jpg)
Using the Second Derivative Test IIExample
Find the local extrema of f(x) = x2/3(x+ 2)
Solu on
I Remember f′(x) =13x−1/3(5x+ 4) which is zero when x = −4/5
I Remember f′′(x) =109x−4/3(5x− 2), which is nega ve when
x = −4/5
I So x = −4/5 is a local maximum.I No ce the Second Deriva ve Test doesn’t catch the local
minimum x = 0 since f is not differen able there.
![Page 102: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/102.jpg)
Using the Second Derivative Test IIExample
Find the local extrema of f(x) = x2/3(x+ 2)
Solu on
I Remember f′(x) =13x−1/3(5x+ 4) which is zero when x = −4/5
I Remember f′′(x) =109x−4/3(5x− 2), which is nega ve when
x = −4/5
I So x = −4/5 is a local maximum.I No ce the Second Deriva ve Test doesn’t catch the local
minimum x = 0 since f is not differen able there.
![Page 103: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/103.jpg)
Using the Second Derivative Test IIExample
Find the local extrema of f(x) = x2/3(x+ 2)
Solu on
I Remember f′(x) =13x−1/3(5x+ 4) which is zero when x = −4/5
I Remember f′′(x) =109x−4/3(5x− 2), which is nega ve when
x = −4/5
I So x = −4/5 is a local maximum.
I No ce the Second Deriva ve Test doesn’t catch the localminimum x = 0 since f is not differen able there.
![Page 104: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/104.jpg)
Using the Second Derivative Test IIExample
Find the local extrema of f(x) = x2/3(x+ 2)
Solu on
I Remember f′(x) =13x−1/3(5x+ 4) which is zero when x = −4/5
I Remember f′′(x) =109x−4/3(5x− 2), which is nega ve when
x = −4/5
I So x = −4/5 is a local maximum.I No ce the Second Deriva ve Test doesn’t catch the local
minimum x = 0 since f is not differen able there.
![Page 105: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/105.jpg)
Using the Second Derivative Test IIGraph
Graph of f(x) = x2/3(x+ 2):
.. x.
y
..
(−4/5, 1.03413)
..(0, 0)
..(2/5, 1.30292)
..(−2, 0)
![Page 106: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/106.jpg)
When the second derivative is zeroRemark
I At inflec on points c, if f′ is differen able at c, then f′′(c) = 0I If f′′(c) = 0, must f have an inflec on point at c?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
All of them have cri cal points at zero with a second deriva ve ofzero. But the first has a local min at 0, the second has a local max at0, and the third has an inflec on point at 0. This is why we say 2DThas nothing to say when f′′(c) = 0.
![Page 107: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/107.jpg)
When the second derivative is zeroRemark
I At inflec on points c, if f′ is differen able at c, then f′′(c) = 0I If f′′(c) = 0, must f have an inflec on point at c?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
All of them have cri cal points at zero with a second deriva ve ofzero. But the first has a local min at 0, the second has a local max at0, and the third has an inflec on point at 0. This is why we say 2DThas nothing to say when f′′(c) = 0.
![Page 108: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/108.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) =
4x3
, f′(0) =
0. min
f′′(x) =
12x2
, f′′(0) =
0
g(x) = −x4g′(x) =
− 4x3
, g′(0) =
0 .max
g′′(x) =
− 12x2
, g′′(0) =
0
h(x) = x3h′(x) =
3x2
, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 109: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/109.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) =
0. min
f′′(x) =
12x2
, f′′(0) =
0
g(x) = −x4g′(x) =
− 4x3
, g′(0) =
0 .max
g′′(x) =
− 12x2
, g′′(0) =
0
h(x) = x3h′(x) =
3x2
, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 110: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/110.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. min
f′′(x) =
12x2
, f′′(0) =
0
g(x) = −x4g′(x) =
− 4x3
, g′(0) =
0 .max
g′′(x) =
− 12x2
, g′′(0) =
0
h(x) = x3h′(x) =
3x2
, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 111: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/111.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. min
f′′(x) = 12x2, f′′(0) =
0
g(x) = −x4g′(x) =
− 4x3
, g′(0) =
0 .max
g′′(x) =
− 12x2
, g′′(0) =
0
h(x) = x3h′(x) =
3x2
, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 112: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/112.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. min
f′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) =
− 4x3
, g′(0) =
0 .max
g′′(x) =
− 12x2
, g′′(0) =
0
h(x) = x3h′(x) =
3x2
, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 113: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/113.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
.
min
f′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) =
− 4x3
, g′(0) =
0 .max
g′′(x) =
− 12x2
, g′′(0) =
0
h(x) = x3h′(x) =
3x2
, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 114: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/114.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. minf′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) =
− 4x3
, g′(0) =
0 .max
g′′(x) =
− 12x2
, g′′(0) =
0
h(x) = x3h′(x) =
3x2
, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 115: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/115.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. minf′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) = − 4x3, g′(0) =
0 .max
g′′(x) =
− 12x2
, g′′(0) =
0
h(x) = x3h′(x) =
3x2
, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 116: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/116.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. minf′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) = − 4x3, g′(0) = 0
.max
g′′(x) =
− 12x2
, g′′(0) =
0
h(x) = x3h′(x) =
3x2
, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 117: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/117.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. minf′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) = − 4x3, g′(0) = 0
.max
g′′(x) = − 12x2, g′′(0) =
0
h(x) = x3h′(x) =
3x2
, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 118: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/118.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. minf′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) = − 4x3, g′(0) = 0
.max
g′′(x) = − 12x2, g′′(0) = 0
h(x) = x3h′(x) =
3x2
, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 119: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/119.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. minf′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) = − 4x3, g′(0) = 0 .
max
g′′(x) = − 12x2, g′′(0) = 0
h(x) = x3h′(x) =
3x2
, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 120: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/120.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. minf′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) = − 4x3, g′(0) = 0 .
maxg′′(x) = − 12x2, g′′(0) = 0
h(x) = x3h′(x) =
3x2
, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 121: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/121.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. minf′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) = − 4x3, g′(0) = 0 .
maxg′′(x) = − 12x2, g′′(0) = 0
h(x) = x3h′(x) = 3x2, h′(0) =
0. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 122: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/122.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. minf′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) = − 4x3, g′(0) = 0 .
maxg′′(x) = − 12x2, g′′(0) = 0
h(x) = x3h′(x) = 3x2, h′(0) = 0
. infl.
h′′(x) =
6x
, h′′(0) =
0
![Page 123: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/123.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. minf′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) = − 4x3, g′(0) = 0 .
maxg′′(x) = − 12x2, g′′(0) = 0
h(x) = x3h′(x) = 3x2, h′(0) = 0
. infl.
h′′(x) = 6x, h′′(0) =
0
![Page 124: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/124.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. minf′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) = − 4x3, g′(0) = 0 .
maxg′′(x) = − 12x2, g′′(0) = 0
h(x) = x3h′(x) = 3x2, h′(0) = 0
. infl.
h′′(x) = 6x, h′′(0) = 0
![Page 125: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/125.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. minf′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) = − 4x3, g′(0) = 0 .
maxg′′(x) = − 12x2, g′′(0) = 0
h(x) = x3h′(x) = 3x2, h′(0) = 0
.
infl.
h′′(x) = 6x, h′′(0) = 0
![Page 126: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/126.jpg)
When first and second derivative are zero
func on deriva ves graph type
f(x) = x4f′(x) = 4x3, f′(0) = 0
. minf′′(x) = 12x2, f′′(0) = 0
g(x) = −x4g′(x) = − 4x3, g′(0) = 0 .
maxg′′(x) = − 12x2, g′′(0) = 0
h(x) = x3h′(x) = 3x2, h′(0) = 0
. infl.h′′(x) = 6x, h′′(0) = 0
![Page 127: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/127.jpg)
When the second derivative is zeroRemark
I At inflec on points c, if f′ is differen able at c, then f′′(c) = 0I If f′′(c) = 0, must f have an inflec on point at c?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
All of them have cri cal points at zero with a second deriva ve ofzero. But the first has a local min at 0, the second has a local max at0, and the third has an inflec on point at 0. This is why we say 2DThas nothing to say when f′′(c) = 0.
![Page 128: Lesson 20: Derivatives and the Shapes of Curves (slides)](https://reader033.vdocument.in/reader033/viewer/2022051610/54813d4cb37959582b8b5d58/html5/thumbnails/128.jpg)
Summary
I Concepts: Mean Value Theorem, monotonicity, concavityI Facts: deriva ves can detect monotonicity and concavityI Techniques for drawing curves: the Increasing/Decreasing Testand the Concavity Test
I Techniques for finding extrema: the First Deriva ve Test andthe Second Deriva ve Test