Lesson 20: Derivatives and the Shape of Curves (Section 041 slides)
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Transcript of Lesson 20: Derivatives and the Shape of Curves (Section 041 slides)
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Section 4.2Derivatives and the Shapes of Curves
V63.0121.041, Calculus I
New York University
November 15, 2010
AnnouncementsI Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7I There is class on November 24
. . . . . .
. . . . . .
Announcements
I Quiz 4 this week inrecitation on 3.3, 3.4, 3.5,3.7
I There is class onNovember 24
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 2 / 32
. . . . . .
Objectives
I Use the derivative of afunction to determine theintervals along which thefunction is increasing ordecreasing (TheIncreasing/DecreasingTest)
I Use the First DerivativeTest to classify criticalpoints of a function as localmaxima, local minima, orneither.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 3 / 32
. . . . . .
Objectives
I Use the second derivativeof a function to determinethe intervals along whichthe graph of the function isconcave up or concavedown (The Concavity Test)
I Use the first and secondderivative of a function toclassify critical points aslocal maxima or localminima, when applicable(The Second DerivativeTest)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 4 / 32
. . . . . .
Outline
Recall: The Mean Value Theorem
MonotonicityThe Increasing/Decreasing TestFinding intervals of monotonicityThe First Derivative Test
ConcavityDefinitionsTesting for ConcavityThe Second Derivative Test
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 5 / 32
. . . . . .
Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be continuous on [a,b]and differentiable 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)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32
. . . . . .
Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be continuous on [a,b]and differentiable 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)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32
. . . . . .
Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be continuous on [a,b]and differentiable 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)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32
. . . . . .
Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be continuous on [a,b]and differentiable 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)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 6 / 32
. . . . . .
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 continuous on[x, y] and differentiable 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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 7 / 32
. . . . . .
Outline
Recall: The Mean Value Theorem
MonotonicityThe Increasing/Decreasing TestFinding intervals of monotonicityThe First Derivative Test
ConcavityDefinitionsTesting for ConcavityThe Second Derivative Test
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 8 / 32
. . . . . .
What does it mean for a function to be increasing?
DefinitionA function 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 function “preserves order.”I I could be bounded or infinite, open, closed, or
half-open/half-closed.I Write your own definition (mutatis mutandis) of decreasing,nonincreasing, nondecreasing
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 9 / 32
. . . . . .
What does it mean for a function to be increasing?
DefinitionA function 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 function “preserves order.”I I could be bounded or infinite, open, closed, or
half-open/half-closed.I Write your own definition (mutatis mutandis) of decreasing,nonincreasing, nondecreasing
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 9 / 32
. . . . . .
The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0 onan interval, then f is decreasing on that interval.
Proof.It works the same as the last theorem. Assume f′(x) > 0 on an intervalI. Pick two points x and y in I with x < y. We must show f(x) < f(y). ByMVT there exists a point c in (x, y) such that
f(y)− f(x) = f′(c)(y− x) > 0.
So f(y) > f(x).
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 10 / 32
. . . . . .
The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0 onan interval, then f is decreasing on that interval.
Proof.It works the same as the last theorem. Assume f′(x) > 0 on an intervalI. Pick two points x and y in I with x < y. We must show f(x) < f(y). ByMVT there exists a point c in (x, y) such that
f(y)− f(x) = f′(c)(y− x) > 0.
So f(y) > f(x).
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 10 / 32
. . . . . .
Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x− 5.
Solutionf′(x) = 2 is always positive, so f is increasing on (−∞,∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solution
Since f′(x) =1
1+ x2is always positive, f(x) is always increasing.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32
. . . . . .
Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x− 5.
Solutionf′(x) = 2 is always positive, so f is increasing on (−∞,∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solution
Since f′(x) =1
1+ x2is always positive, f(x) is always increasing.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32
. . . . . .
Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x− 5.
Solutionf′(x) = 2 is always positive, so f is increasing on (−∞,∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solution
Since f′(x) =1
1+ x2is always positive, f(x) is always increasing.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32
. . . . . .
Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x− 5.
Solutionf′(x) = 2 is always positive, so f is increasing on (−∞,∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solution
Since f′(x) =1
1+ x2is always positive, f(x) is always increasing.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 11 / 32
. . . . . .
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative 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,∞)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
. . . . . .
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative 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,∞)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
. . . . . .
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative 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,∞)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
. . . . . .
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative 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,∞)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
. . . . . .
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative 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,∞)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
. . . . . .
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative 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,∞)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 12 / 32
. . . . . .
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solution
f′(x) = 23x
−1/3(x+ 2) + x2/3 = 13x
−1/3 (5x+ 4)
The critical points are 0 and and −4/5... x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.
−
.
+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
. . . . . .
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solution
f′(x) = 23x
−1/3(x+ 2) + x2/3 = 13x
−1/3 (5x+ 4)
The critical points are 0 and and −4/5... x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.
−
.
+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
. . . . . .
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solution
f′(x) = 23x
−1/3(x+ 2) + x2/3 = 13x
−1/3 (5x+ 4)
The critical points are 0 and and −4/5... x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.
−
.
+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
. . . . . .
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solution
f′(x) = 23x
−1/3(x+ 2) + x2/3 = 13x
−1/3 (5x+ 4)
The critical points are 0 and and −4/5... x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.
−
.
+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
. . . . . .
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solution
f′(x) = 23x
−1/3(x+ 2) + x2/3 = 13x
−1/3 (5x+ 4)
The critical points are 0 and and −4/5... x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.
−
.
+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
. . . . . .
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solution
f′(x) = 23x
−1/3(x+ 2) + x2/3 = 13x
−1/3 (5x+ 4)
The critical points are 0 and and −4/5... x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.
−
.
+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
. . . . . .
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solution
f′(x) = 23x
−1/3(x+ 2) + x2/3 = 13x
−1/3 (5x+ 4)
The critical points are 0 and and −4/5... x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.
−
.
+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
. . . . . .
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solution
f′(x) = 23x
−1/3(x+ 2) + x2/3 = 13x
−1/3 (5x+ 4)
The critical points are 0 and and −4/5... x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.
−
.
+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
. . . . . .
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solution
f′(x) = 23x
−1/3(x+ 2) + x2/3 = 13x
−1/3 (5x+ 4)
The critical points are 0 and and −4/5... x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.
−
.
+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 13 / 32
. . . . . .
The First Derivative Test
Theorem (The First Derivative Test)
Let f be continuous on [a,b] and c a critical point of f in (a,b).I If f′ changes from positive to negative at c, then c is a local
maximum.I If f′ changes from negative to positive 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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 14 / 32
. . . . . .
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative 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,∞)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 15 / 32
. . . . . .
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solution
I f′(x) = 2x, which is positive when x > 0 and negative 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,∞)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 15 / 32
. . . . . .
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solution
f′(x) = 23x
−1/3(x+ 2) + x2/3 = 13x
−1/3 (5x+ 4)
The critical points are 0 and and −4/5... x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.
−
.
+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 16 / 32
. . . . . .
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solution
f′(x) = 23x
−1/3(x+ 2) + x2/3 = 13x
−1/3 (5x+ 4)
The critical points are 0 and and −4/5... x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.
−
.
+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 16 / 32
. . . . . .
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3(x+ 2).
Solution
f′(x) = 23x
−1/3(x+ 2) + x2/3 = 13x
−1/3 (5x+ 4)
The critical points are 0 and and −4/5... x−1/3..0.×.− . +.
5x+ 4
..
−4/5
.
0
.
−
.
+
.
f′(x)
.
f(x)
..
−4/5
.
0
..
0
.
×
..
+
.
↗
..
−
.
↘
..
+
.
↗
.
max
.
min
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 16 / 32
. . . . . .
Outline
Recall: The Mean Value Theorem
MonotonicityThe Increasing/Decreasing TestFinding intervals of monotonicityThe First Derivative Test
ConcavityDefinitionsTesting for ConcavityThe Second Derivative Test
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 17 / 32
. . . . . .
Concavity
DefinitionThe graph of f is called concave upwards on an interval if it lies aboveall its tangents on that interval. The graph of f is called concavedownwards on an interval if it lies below all its tangents on thatinterval.
.
concave up
.
concave downWe sometimes say a concave up graph “holds water” and a concavedown graph “spills water”.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 18 / 32
. . . . . .
Synonyms for concavity
Remark
I “concave up” = “concave upwards” = “convex”I “concave down” = “concave downwards” = “concave”
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 19 / 32
. . . . . .
Inflection points indicate a change in concavity
DefinitionA point P on a curve y = f(x) is called an inflection point if f iscontinuous at P and the curve changes from concave upward toconcave downward at P (or vice versa).
..concavedown
.
concave up
..inflection point
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 20 / 32
. . . . . .
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.
Proof.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).
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 21 / 32
. . . . . .
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.
Proof.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).V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 21 / 32
. . . . . .
Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2.
Solution
I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x+ 2.I This is negative when x < −1/3, positive when x > −1/3, and 0
when x = −1/3
I So f is concave down on the open interval (−∞,−1/3), concave upon the open interval (−1/3,∞), and has an inflection point at thepoint (−1/3, 2/27)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32
. . . . . .
Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2.
Solution
I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x+ 2.
I This is negative when x < −1/3, positive when x > −1/3, and 0when x = −1/3
I So f is concave down on the open interval (−∞,−1/3), concave upon the open interval (−1/3,∞), and has an inflection point at thepoint (−1/3, 2/27)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32
. . . . . .
Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2.
Solution
I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x+ 2.I This is negative when x < −1/3, positive when x > −1/3, and 0
when x = −1/3
I So f is concave down on the open interval (−∞,−1/3), concave upon the open interval (−1/3,∞), and has an inflection point at thepoint (−1/3, 2/27)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32
. . . . . .
Finding Intervals of Concavity I
Example
Find the intervals of concavity for the graph of f(x) = x3 + x2.
Solution
I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x+ 2.I This is negative when x < −1/3, positive when x > −1/3, and 0
when x = −1/3
I So f is concave down on the open interval (−∞,−1/3), concave upon the open interval (−1/3,∞), and has an inflection point at thepoint (−1/3, 2/27)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 22 / 32
. . . . . .
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solution
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
. . . . . .
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solution
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
. . . . . .
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solution
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
. . . . . .
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solution
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
. . . . . .
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solution
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
. . . . . .
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solution
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
. . . . . .
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solution
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
. . . . . .
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solution
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
. . . . . .
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solution
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
. . . . . .
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).
Solution
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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 23 / 32
. . . . . .
The Second Derivative Test
Theorem (The Second Derivative Test)
Let f, f′, and f′′ be continuous on [a,b]. Let c be be a point in (a,b) withf′(c) = 0.
I If f′′(c) < 0, then c is a local maximum.I If f′′(c) > 0, then c is a local minimum.
Remarks
I If f′′(c) = 0, the second derivative test is inconclusive (this doesnot mean c is neither; we just don’t know yet).
I We look for zeroes of f′ and plug them into f′′ to determine if their fvalues are local extreme values.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 24 / 32
. . . . . .
The Second Derivative Test
Theorem (The Second Derivative Test)
Let f, f′, and f′′ be continuous on [a,b]. Let c be be a point in (a,b) withf′(c) = 0.
I If f′′(c) < 0, then c is a local maximum.I If f′′(c) > 0, then c is a local minimum.
Remarks
I If f′′(c) = 0, the second derivative test is inconclusive (this doesnot mean c is neither; we just don’t know yet).
I We look for zeroes of f′ and plug them into f′′ to determine if their fvalues are local extreme values.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 24 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+
..+ .. +.↗
.↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+
..+ .. +.↗
.↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+
.. +.↗
.↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +
.↗
.↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +
.↗
.↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗
.↗
.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Proof of the Second Derivative Test
Proof.Suppose f′(c) = 0 and f′′(c) > 0.
I Since f′′ is continuous,f′′(x) > 0 for all xsufficiently close to c.
I Since f′′ = (f′)′, we know f′
is increasing near c.
.. f′′ = (f′)′.f′
...c.+..+ .. +.
↗.
↗.
f′
.
f
...
c
.
0
..
−
..
+
.
↘
.
↗
.
min
I Since f′(c) = 0 and f′ is increasing, f′(x) < 0 for x close to c andless than c, and f′(x) > 0 for x close to c and more than c.
I This means f′ changes sign from negative to positive at c, whichmeans (by the First Derivative Test) that f has a local minimumat c.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 25 / 32
. . . . . .
Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2.
Solution
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
. . . . . .
Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2.
Solution
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
. . . . . .
Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2.
Solution
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
. . . . . .
Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2.
Solution
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
. . . . . .
Using the Second Derivative Test I
Example
Find the local extrema of f(x) = x3 + x2.
Solution
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.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 26 / 32
. . . . . .
Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3(x+ 2)
Solution
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 negative when
x = −4/5
I So x = −4/5 is a local maximum.I Notice the Second Derivative Test doesn’t catch the local
minimum x = 0 since f is not differentiable there.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
. . . . . .
Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3(x+ 2)
Solution
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 negative when
x = −4/5
I So x = −4/5 is a local maximum.I Notice the Second Derivative Test doesn’t catch the local
minimum x = 0 since f is not differentiable there.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
. . . . . .
Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3(x+ 2)
Solution
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 negative when
x = −4/5
I So x = −4/5 is a local maximum.I Notice the Second Derivative Test doesn’t catch the local
minimum x = 0 since f is not differentiable there.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
. . . . . .
Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3(x+ 2)
Solution
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 negative when
x = −4/5
I So x = −4/5 is a local maximum.
I Notice the Second Derivative Test doesn’t catch the localminimum x = 0 since f is not differentiable there.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
. . . . . .
Using the Second Derivative Test II
Example
Find the local extrema of f(x) = x2/3(x+ 2)
Solution
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 negative when
x = −4/5
I So x = −4/5 is a local maximum.I Notice the Second Derivative Test doesn’t catch the local
minimum x = 0 since f is not differentiable there.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 27 / 32
. . . . . .
Using the Second Derivative Test II: Graph
Graph of f(x) = x2/3(x+ 2):
.. x.
y
..
(−4/5,1.03413)
..(0,0)
..
(2/5,1.30292)
..(−2,0)
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 28 / 32
. . . . . .
When the second derivative is zero
Remark
I At inflection points c, if f′ is differentiable at c, then f′′(c) = 0I If f′′(c) = 0, must f have an inflection point at c?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
All of them have critical points at zero with a second derivative of zero.But the first has a local min at 0, the second has a local max at 0, andthe third has an inflection point at 0. This is why we say 2DT hasnothing to say when f′′(c) = 0.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 29 / 32
. . . . . .
When the second derivative is zero
Remark
I At inflection points c, if f′ is differentiable at c, then f′′(c) = 0I If f′′(c) = 0, must f have an inflection point at c?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
All of them have critical points at zero with a second derivative of zero.But the first has a local min at 0, the second has a local max at 0, andthe third has an inflection point at 0. This is why we say 2DT hasnothing to say when f′′(c) = 0.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 29 / 32
. . . . . .
When first and second derivative are zero
function derivatives 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
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 30 / 32
. . . . . .
When the second derivative is zero
Remark
I At inflection points c, if f′ is differentiable at c, then f′′(c) = 0I If f′′(c) = 0, must f have an inflection point at c?
Consider these examples:
f(x) = x4 g(x) = −x4 h(x) = x3
All of them have critical points at zero with a second derivative of zero.But the first has a local min at 0, the second has a local max at 0, andthe third has an inflection point at 0. This is why we say 2DT hasnothing to say when f′′(c) = 0.
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 31 / 32
. . . . . .
Summary
I Concepts: Mean Value Theorem, monotonicity, concavityI Facts: derivatives can detect monotonicity and concavityI Techniques for drawing curves: the Increasing/Decreasing Test
and the Concavity TestI Techniques for finding extrema: the First Derivative Test and theSecond Derivative Test
V63.0121.041, Calculus I (NYU) Section 4.2 The Shapes of Curves November 15, 2010 32 / 32