lesson 18: derivatives and the shapes of curves
DESCRIPTION
TRANSCRIPT
Section 4.2Derivatives and the Shapes of Curves
V63.0121.006/016, Calculus I
New York University
March 30, 2010
Announcements
I Quiz 3 on Friday (Sections 2.6–3.5)
I Midterm Exam scores have been updated
I If your Midterm Letter Grade on Blackboard differs from yourmidterm grade on Albert, trust Blackboard
Announcements
I Quiz 3 on Friday (Sections 2.6–3.5)
I Midterm Exam scores have been updated
I If your Midterm Letter Grade on Blackboard differs from yourmidterm grade on Albert, trust Blackboard
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 2 / 28
Outline
Recall: The Mean Value Theorem
MonotonicityThe Increasing/Decreasing TestFinding intervals of monotonicityThe First Derivative Test
ConcavityDefinitionsTesting for ConcavityThe Second Derivative Test
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 3 / 28
Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be continuous on [a, b] anddifferentiable on (a, b). Thenthere exists a point c in (a, b)such that
f (b)− f (a)
b − a= f ′(c).
a
b
c
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 4 / 28
Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be continuous on [a, b] anddifferentiable on (a, b). Thenthere exists a point c in (a, b)such that
f (b)− f (a)
b − a= f ′(c).
a
b
c
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 4 / 28
Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be continuous on [a, b] anddifferentiable on (a, b). Thenthere exists a point c in (a, b)such that
f (b)− f (a)
b − a= f ′(c).
a
b
c
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 4 / 28
Why the MVT is the MITCMost Important Theorem In Calculus!
Theorem
Let 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)
y − x= f ′(z) = 0.
So f (y) = f (x). Since this is true for all x and y in (a, b), then f isconstant.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 5 / 28
Outline
Recall: The Mean Value Theorem
MonotonicityThe Increasing/Decreasing TestFinding intervals of monotonicityThe First Derivative Test
ConcavityDefinitionsTesting for ConcavityThe Second Derivative Test
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 6 / 28
What does it mean for a function to be increasing?
Definition
A function f is increasing on (a, b) if
f (x) < f (y)
whenever x and y are two points in (a, b) with x < y .
I An increasing function “preserves order.”
I Write your own definition (mutatis mutandis) of decreasing,nonincreasing, nondecreasing
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 7 / 28
What does it mean for a function to be increasing?
Definition
A function f is increasing on (a, b) if
f (x) < f (y)
whenever x and y are two points in (a, b) with x < y .
I An increasing function “preserves order.”
I Write your own definition (mutatis mutandis) of decreasing,nonincreasing, nondecreasing
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 7 / 28
The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f ′ > 0 on (a, b), then f is increasing on (a, b). If f ′ < 0 on (a, b), thenf is decreasing on (a, b).
Proof.
It works the same as the last theorem. Pick two points x and y in (a, b)with x < y . We must show f (x) < f (y). By MVT there exists a point cin (x , y) such that
f (y)− f (x)
y − x= f ′(c) > 0.
Sof (y)− f (x) = f ′(c)(y − x) > 0.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 8 / 28
The Increasing/Decreasing Test
Theorem (The Increasing/Decreasing Test)
If f ′ > 0 on (a, b), then f is increasing on (a, b). If f ′ < 0 on (a, b), thenf is decreasing on (a, b).
Proof.
It works the same as the last theorem. Pick two points x and y in (a, b)with x < y . We must show f (x) < f (y). By MVT there exists a point cin (x , y) such that
f (y)− f (x)
y − x= f ′(c) > 0.
Sof (y)− f (x) = f ′(c)(y − x) > 0.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 8 / 28
Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f (x) = 2x − 5.
Solution
f ′(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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 9 / 28
Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f (x) = 2x − 5.
Solution
f ′(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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 9 / 28
Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f (x) = 2x − 5.
Solution
f ′(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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 9 / 28
Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f (x) = 2x − 5.
Solution
f ′(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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 9 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 10 / 28
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 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
x−1/30
×− +
5x + 4−4/5
0− +
f ′(x)
f (x)−4/5
0
0
×
+
↗−↘
+
↗max min
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
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 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
x−1/30
×− +
5x + 4−4/5
0− +
f ′(x)
f (x)−4/5
0
0
×
+
↗−↘
+
↗max min
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
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 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
x−1/30
×− +
5x + 4−4/5
0− +
f ′(x)
f (x)−4/5
0
0
×
+
↗−↘
+
↗max min
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
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 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
x−1/30
×− +
5x + 4−4/5
0− +
f ′(x)
f (x)−4/5
0
0
×+
↗−↘
+
↗max min
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
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 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
x−1/30
×− +
5x + 4−4/5
0− +
f ′(x)
f (x)−4/5
0
0
×+
↗
−
↘+
↗max min
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
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 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
x−1/30
×− +
5x + 4−4/5
0− +
f ′(x)
f (x)−4/5
0
0
×+
↗
−
↘
+
↗max min
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
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 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
x−1/30
×− +
5x + 4−4/5
0− +
f ′(x)
f (x)−4/5
0
0
×+
↗−
↘
+
↗max min
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
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 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
x−1/30
×− +
5x + 4−4/5
0− +
f ′(x)
f (x)−4/5
0
0
×+
↗−↘
+
↗max min
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
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 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
x−1/30
×− +
5x + 4−4/5
0− +
f ′(x)
f (x)−4/5
0
0
×+
↗−↘
+
↗
max min
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 11 / 28
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 ′(x) > 0 on (a, c) and f ′(x) < 0 on (c , b), then c is a localmaximum.
I If f ′(x) < 0 on (a, c) and f ′(x) > 0 on (c , b), then c is a localminimum.
I If f ′(x) has the same sign on (a, c) and (c , b), then c is not a localextremum.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 12 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 13 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 13 / 28
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 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
x−1/30
×− +
5x + 4−4/5
0− +
f ′(x)
f (x)−4/5
0
0
×+
↗−↘
+
↗
max min
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 14 / 28
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 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
x−1/30
×− +
5x + 4−4/5
0− +
f ′(x)
f (x)−4/5
0
0
×+
↗−↘
+
↗max
min
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 14 / 28
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 = 1
3x−1/3 (5x + 4)
The critical points are 0 and and −4/5.
x−1/30
×− +
5x + 4−4/5
0− +
f ′(x)
f (x)−4/5
0
0
×+
↗−↘
+
↗max min
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 14 / 28
Outline
Recall: The Mean Value Theorem
MonotonicityThe Increasing/Decreasing TestFinding intervals of monotonicityThe First Derivative Test
ConcavityDefinitionsTesting for ConcavityThe Second Derivative Test
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 15 / 28
Concavity
Definition
The graph of f is called concave up on an interval I if it lies above all itstangents on I . The graph of f is called concave down on I if it lies belowall its tangents on I .
concave up concave downWe sometimes say a concave up graph “holds water” and a concave downgraph “spills water”.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 16 / 28
Inflection points indicate a change in concavity
Definition
A 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 to concavedownward at P (or vice versa).
concavedown
concave up
inflection point
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 17 / 28
Theorem (Concavity Test)
I If f ′′(x) > 0 for all x in an interval I , then the graph of f is concaveupward on I .
I If f ′′(x) < 0 for all x in I , then the graph of f is concave downwardon I .
Proof.
Suppose f ′′(x) > 0 on I . This means f ′ is increasing on I . Let a and x bein I . The tangent line through (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 withf (x)− f (a)
x − a= f ′(c). So
f (x) = f (a) + f ′(c)(x − a) ≥ f (a) + f ′(a)(x − a) = L(x) .
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 18 / 28
Theorem (Concavity Test)
I If f ′′(x) > 0 for all x in an interval I , then the graph of f is concaveupward on I .
I If f ′′(x) < 0 for all x in I , then the graph of f is concave downwardon I .
Proof.
Suppose f ′′(x) > 0 on I . This means f ′ is increasing on I . Let a and x bein I . The tangent line through (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 withf (x)− f (a)
x − a= f ′(c). So
f (x) = f (a) + f ′(c)(x − a) ≥ f (a) + f ′(a)(x − a) = L(x) .
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 18 / 28
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 whenx = −1/3
I So f is concave down on (−∞,−1/3), concave up on (−1/3,∞), andhas an inflection point at (−1/3, 2/27)
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 19 / 28
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 whenx = −1/3
I So f is concave down on (−∞,−1/3), concave up on (−1/3,∞), andhas an inflection point at (−1/3, 2/27)
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 19 / 28
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 whenx = −1/3
I So f is concave down on (−∞,−1/3), concave up on (−1/3,∞), andhas an inflection point at (−1/3, 2/27)
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 19 / 28
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 whenx = −1/3
I So f is concave down on (−∞,−1/3), concave up on (−1/3,∞), andhas an inflection point at (−1/3, 2/27)
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 19 / 28
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f (x) = x2/3(x + 2).
Solution
I f ′′(x) =10
9x−1/3 − 4
9x−4/3 =
2
9x−4/3(5x − 2)
I The second derivative f ′′(x) is not defined at 0
I Otherwise, x−4/3 is always positive, so the concavity is determined bythe 5x − 2 factor
I So f is concave down on (−∞, 0], concave down on [0, 2/5), concaveup on (2/5,∞), and has an inflection point when x = 2/5
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f (x) = x2/3(x + 2).
Solution
I f ′′(x) =10
9x−1/3 − 4
9x−4/3 =
2
9x−4/3(5x − 2)
I The second derivative f ′′(x) is not defined at 0
I Otherwise, x−4/3 is always positive, so the concavity is determined bythe 5x − 2 factor
I So f is concave down on (−∞, 0], concave down on [0, 2/5), concaveup on (2/5,∞), and has an inflection point when x = 2/5
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f (x) = x2/3(x + 2).
Solution
I f ′′(x) =10
9x−1/3 − 4
9x−4/3 =
2
9x−4/3(5x − 2)
I The second derivative f ′′(x) is not defined at 0
I Otherwise, x−4/3 is always positive, so the concavity is determined bythe 5x − 2 factor
I So f is concave down on (−∞, 0], concave down on [0, 2/5), concaveup on (2/5,∞), and has an inflection point when x = 2/5
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f (x) = x2/3(x + 2).
Solution
I f ′′(x) =10
9x−1/3 − 4
9x−4/3 =
2
9x−4/3(5x − 2)
I The second derivative f ′′(x) is not defined at 0
I Otherwise, x−4/3 is always positive, so the concavity is determined bythe 5x − 2 factor
I So f is concave down on (−∞, 0], concave down on [0, 2/5), concaveup on (2/5,∞), and has an inflection point when x = 2/5
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f (x) = x2/3(x + 2).
Solution
I f ′′(x) =10
9x−1/3 − 4
9x−4/3 =
2
9x−4/3(5x − 2)
I The second derivative f ′′(x) is not defined at 0
I Otherwise, x−4/3 is always positive, so the concavity is determined bythe 5x − 2 factor
I So f is concave down on (−∞, 0], concave down on [0, 2/5), concaveup on (2/5,∞), and has an inflection point when x = 2/5
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 20 / 28
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)with f ′(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 does notmean 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 21 / 28
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)with f ′(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 does notmean 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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 21 / 28
Proof of the Second Derivative Test
Proof.
Suppose f ′(c) = 0 and f ′′(c) > 0. Since f ′′ is continuous, f ′′(x) > 0 forall x sufficiently close to c. So f ′ is increasing on an interval containing c .Since f ′(c) = 0 and f ′ is increasing, f ′(x) < 0 for x close to c and lessthan c , and f ′(x) > 0 for x close to c and more than c. This means f ′
changes sign from negative to positive at c , which means (by the FirstDerivative Test) that f has a local minimum at c .
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 22 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 23 / 28
Using the Second Derivative Test II
Example
Find the local extrema of f (x) = x2/3(x + 2)
Solution
I Remember f ′(x) =1
3x−1/3(5x + 4) which is zero when x = −4/5
I Remember f ′′(x) =10
9x−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 minimumx = 0 since f is not differentiable there.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
Using the Second Derivative Test II
Example
Find the local extrema of f (x) = x2/3(x + 2)
Solution
I Remember f ′(x) =1
3x−1/3(5x + 4) which is zero when x = −4/5
I Remember f ′′(x) =10
9x−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 minimumx = 0 since f is not differentiable there.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
Using the Second Derivative Test II
Example
Find the local extrema of f (x) = x2/3(x + 2)
Solution
I Remember f ′(x) =1
3x−1/3(5x + 4) which is zero when x = −4/5
I Remember f ′′(x) =10
9x−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 minimumx = 0 since f is not differentiable there.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
Using the Second Derivative Test II
Example
Find the local extrema of f (x) = x2/3(x + 2)
Solution
I Remember f ′(x) =1
3x−1/3(5x + 4) which is zero when x = −4/5
I Remember f ′′(x) =10
9x−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 minimumx = 0 since f is not differentiable there.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
Using the Second Derivative Test II
Example
Find the local extrema of f (x) = x2/3(x + 2)
Solution
I Remember f ′(x) =1
3x−1/3(5x + 4) which is zero when x = −4/5
I Remember f ′′(x) =10
9x−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 minimumx = 0 since f is not differentiable there.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 24 / 28
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, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 25 / 28
When the second derivative is zero
I At inflection points c , if f ′ is differentiable at c, then f ′′(c) = 0
I Is it necessarily true, though?
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, and thethird has an inflection point at 0. This is why we say 2DT has nothing tosay when f ′′(c) = 0.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 26 / 28
When the second derivative is zero
I At inflection points c , if f ′ is differentiable at c, then f ′′(c) = 0
I Is it necessarily true, though?
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, and thethird has an inflection point at 0. This is why we say 2DT has nothing tosay when f ′′(c) = 0.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 26 / 28
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
maxg ′′(x) = −12x2, g ′′(0) = 0
h(x) = x3h′(x) = 3x2, h′(0) = 0
infl.h′′(x) = 6x , h′′(0) = 0
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 27 / 28
When the second derivative is zero
I At inflection points c , if f ′ is differentiable at c, then f ′′(c) = 0
I Is it necessarily true, though?
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, and thethird has an inflection point at 0. This is why we say 2DT has nothing tosay when f ′′(c) = 0.
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 28 / 28
What have we learned today?
I Concepts: Mean Value Theorem, monotonicity, concavity
I Facts: derivatives can detect monotonicity and concavity
I Techniques for drawing curves: the Increasing/Decreasing Test andthe Concavity Test
I Techniques for finding extrema: the First Derivative Test and theSecond Derivative Test
Next time: Graphing functions!
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 29 / 28
What have we learned today?
I Concepts: Mean Value Theorem, monotonicity, concavity
I Facts: derivatives can detect monotonicity and concavity
I Techniques for drawing curves: the Increasing/Decreasing Test andthe Concavity Test
I Techniques for finding extrema: the First Derivative Test and theSecond Derivative Test
Next time: Graphing functions!
V63.0121, Calculus I (NYU) Section 4.2 The Shapes of Curves March 30, 2010 29 / 28