Lesson 18: Derivatives and the Shapes of Curves

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



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





f (b)− f (a)

b − a= f ′(c).

a

b

c





f (b)− f (a)

b − a= f ′(c).

a

b

c


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.


Outline





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


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


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.


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.


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.



Example


Solution


Example


Solution

Since f ′(x) =1




Example


Solution


Example


Solution

Since f ′(x) =1




Example


Solution


Example


Solution

Since f ′(x) =1



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,∞)



Example


Solution



f ′−0

0 +

min





Example


Solution



f ′−0

0 +

min





Example


Solution



f ′

f

−↘ 0

0 +

↗

min





Example


Solution



f ′

f

−↘ 0

0 +

↗

min





Example


Solution



f ′

f

−↘ 0

0 +

↗

min




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



Example


Solution

f ′(x) = 23x−1/3(x + 2) + x2/3 = 1

3x−1/3 (5x + 4)


x−1/30

×− +

5x + 4−4/5

0− +

f ′(x)

f (x)−4/5

0

0

×

+

↗−↘

+

↗max min



Example


Solution

f ′(x) = 23x−1/3(x + 2) + x2/3 = 1

3x−1/3 (5x + 4)


x−1/30

×− +

5x + 4−4/5

0− +

f ′(x)

f (x)−4/5

0

0

×

+

↗−↘

+

↗max min



Example


Solution

f ′(x) = 23x−1/3(x + 2) + x2/3 = 1

3x−1/3 (5x + 4)


x−1/30

×− +

5x + 4−4/5

0− +

f ′(x)

f (x)−4/5

0

0

×+

↗−↘

+

↗max min



Example


Solution

f ′(x) = 23x−1/3(x + 2) + x2/3 = 1

3x−1/3 (5x + 4)


x−1/30

×− +

5x + 4−4/5

0− +

f ′(x)

f (x)−4/5

0

0

×+

↗

−

↘+

↗max min



Example


Solution

f ′(x) = 23x−1/3(x + 2) + x2/3 = 1

3x−1/3 (5x + 4)


x−1/30

×− +

5x + 4−4/5

0− +

f ′(x)

f (x)−4/5

0

0

×+

↗

−

↘

+

↗max min



Example


Solution

f ′(x) = 23x−1/3(x + 2) + x2/3 = 1

3x−1/3 (5x + 4)


x−1/30

×− +

5x + 4−4/5

0− +

f ′(x)

f (x)−4/5

0

0

×+

↗−

↘

+

↗max min



Example


Solution

f ′(x) = 23x−1/3(x + 2) + x2/3 = 1

3x−1/3 (5x + 4)


x−1/30

×− +

5x + 4−4/5

0− +

f ′(x)

f (x)−4/5

0

0

×+

↗−↘

+

↗max min



Example


Solution

f ′(x) = 23x−1/3(x + 2) + x2/3 = 1

3x−1/3 (5x + 4)


x−1/30

×− +

5x + 4−4/5

0− +

f ′(x)

f (x)−4/5

0

0

×+

↗−↘

+

↗

max min


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.



Example


Solution



f ′

f

−↘ 0

0 +

↗

min





Example


Solution



f ′

f

−↘ 0

0 +

↗min





Example


Solution

f ′(x) = 23x−1/3(x + 2) + x2/3 = 1

3x−1/3 (5x + 4)


x−1/30

×− +

5x + 4−4/5

0− +

f ′(x)

f (x)−4/5

0

0

×+

↗−↘

+

↗

max min



Example


Solution

f ′(x) = 23x−1/3(x + 2) + x2/3 = 1

3x−1/3 (5x + 4)


x−1/30

×− +

5x + 4−4/5

0− +

f ′(x)

f (x)−4/5

0

0

×+

↗−↘

+

↗max

min



Example


Solution

f ′(x) = 23x−1/3(x + 2) + x2/3 = 1

3x−1/3 (5x + 4)


x−1/30

×− +

5x + 4−4/5

0− +

f ′(x)

f (x)−4/5

0

0

×+

↗−↘

+

↗max min


Outline





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”.


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


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) .


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) .


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)



Example


Solution






Example


Solution






Example


Solution





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



Example


Solution

I f ′′(x) =10

9x−1/3 − 4

9x−4/3 =

2

9x−4/3(5x − 2)






Example


Solution

I f ′′(x) =10

9x−1/3 − 4

9x−4/3 =

2

9x−4/3(5x − 2)






Example


Solution

I f ′′(x) =10

9x−1/3 − 4

9x−4/3 =

2

9x−4/3(5x − 2)






Example


Solution

I f ′′(x) =10

9x−1/3 − 4

9x−4/3 =

2

9x−4/3(5x − 2)





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.


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.


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 .


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.



Example


Solution







Example


Solution







Example


Solution







Example


Solution






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.



Example


Solution





x = −4/5





Example


Solution





x = −4/5





Example


Solution





x = −4/5





Example


Solution





x = −4/5




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)


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.






f (x) = x4 g(x) = −x4 h(x) = x3



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






f (x) = x4 g(x) = −x4 h(x) = x3



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!


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!


Education

Lesson 18: Derivatives and the Shapes of Curves