Lesson20 -derivatives_and_the_shape_of_curves_021_slides

..

Section 4.2Derivatives and the Shapes of Curves

V63.0121.021, Calculus I

New York University

November 16, 2010

AnnouncementsI Quiz 4 this week in recitation on 3.3, 3.4, 3.5, 3.7I There is class on November 23. The homework is due on

November 24. Turn in homework to my mailbox or bring to classon November 23.

. . . . . .

. . . . . .

Announcements

I Quiz 4 this week inrecitation on 3.3, 3.4, 3.5,3.7

I There is class onNovember 23. Thehomework is due onNovember 24. Turn inhomework to my mailboxor bring to class onNovember 23.

V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 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.


. . . . . .

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)


. . . . . .

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]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)


. . . . . .




f(b)− f(a)b− a

= f′(c). ...a

..b

..

c


f(b) = f(a) + f′(c)(b− a)


. . . . . .




f(b)− f(a)b− a

= f′(c). ...a

..b

..

c


f(b) = f(a) + f′(c)(b− a)


. . . . . .




f(b)− f(a)b− a

= f′(c). ...a

..b

..

c


f(b) = f(a) + f′(c)(b− a)


. . . . . .

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.


. . . . . .

Outline





. . . . . .

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


. . . . . .

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


. . . . . .

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).


. . . . . .

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).


. . . . . .

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.


. . . . . .


Example



Example


Solution

Since f′(x) =1



. . . . . .


Example



Example


Solution

Since f′(x) =1



. . . . . .


Example



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

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


[0,∞)


. . . . . .


Example


Solution


.. f′.− ..0.0. +

.

min


[0,∞)


. . . . . .


Example


Solution


.. 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,∞)


. . . . . .


Example


Solution


.. 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,∞)


. . . . . .


Example


Solution


.. f′.f

.− .↘

..0.0. +.

↗

.

min


[0,∞)


. . . . . .

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


. . . . . .


Example


Solution

f′(x) = 23x

−1/3(x+ 2) + x2/3 = 13x

−1/3 (5x+ 4)


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 = 13x

−1/3 (5x+ 4)


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 = 13x

−1/3 (5x+ 4)


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 = 13x

−1/3 (5x+ 4)


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 = 13x

−1/3 (5x+ 4)


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 = 13x

−1/3 (5x+ 4)


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 = 13x

−1/3 (5x+ 4)


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 = 13x

−1/3 (5x+ 4)


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′ 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.


. . . . . .


Example


Solution


.. f′.f

.− .↘

..0.0. +.

↗

.

min


[0,∞)


. . . . . .


Example


Solution


.. f′.f

.− .↘

..0.0. +.

↗.

min


[0,∞)


. . . . . .


Example


Solution

f′(x) = 23x

−1/3(x+ 2) + x2/3 = 13x

−1/3 (5x+ 4)


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 = 13x

−1/3 (5x+ 4)


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 = 13x

−1/3 (5x+ 4)


5x+ 4

..

−4/5

.

0

.

−

.

+

.

f′(x)

.

f(x)

..

−4/5

.

0

..

0

.

×

..

+

.

↗

..

−

.

↘

..

+

.

↗

.

max

.

min


. . . . . .

Outline





. . . . . .

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”.


. . . . . .

Synonyms for concavity

Remark

I “concave up” = “concave upwards” = “convex”I “concave down” = “concave downwards” = “concave”


. . . . . .

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


. . . . . .

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).


. . . . . .

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).V63.0121.021, Calculus I (NYU) Section 4.2 The Shapes of Curves November 16, 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)


. . . . . .


Example


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



. . . . . .


Example


Solution


when x = −1/3



. . . . . .


Example


Solution


when x = −1/3



. . . . . .

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


. . . . . .


Example


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


. . . . . .


Example


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


. . . . . .


Example


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


. . . . . .


Example


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


. . . . . .


Example


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


. . . . . .


Example


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


. . . . . .


Example


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


. . . . . .


Example


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


. . . . . .


Example


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


. . . . . .

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.


. . . . . .

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.


. . . . . .

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.


. . . . . .






.. f′′ = (f′)′.f′

...c.+

..+ .. +.↗

.↗

.

f′

.

f

...

c

.

0

..

−

..

+

.

↘

.

↗

.

min




. . . . . .






.. f′′ = (f′)′.f′

...c.+..+

.. +.↗

.↗

.

f′

.

f

...

c

.

0

..

−

..

+

.

↘

.

↗

.

min




. . . . . .






.. f′′ = (f′)′.f′

...c.+..+ .. +

.↗

.↗

.

f′

.

f

...

c

.

0

..

−

..

+

.

↘

.

↗

.

min




. . . . . .






.. f′′ = (f′)′.f′

...c.+..+ .. +

.↗

.↗

.

f′

.

f

...

c

.

0

..

−

..

+

.

↘

.

↗

.

min




. . . . . .






.. f′′ = (f′)′.f′

...c.+..+ .. +.

↗

.↗

.

f′

.

f

...

c

.

0

..

−

..

+

.

↘

.

↗

.

min




. . . . . .






.. f′′ = (f′)′.f′

...c.+..+ .. +.

↗.

↗.

f′

.

f

...

c

.

0

..

−

..

+

.

↘

.

↗

.

min




. . . . . .






.. f′′ = (f′)′.f′

...c.+..+ .. +.

↗.

↗.

f′

.

f

...

c

.

0

..

−

..

+

.

↘

.

↗

.

min




. . . . . .






.. f′′ = (f′)′.f′

...c.+..+ .. +.

↗.

↗.

f′

.

f

...

c

.

0

..

−

..

+

.

↘

.

↗

.

min




. . . . . .






.. f′′ = (f′)′.f′

...c.+..+ .. +.

↗.

↗.

f′

.

f

...

c

.

0

..

−

..

+

.

↘

.

↗

.

min




. . . . . .






.. f′′ = (f′)′.f′

...c.+..+ .. +.

↗.

↗.

f′

.

f

...

c

.

0

..

−

..

+

.

↘

.

↗

.

min




. . . . . .






.. f′′ = (f′)′.f′

...c.+..+ .. +.

↗.

↗.

f′

.

f

...

c

.

0

..

−

..

+

.

↘

.

↗

.

min




. . . . . .






.. f′′ = (f′)′.f′

...c.+..+ .. +.

↗.

↗.

f′

.

f

...

c

.

0

..

−

..

+

.

↘

.

↗

.

min




. . . . . .






.. f′′ = (f′)′.f′

...c.+..+ .. +.

↗.

↗.

f′

.

f

...

c

.

0

..

−

..

+

.

↘

.

↗

.

min




. . . . . .

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.


. . . . . .


Example


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.


. . . . . .


Example


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

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.


. . . . . .


Example


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.


. . . . . .

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.


. . . . . .


Example


Solution



x = −4/5




. . . . . .


Example


Solution



x = −4/5




. . . . . .


Example


Solution



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.


. . . . . .


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

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.


. . . . . .


Remark



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

.max

g′′(x) = −12x2, g′′(0) = 0

h(x) = x3h′(x) = 3x2, h′(0) = 0

.infl.

h′′(x) = 6x, h′′(0) = 0


. . . . . .


Remark



f(x) = x4 g(x) = −x4 h(x) = x3



. . . . . .

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


Technology

Lesson20 -derivatives_and_the_shape_of_curves_021_slides