Lesson 20: Derivatives and the Shape of Curves (Section 041 slides)

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.

. . . . . .

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

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

f(b) = f(a) + f′(c)(b− a)

. . . . . .

f(b)− f(a)b− a

= f′(c). ...a

f(b) = f(a) + f′(c)(b− a)

. . . . . .

f(b)− f(a)b− a

= f′(c). ...a

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

Solution

Since f′(x) =1

. . . . . .

Example

Solution

Since f′(x) =1

. . . . . .

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. +

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. +

[0,∞)

. . . . . .

Example

Solution

.. f′.− ..0.0. +

[0,∞)

. . . . . .

Example

Solution

.. f′.f

.− .↘

..0.0. +.

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. +.

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. +.

[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.×.− . +.

−4/5

f′(x)

−4/5

. . . . . .

Example

Solution

f′(x) = 23x

−1/3(x+ 2) + x2/3 = 13x

−1/3 (5x+ 4)

−4/5

f′(x)

−4/5

. . . . . .

Example

Solution

f′(x) = 23x

−1/3(x+ 2) + x2/3 = 13x

−1/3 (5x+ 4)

−4/5

f′(x)

−4/5

. . . . . .

Example

Solution

f′(x) = 23x

−1/3(x+ 2) + x2/3 = 13x

−1/3 (5x+ 4)

−4/5

f′(x)

−4/5

. . . . . .

Example

Solution

f′(x) = 23x

−1/3(x+ 2) + x2/3 = 13x

−1/3 (5x+ 4)

−4/5

f′(x)

−4/5

. . . . . .

Example

Solution

f′(x) = 23x

−1/3(x+ 2) + x2/3 = 13x

−1/3 (5x+ 4)

−4/5

f′(x)

−4/5

. . . . . .

Example

Solution

f′(x) = 23x

−1/3(x+ 2) + x2/3 = 13x

−1/3 (5x+ 4)

−4/5

f′(x)

−4/5

. . . . . .

Example

Solution

f′(x) = 23x

−1/3(x+ 2) + x2/3 = 13x

−1/3 (5x+ 4)

−4/5

f′(x)

−4/5

. . . . . .

Example

Solution

f′(x) = 23x

−1/3(x+ 2) + x2/3 = 13x

−1/3 (5x+ 4)

−4/5

f′(x)

−4/5

. . . . . .

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. +.

[0,∞)

. . . . . .

Example

Solution

.. f′.f

.− .↘

..0.0. +.

[0,∞)

. . . . . .

Example

Solution

f′(x) = 23x

−1/3(x+ 2) + x2/3 = 13x

−1/3 (5x+ 4)

−4/5

f′(x)

−4/5

. . . . . .

Example

Solution

f′(x) = 23x

−1/3(x+ 2) + x2/3 = 13x

−1/3 (5x+ 4)

−4/5

f′(x)

−4/5

. . . . . .

Example

Solution

f′(x) = 23x

−1/3(x+ 2) + x2/3 = 13x

−1/3 (5x+ 4)

−4/5

f′(x)

−4/5

. . . . . .

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).V63.0121.041, 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)

. . . . . .

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

f′′(x)

−−

. . . . . .

Example

Solution

We have f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2).

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

. . . . . .

Example

Solution

We have f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2).

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

. . . . . .

Example

Solution

We have f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2).

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

. . . . . .

Example

Solution

We have f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2).

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

. . . . . .

Example

Solution

We have f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2).

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

. . . . . .

Example

Solution

We have f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2).

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

. . . . . .

Example

Solution

We have f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2).

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

. . . . . .

Example

Solution

We have f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2).

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

. . . . . .

Example

Solution

We have f′′(x) =109x−1/3 − 4

9x−4/3 =

29x−4/3(5x− 2).

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

. . . . . .

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.+

..+ .. +.↗

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′)′.f′

...c.+..+

.. +.↗

. . . . . .

.. f′′ = (f′)′.f′

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

. . . . . .

.. f′′ = (f′)′.f′

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

. . . . . .

.. f′′ = (f′)′.f′

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

. . . . . .

.. f′′ = (f′)′.f′

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

. . . . . .

.. f′′ = (f′)′.f′

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

. . . . . .

.. f′′ = (f′)′.f′

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

. . . . . .

.. f′′ = (f′)′.f′

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

. . . . . .

.. f′′ = (f′)′.f′

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

. . . . . .

.. f′′ = (f′)′.f′

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

. . . . . .

.. f′′ = (f′)′.f′

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

. . . . . .

.. f′′ = (f′)′.f′

...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+ 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):

(−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

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

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)