Lesson 20: Derivatives and the Shapes of Curves (slides)

Sec on 4.3Deriva ves and the Shapes of Curves

V63.0121.001: Calculus IProfessor Ma hew Leingang

New York University

April 11, 2011

Announcements

I Quiz 4 on Sec ons 3.3,3.4, 3.5, and 3.7 thisweek (April 14/15)

I Quiz 5 on Sec ons4.1–4.4 April 28/29

I Final Exam Thursday May12, 2:00–3:50pm

ObjectivesI Use the deriva ve of a func onto determine the intervals alongwhich the func on is increasingor decreasing (TheIncreasing/Decreasing Test)

I Use the First Deriva ve Test toclassify cri cal points of afunc on as local maxima, localminima, or neither.

ObjectivesI Use the second deriva ve of afunc on to determine theintervals along which the graphof the func on is concave up orconcave down (The ConcavityTest)

I Use the first and secondderiva ve of a func on toclassify cri cal points as localmaxima or local minima, whenapplicable (The SecondDeriva ve Test)

OutlineRecall: The Mean Value Theorem

MonotonicityThe Increasing/Decreasing TestFinding intervals of monotonicityThe First Deriva ve Test

ConcavityDefini onsTes ng for ConcavityThe Second Deriva ve Test

Recall: The Mean Value TheoremTheorem (The Mean Value Theorem)

Let f be con nuous on [a, b]and differen able on (a, b).Then there exists a point c in(a, b) such that

f(b)− f(a)b− a

= f′(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)....

f(b) = f(a) + f′(c)(b− a)

f(b)− f(a)b− a

= f′(c)....

f(b) = f(a) + f′(c)(b− a)

f(b)− f(a)b− a

= f′(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 con nuous on[x, y] and differen able 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.

Increasing FunctionsDefini onA func on 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 func on “preserves order.”I I could be bounded or infinite, open, closed, orhalf-open/half-closed.

I Write your own defini on (muta s mutandis) of decreasing,nonincreasing, nondecreasing

Increasing FunctionsDefini onA func on 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 func on “preserves order.”I I could be bounded or infinite, open, closed, orhalf-open/half-closed.

I Write your own defini on (muta s mutandis) of decreasing,nonincreasing, nondecreasing

The Increasing/Decreasing TestTheorem (The Increasing/Decreasing Test)

If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0on an interval, then f is decreasing on that interval.

Proof.It works the same as the last theorem. Assume f′(x) > 0 on aninterval I. Pick two points x and y in I with x < y. We must showf(x) < f(y). By MVT there exists a point c in (x, y) such that

f(y)− f(x) = f′(c)(y− x) > 0.

So f(y) > f(x).

Proof.It works the same as the last theorem. Assume f′(x) > 0 on aninterval I. Pick two points x and y in I with x < y. We must showf(x) < f(y).

By MVT there exists a point c in (x, y) such that

f(y)− f(x) = f′(c)(y− x) > 0.

So f(y) > f(x).

f(y)− f(x) = f′(c)(y− x) > 0.

So f(y) > f(x).

f(y)− f(x) = f′(c)(y− x) > 0.

So f(y) > f(x).

Finding intervals of monotonicity IExample

Find the intervals of monotonicity of f(x) = 2x− 5.

Solu onf′(x) = 2 is always posi ve, so f is increasing on (−∞,∞).

Example

Describe the monotonicity of f(x) = arctan(x).

Solu on

Since f′(x) =1

1+ x2is always posi ve, f(x) is always increasing.

Example

Solu on

Since f′(x) =1

Example

Solu on

Since f′(x) =1

Example

Solu on

Since f′(x) =1

Finding intervals of monotonicity IIExample

Find the intervals of monotonicity of f(x) = x2 − 1.

Solu on

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,∞)

Solu on

I f′(x) = 2x, which is posi ve when x > 0 and nega ve when x is.

Solu on

I f′(x) = 2x, which is posi ve when x > 0 and nega ve 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,∞)

Solu on

I f′(x) = 2x, which is posi ve when x > 0 and nega ve when x is.I We can draw a number line:

.. f′.f

.− .↘

..0.0. +.

Solu on

.. 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,∞)

Solu on

.. f′.f

.− .↘

..0.0. +.

Finding intervals of monotonicity IIIExample

Find the intervals of monotonicity of f(x) = x2/3(x+ 2).

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

The cri cal points are 0 andand−4/5.

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.−

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

f′(x)

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

f′(x)

−4/5

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

f′(x)

−4/5

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

f′(x)

−4/5

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

f′(x)

−4/5

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

f′(x)

−4/5

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

f′(x)

−4/5

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

f′(x)

−4/5

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

f′(x)

−4/5

The First Derivative Test

Theorem (The First Deriva ve Test)

Let f be con nuous on [a, b] and c a cri cal point of f in (a, b).I If f′ changes from posi ve to nega ve at c, then c is a local

maximum.I If f′ changes from nega ve to posi ve 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.

Solu on

.. f′.f

.− .↘

..0.0. +.

Solu on

.. f′.f

.− .↘

..0.0. +.

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

f′(x)

−4/5

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

f′(x)

−4/5

Solu on

f′(x) = 23x

−1/3(x+ 2) + x2/3

−1/3 (5x+ 4)

.. x−1/3..0.×.− . +.

−4/5

f′(x)

−4/5

ConcavityDefini onThe graph of f is called concave upwards on an interval if it liesabove all its tangents on that interval. The graph of f is calledconcave downwards on an interval if it lies below all its tangents onthat interval.

concave up

concave down

ConcavityDefini onThe graph of f is called concave upwards on an interval if it liesabove all its tangents on that interval. The graph of f is calledconcave downwards on an interval if it lies below all its tangents onthat interval.

concave up

concave down

ConcavityDefini on

concave up

concave downWe some mes 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 mean change in concavityDefini onA point P on a curve y = f(x) is called an inflec on point if f iscon nuous at P and the curve changes from concave upward toconcave downward at P (or vice versa).

..concavedown

concaveup

..inflec on point

Testing for Concavity

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.

Testing for ConcavityProof.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).

Testing for ConcavityProof.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)

f(x) = f(a) + f′(c)(x− a)

L(x) = f(a) + f′(a)(x− a)

f(x) = f(a) + f′(c)(x− a)

L(x) = f(a) + f′(a)(x− a)

f(x) = f(a) + f′(c)(x− a)

Finding Intervals of Concavity IExample

Find the intervals of concavity for the graph of f(x) = x3 + x2.

Solu on

I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x+ 2.I This is nega ve when x < −1/3, posi ve when x > −1/3, and 0

when x = −1/3

I So f is concave down on the open interval (−∞,−1/3), concaveup on the open interval (−1/3,∞), and has an inflec on pointat the point (−1/3, 2/27)

Solu on

I We have f′(x) = 3x2 + 2x, so f′′(x) = 6x+ 2.

I This is nega ve when x < −1/3, posi ve when x > −1/3, and 0when x = −1/3

Solu on

when x = −1/3

Solu on

when x = −1/3

Finding Intervals of Concavity IIExample

Find the intervals of concavity of the graph of f(x) = x2/3(x+ 2).

Solu on

We have

f′′(x) =109x−1/3 − 4

9x−4/3

=29x−4/3(5x− 2)

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

Solu on

We have

f′′(x) =109x−1/3 − 4

9x−4/3

=29x−4/3(5x− 2)

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

Solu on

We have

f′′(x) =109x−1/3 − 4

9x−4/3

=29x−4/3(5x− 2)

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

Solu on

We have

f′′(x) =109x−1/3 − 4

9x−4/3

=29x−4/3(5x− 2)

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

Solu on

We have

f′′(x) =109x−1/3 − 4

9x−4/3

=29x−4/3(5x− 2)

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

Solu on

We have

f′′(x) =109x−1/3 − 4

9x−4/3

=29x−4/3(5x− 2)

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

Solu on

We have

f′′(x) =109x−1/3 − 4

9x−4/3

=29x−4/3(5x− 2)

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

Solu on

We have

f′′(x) =109x−1/3 − 4

9x−4/3

=29x−4/3(5x− 2)

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

Solu on

We have

f′′(x) =109x−1/3 − 4

9x−4/3

=29x−4/3(5x− 2)

.. x−4/3..0.×.+ . +.

5x− 2

f′′(x)

−−

Solu on

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 TestTheorem (The Second Deriva ve Test)

Let f, f′, and f′′ be con nuous 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 of f.I If f′′(c) > 0, then c is a local minimum of f.

Remarks

I If f′′(c) = 0, the second deriva ve test is inconclusiveI We look for zeroes of f′ and plug them into f′′ to determine iftheir f values are local extreme values.

The Second Derivative TestTheorem (The Second Deriva ve Test)

Let f, f′, and f′′ be con nuous 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 of f.I If f′′(c) > 0, then c is a local minimum of f.

Remarks

I If f′′(c) = 0, the second deriva ve test is inconclusiveI We look for zeroes of f′ and plug them into f′′ to determine iftheir f values are local extreme values.

Proof of the Second Derivative TestProof.Suppose f′(c) = 0 and f′′(c) > 0.

.. f′′ = (f′)′.f′

...c.+

..+ .. +.↗

I Since f′′ is con nuous,f′′(x) > 0 for all xsufficiently close to c. .. f′′ = (f′)′.

f′...c.+

..+ .. +.↗

f′...c.+..+

.. +.↗

f′...c.+..+ .. +

I Since f′′ is con nuous,f′′(x) > 0 for all xsufficiently close to c.

I Since f′′ = (f′)′, we knowf′ is increasing near c.

.. f′′ = (f′)′.f′

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

.. f′′ = (f′)′.f′

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

.. f′′ = (f′)′.f′

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

I Since f′(c) = 0 and f′ isincreasing, f′(x) < 0 for xclose to c and less than c,and f′(x) > 0 for x closeto c and more than c.

.. f′′ = (f′)′.f′

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

.. f′′ = (f′)′.f′

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

.. f′′ = (f′)′.f′

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

I This means f′ changessign from nega ve toposi ve at c, whichmeans (by the FirstDeriva ve Test) that f hasa local minimum at c.

.. f′′ = (f′)′.f′

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

.. f′′ = (f′)′.f′

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

.. f′′ = (f′)′.f′

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

.. f′′ = (f′)′.f′

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

Using the Second Derivative Test IExample

Find the local extrema of f(x) = x3 + x2.

Solu on

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.

Solu on

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.

Solu on

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.

Solu on

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.

Solu on

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 IIExample

Find the local extrema of f(x) = x2/3(x+ 2)

Solu on

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 nega ve when

x = −4/5

I So x = −4/5 is a local maximum.I No ce the Second Deriva ve Test doesn’t catch the local

minimum x = 0 since f is not differen able there.

Solu on

x = −4/5

Solu on

x = −4/5

Solu on

x = −4/5

I So x = −4/5 is a local maximum.

I No ce the Second Deriva ve Test doesn’t catch the localminimum x = 0 since f is not differen able there.

Solu on

x = −4/5

Using the Second Derivative Test IIGraph

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 zeroRemark

I At inflec on points c, if f′ is differen able at c, then f′′(c) = 0I If f′′(c) = 0, must f have an inflec on point at c?

Consider these examples:

f(x) = x4 g(x) = −x4 h(x) = x3

All of them have cri cal points at zero with a second deriva ve ofzero. But the first has a local min at 0, the second has a local max at0, and the third has an inflec on point at 0. This is why we say 2DThas nothing to say when f′′(c) = 0.

f(x) = x4 g(x) = −x4 h(x) = x3

When first and second derivative are zero

func on deriva ves graph type

f(x) = x4f′(x) =

, f′(0) =

0. min

f′′(x) =

, f′′(0) =

g(x) = −x4g′(x) =

− 4x3

, g′(0) =

0 .max

g′′(x) =

− 12x2

, g′′(0) =

h(x) = x3h′(x) =

, h′(0) =

0. infl.

h′′(x) =

, h′′(0) =

f(x) = x4f′(x) = 4x3, f′(0) =

0. min

f′′(x) =

, f′′(0) =

g(x) = −x4g′(x) =

− 4x3

, g′(0) =

0 .max

g′′(x) =

− 12x2

, g′′(0) =

h(x) = x3h′(x) =

, h′(0) =

0. infl.

h′′(x) =

, h′′(0) =

f(x) = x4f′(x) = 4x3, f′(0) = 0

f′′(x) =

, f′′(0) =

g(x) = −x4g′(x) =

− 4x3

, g′(0) =

0 .max

g′′(x) =

− 12x2

, g′′(0) =

h(x) = x3h′(x) =

, h′(0) =

0. infl.

h′′(x) =

, h′′(0) =

f(x) = x4f′(x) = 4x3, f′(0) = 0

f′′(x) = 12x2, f′′(0) =

g(x) = −x4g′(x) =

− 4x3

, g′(0) =

0 .max

g′′(x) =

− 12x2

, g′′(0) =

h(x) = x3h′(x) =

, h′(0) =

0. infl.

h′′(x) =

, h′′(0) =

f(x) = x4f′(x) = 4x3, f′(0) = 0

f′′(x) = 12x2, f′′(0) = 0

g(x) = −x4g′(x) =

− 4x3

, g′(0) =

0 .max

g′′(x) =

− 12x2

, g′′(0) =

h(x) = x3h′(x) =

, h′(0) =

0. infl.

h′′(x) =

, h′′(0) =

f(x) = x4f′(x) = 4x3, f′(0) = 0

f′′(x) = 12x2, f′′(0) = 0

g(x) = −x4g′(x) =

− 4x3

, g′(0) =

0 .max

g′′(x) =

− 12x2

, g′′(0) =

h(x) = x3h′(x) =

, h′(0) =

0. infl.

h′′(x) =

, h′′(0) =

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) =

h(x) = x3h′(x) =

, h′(0) =

0. infl.

h′′(x) =

, h′′(0) =

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) =

h(x) = x3h′(x) =

, h′(0) =

0. infl.

h′′(x) =

, h′′(0) =

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) =

h(x) = x3h′(x) =

, h′(0) =

0. infl.

h′′(x) =

, h′′(0) =

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) =

h(x) = x3h′(x) =

, h′(0) =

0. infl.

h′′(x) =

, h′′(0) =

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) =

, h′(0) =

0. infl.

h′′(x) =

, h′′(0) =

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) =

, h′(0) =

0. infl.

h′′(x) =

, h′′(0) =

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) =

, h′(0) =

0. infl.

h′′(x) =

, h′′(0) =

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) =

, h′′(0) =

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) =

, h′′(0) =

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) =

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) = 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

h′′(x) = 6x, h′′(0) = 0

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

Summary

I Concepts: Mean Value Theorem, monotonicity, concavityI Facts: deriva ves can detect monotonicity and concavityI Techniques for drawing curves: the Increasing/Decreasing Testand the Concavity Test

I Techniques for finding extrema: the First Deriva ve Test andthe Second Deriva ve Test

Lesson 20: Derivatives and the Shapes of Curves (slides)

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Transcript of Lesson 20: Derivatives and the Shapes of Curves (slides)

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