SLecture6 Maxima Minima continued

MA 105 Calculus: Lecture 6Maxima/Minima (Contd.)

How Derivatives Affect the Shape of a Graph

S. Sivaji Ganesh

Mathematics DepartmentIIT Bombay

August 4, 2009

S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 1 / 30

What does f ′ say about f ?



Definition



Definition1 A function f is called increasing



Definition1 A function f is called increasing on an interval I



Definition1 A function f is called increasing on an interval I if




f (x1) < f (x2)




f (x1) < f (x2) whenever x1, x2 are in I and




f (x1) < f (x2) whenever x1, x2 are in I and x1 < x2.





2 A function f is called decreasing





2 A function f is called decreasing on an interval I





2 A function f is called decreasing on an interval I if






f (x1) > f (x2)






f (x1) > f (x2) whenever x1, x2 are in I and






f (x1) > f (x2) whenever x1, x2 are in I and x1 < x2.







Say True/False







Say True/False1 If f is increasing on an interval, then







Say True/False1 If f is increasing on an interval, then f ′(x) > 0 on that interval.








False.








False. Can we correct this statement and make it correct?








False. Can we correct this statement and make it correct?2 If f is decreasing on an interval, then








False. Can we correct this statement and make it correct?2 If f is decreasing on an interval, then f ′(x) < 0 on that interval.









Must be False.S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30








Must be False. why?S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30








Must be False. why? because the previous one is false!S. Sivaji Ganesh (IIT Bombay) MA 105 Calculus: Lecture 6 August 4, 2009 2 / 30


Increasing/Decreasing test




If f ′(x) > 0 on an interval, then




If f ′(x) > 0 on an interval, then f is increasing on that interval.





If f ′(x) < 0 on an interval, then





If f ′(x) < 0 on an interval, then f is decreasing on that interval.






Proof Let x1, x2 be any two points in the interval






Proof Let x1, x2 be any two points in the interval with x1 < x2.






Proof Let x1, x2 be any two points in the interval with x1 < x2. By MVT,






Proof Let x1, x2 be any two points in the interval with x1 < x2. By MVT,there is a point c in (x1, x2)






Proof Let x1, x2 be any two points in the interval with x1 < x2. By MVT,there is a point c in (x1, x2) such that







f (x2) − f (x1) = f ′(c)(x2 − x1).







f (x2) − f (x1) = f ′(c)(x2 − x1).

The above two statements now follow.







f (x2) − f (x1) = f ′(c)(x2 − x1).

The above two statements now follow.Exercise Find where the function f (x) = 3x4 − 4x3 − 12x2 + 5 isincreasing and







f (x2) − f (x1) = f ′(c)(x2 − x1).

The above two statements now follow.Exercise Find where the function f (x) = 3x4 − 4x3 − 12x2 + 5 isincreasing and where it is decreasing.


Increasing/Decreasing



Graph of f (x) = 3x4 − 4x3 − 12x2 + 5



Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note that



Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0)



Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2),



Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2), thus f has a localmaximum at 0.



Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2), thus f has a localmaximum at 0. Note that f ′ change sign from +ve to -ve at 0.



Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2), thus f has a localmaximum at 0. Note that f ′ change sign from +ve to -ve at 0.f decreases on (0,2)



Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2), thus f has a localmaximum at 0. Note that f ′ change sign from +ve to -ve at 0.f decreases on (0,2) and increases on (2,3),



Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2), thus f has a localmaximum at 0. Note that f ′ change sign from +ve to -ve at 0.f decreases on (0,2) and increases on (2,3), thus f has a localminimum at 2.



Graph of f (x) = 3x4 − 4x3 − 12x2 + 5

Note thatf increases on (−1,0) and decreases on (0,2), thus f has a localmaximum at 0. Note that f ′ change sign from +ve to -ve at 0.f decreases on (0,2) and increases on (2,3), thus f has a localminimum at 2. Note that f ′ change sign from -ve to +ve at 2.


How to catch local Minima/Maxima points using First Derivat e?



We know, by Fermat’s theorem,



We know, by Fermat’s theorem, that



We know, by Fermat’s theorem, that if f has a local extremum at c,



We know, by Fermat’s theorem, that if f has a local extremum at c,then c must be a critical point.



We know, by Fermat’s theorem, that if f has a local extremum at c,then c must be a critical point. We also know that



We know, by Fermat’s theorem, that if f has a local extremum at c,then c must be a critical point. We also know that we may not have alocal extremum at every critical point.




Question: At what critical points we have a local extremum?




Question: At what critical points we have a local extremum? At whatcritical points we do not have a local extremum?




Question: At what critical points we have a local extremum? At whatcritical points we do not have a local extremum?

The next result answers these questions.



The First Derivative Test



The First Derivative TestLet f be a continuous function and



The First Derivative TestLet f be a continuous function and c be a critical point of f .




1 If f ′ changes sign from positive to negative at c,




1 If f ′ changes sign from positive to negative at c, then f has a localmaximum at c.





2 If f ′ changes sign from negative to positive at c,





2 If f ′ changes sign from negative to positive at c, then f has a localminimum at c.






3 If f ′ does not change sign at c,






3 If f ′ does not change sign at c, then f has no local extremum at c.







Proof follows from Increasing/Decreasing test.







Proof follows from Increasing/Decreasing test.Remark/Questions:








1 Can we drop the hypothesis








1 Can we drop the hypothesis of continuity on f ?








1 Can we drop the hypothesis of continuity on f ? No.








1 Can we drop the hypothesis of continuity on f ? No.2 We require f to be differentiable








1 Can we drop the hypothesis of continuity on f ? No.2 We require f to be differentiable in an open interval containing c,








1 Can we drop the hypothesis of continuity on f ? No.2 We require f to be differentiable in an open interval containing c,

and we do not insist on differentiability of f at c.



Graph of f (x) = 3x4 − 4x3 − 12x2 + 5,



Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are



Graph of f (x) = 3x4 − 4x3 − 12x2 + 5, critical points are −1,0,2.




Note that




Note thatf ′ changes sign from +ve to -ve at 0.




Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0)




Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0) and decreases on (0,2),




Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0) and decreases on (0,2), thus f has a local maximum at 0.




Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0) and decreases on (0,2), thus f has a local maximum at 0.f ′ changes sign from -ve to +ve at 2.




Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0) and decreases on (0,2), thus f has a local maximum at 0.f ′ changes sign from -ve to +ve at 2. Note that f decreases on(0,2)




Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0) and decreases on (0,2), thus f has a local maximum at 0.f ′ changes sign from -ve to +ve at 2. Note that f decreases on(0,2) and increases on (2,3),




Note thatf ′ changes sign from +ve to -ve at 0. Note that f increases on(−1,0) and decreases on (0,2), thus f has a local maximum at 0.f ′ changes sign from -ve to +ve at 2. Note that f decreases on(0,2) and increases on (2,3), thus f has a local minimum at 2.


The First Derivative Test Illustrated



2 is a critical point,



2 is a critical point, f ′ changes sign from



2 is a critical point, f ′ changes sign from +ve to -ve at 2.



1 is a critical point, f ′ changes sign from



1 is a critical point, f ′ changes sign from -ve to +ve at 1.



2 is a critical point, f ′ keeps



2 is a critical point, f ′ keeps +ve sign around x = 2.



2 is a critical point, f ′ keeps



2 is a critical point, f ′ keeps -ve sign around x = 2.



Observe



Observe how



Observe how f ′ changes its sign.


What does f ′′ say about f ?



DefinitionLet f be differentiable on an interval I,



DefinitionLet f be differentiable on an interval I, and



DefinitionLet f be differentiable on an interval I, and c ∈ I.




1 Denote by lc(x) := f (c) + f ′(c)(x − c),




1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line




1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))




1 Denote by lc(x) := f (c) + f ′(c)(x − c), the tangent line at (c, f (c))to the graph of f .





2 If the graph of f





2 If the graph of f lies above all of its tangents





2 If the graph of f lies above all of its tangents on an interval I,





2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I.





2 If the graph of f lies above all of its tangents on an interval I, thenit is called concave upward (CU) on I. That is,






f (x) > lc(x)






f (x) > lc(x) for all x ∈ I \ {c}






f (x) > lc(x) for all x ∈ I \ {c} and for all c ∈ I.







3 If the graph of f







3 If the graph of f lies below all of its tangents







3 If the graph of f lies below all of its tangents on an interval I,







3 If the graph of f lies below all of its tangents on an interval I, then itis called concave downward (CD) on I.







3 If the graph of f lies below all of its tangents on an interval I, then itis called concave downward (CD) on I. That is,








f (x) < lc(x)








f (x) < lc(x) for all x ∈ I \ {c}








f (x) < lc(x) for all x ∈ I \ {c} and for all c ∈ I.


On the definition of CU and CD



Note that



Note that for each fixed c,



Note that for each fixed c, f (c) = lc(c)



Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x)



Note that for each fixed c, f (c) = lc(c) and hence we hadf (x) > (resp. <)lc(x) for all x ∈ I \ {c} for CU (resp. CD) definition.




1 Draw the graph of f on the interval I.




1 Draw the graph of f on the interval I.2 Fix any c ∈ I.




1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.




1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.




1 Draw the graph of f on the interval I.2 Fix any c ∈ I. Draw tangent line (x , lc(x)) for x ∈ I \ {c}.3 Do it for each c ∈ I.4 Now have a look at this picture.





Then,





Then,

f is (CU) on I if





Then,

f is (CU) on I if graph of f lies above all those lines drawn in (2)and (3) above.





Then,


f is (CD) on I if





Then,


f is (CD) on I if graph of f lies below all those lines drawn in (2)and (3) above.





Then,



Caution: Some books call





Then,



Caution: Some books call concave upward as convex ,





Then,



Caution: Some books call concave upward as convex , and





Then,



Caution: Some books call concave upward as convex , and concavedownward as concave .





Then,



Caution: Some books call concave upward as convex , and concavedownward as concave . We defined the concepts of CU and CD





Then,



Caution: Some books call concave upward as convex , and concavedownward as concave . We defined the concepts of CU and CD onlyon an interval.


CU, CD Illustrated



Concavity Test



Concavity Test1 If f ′′(x) > 0 for all x in an interval I,



Concavity Test1 If f ′′(x) > 0 for all x in an interval I, then the graph of f is concave

upward on I.




upward on I.2 If f ′′(x) < 0 for all x in an interval I,




upward on I.2 If f ′′(x) < 0 for all x in an interval I, then the graph of f is concave

downward on I.





downward on I.

Proof is a consequence of MVT.





downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.





downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I





downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means





downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is





downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.





downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.2 How do we prove f is CU on I?





downward on I.

Proof is a consequence of MVT. Let us prove (1), for instance.1 f ′′(x) > 0 for all x ∈ I means f ′ is increasing.2 How do we prove f is CU on I? we prove graph of f lies above

tangent line at any point.





downward on I.


tangent line at any point.3 Let a ∈ I.





downward on I.


tangent line at any point.3 Let a ∈ I. let x > a.





downward on I.


tangent line at any point.3 Let a ∈ I. let x > a. Then f (x) − f (a) = f ′(c)(x − a) for some

c ∈ [a, x ]





downward on I.



c ∈ [a, x ] by MVT.





downward on I.



c ∈ [a, x ] by MVT.4 f (x) − f (a) = f ′(c)(x − a) > f ′(a)(x − a).





downward on I.



c ∈ [a, x ] by MVT.4 f (x) − f (a) = f ′(c)(x − a) > f ′(a)(x − a). QED+5 repeat arguments in (3) and (4),





downward on I.



c ∈ [a, x ] by MVT.4 f (x) − f (a) = f ′(c)(x − a) > f ′(a)(x − a). QED+5 repeat arguments in (3) and (4), for x < a.





downward on I.



c ∈ [a, x ] by MVT.4 f (x) − f (a) = f ′(c)(x − a) > f ′(a)(x − a). QED+5 repeat arguments in (3) and (4), for x < a. QED



DefinitionA point P on a curve y = f (x) is called an inflection point if



DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there



DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there and



DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there and the curve changes from concave upward toconcave downward



DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there and the curve changes from concave upward toconcave downward or



DefinitionA point P on a curve y = f (x) is called an inflection point if f iscontinuous there and the curve changes from concave upward toconcave downward or concave downward to concave upward at P.




Interpretation: If f (t) is the distance travelled by the car, secondderivative is




Interpretation: If f (t) is the distance travelled by the car, secondderivative is its acceleration/deceleration.




Interpretation: If f (t) is the distance travelled by the car, secondderivative is its acceleration/deceleration. An inflection point is the timeat which car changes from acceleration to deceleration or vice versa.





Question: If f ′′(c) = 0, then is it true that (c, f (c)) is an inflection point?





Question: If f ′′(c) = 0, then is it true that (c, f (c)) is an inflection point?No.



Graph of f (x) = x4.



Graph of f (x) = x4. Is x = 0 an inflection point?



Sketch a possible graph of a function f that satisfies the followingconditions:



Sketch a possible graph of a function f that satisfies the followingconditions:

1 f ′(x) > 0 on (−∞,1), f ′(x) < 0 on (1,∞).2 f ′′(x) > 0 on (−∞,−2) and (2,∞), f ′′(x) < 0 on (−2,2)

3 limx→−∞

f (x) = −2, limx→∞

f (x) = 0.


How to catch local Minima/Maxima points using Second Deriva te?



Second Derivative Test



Second Derivative Test1 If f ′(c) = 0 and f ′′(c) > 0,



Second Derivative Test1 If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.



Second Derivative Test1 If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.2 If f ′(c) = 0 and f ′′(c) < 0,



Second Derivative Test1 If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.2 If f ′(c) = 0 and f ′′(c) < 0, then f has a local maximum at c.




Remark: The Second derivative test is silent if




Remark: The Second derivative test is silent if f ′′(c) = 0




Remark: The Second derivative test is silent if f ′′(c) = 0Proof is a consequence of Concavity test .


Why Second Derivate Test could be true?



Let’s prove:



Let’s prove: If f ′(c) = 0 and f ′′(c) > 0,



Let’s prove: If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.



Let’s prove: If f ′(c) = 0 and f ′′(c) > 0, then f has a local minimum at c.Proof: Recall

TheoremSuppose f is differentiable at a.




TheoremSuppose f is differentiable at a. Then there exists a function φ suchthat





f (x) = f (a) + (x − a)f ′(a) + (x − a)φ(x),





f (x) = f (a) + (x − a)f ′(a) + (x − a)φ(x),

andlimx→a

φ(x) = 0.





f (x) = f (a) + (x − a)f ′(a) + (x − a)φ(x),

andlimx→a

φ(x) = 0.

A similar result is true:


Why Second Derivate Test could be true? (contd.)

TheoremSuppose f is twice differentiable at a.



TheoremSuppose f is twice differentiable at a. Then there exists a function ψsuch that




f (x) = f (a) + (x − a)f ′(a) +(x − a)2

2f ′′(a) +

(x − a)2

2ψ(x),




f (x) = f (a) + (x − a)f ′(a) +(x − a)2

2f ′′(a) +

(x − a)2

2ψ(x),

andlimx→a

ψ(x) = 0.



Now, at the critical point c,




f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)




f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.




f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.

The required conclusion follows by looking at the three terms on RHSof the first equation:

1 first term is 0




f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.


1 first term is 02 The third term is negligible,




f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.


1 first term is 02 The third term is negligible, compared to the second term,




f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.


1 first term is 02 The third term is negligible, compared to the second term, when

we are near c.




f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.



we are near c.3 Thus the sign of RHS is determined by that of




f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.



we are near c.3 Thus the sign of RHS is determined by that of f ′′(c).




f (x) − f (c) = (x − c)f ′(c) +(x − c)2

2f ′′(c) +

(x − c)2

2ψ(x)

andlimx→c

ψ(x) = 0.



we are near c.3 Thus the sign of RHS is determined by that of f ′′(c). QED


The Second Derivative Test Illustrated



2 is a critical point, f ′′(2) < 0



2 is a critical point, f ′′(2) < 0 and f has a local maximum at 2



1 is a critical point, f ′′(1) > 0



1 is a critical point, f ′′(1) > 0 and f has a local maximum at 1



2 is a critical point, f ′′(2) = 0



2 is a critical point, f ′′(2) = 0 and f does not have a local extremum at2.


Problems


Problems

1 Sketch a possible graph of a function f that satisfies the followingconditions:


Problems

1 Sketch a possible graph of a function f that satisfies the followingconditions:

1 f ′(0) = f ′(2) = f ′(4) = 0,2 f ′(x) > 0 on (−∞, 0) and (2, 4),3 f ′(x) < 0 on (0, 2) and (4,∞),4 f ′′(x) > 0 on (1, 3),5 f ′′(x) < 0 on (−∞, 1) and (3,∞).

2 Show that the function g(x) = x |x | has an inflection point at (0,0)but g′′(0) does not exist.

3 Prove that if (c, f (c)) is an inflection point of the graph of f andf ′′(c) exists in an open interval that contains c, then f ′′(c) = 0.

4 Show that tan x > x for 0 < x < π

2 .


SLecture6 Maxima Minima continued

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