Calculus misconception cleared

by Rishi Tue Apr 30, 2013 10:50 pm

So far I thought (x0,f(x0)) is an inflection point if the second derivative f"(x0) = 0.

It seems that it is a misconception.

1. (xo, y0) being an inflection point does not imply that f"(x0) = 0.

2. Conversely f"(x0)=0 does not imply that x0,f(x0)) is an inflection point.

This I got it from another MOOC calculus course.

Looks like I have to watch all the Calculus videos from different teachers to get the concepts completely correct in my head.

by Marathadi-Saamiyaar Tue Apr 30, 2013 10:52 pm

Rishi wrote:So far I thought (x0,f(x0)) is an inflection point if the second derivative f"(x0) = 0.

It seems that it is a misconception.

1. (xo, y0) being an inflection point does not imply that f"(x0) = 0.

2. Conversely f"(x0)=0 does not imply that x0,f(x0)) is an inflection point.

This I got it from another MOOC calculus course.

Looks like I have to watch all the Calculus videos from different teachers to get the concepts completely correct in my head.

IyerE:

OrEy late GnanOdhayam.

by MaxEntropy_Man Tue Apr 30, 2013 10:56 pm

yes that's a classic example of a necessary but not sufficient condition. in addition to that you need the first higher order derivative (above the second derivative) that is non-zero to be an odd-order derivative.

by Rishi Tue Apr 30, 2013 11:00 pm

MaxEntropy_Man wrote:yes that's a classic example of a necessary but not sufficient condition. in addition to that you need the first higher order derivative (above the second derivative) that is non-zero to be an odd-order derivative.

Why do you say necessary?

I saw an example of an inflection point which was not differentiable at all.

by MaxEntropy_Man Tue Apr 30, 2013 11:03 pm

Rishi wrote:
MaxEntropy_Man wrote:yes that's a classic example of a necessary but not sufficient condition. in addition to that you need the first higher order derivative (above the second derivative) that is non-zero to be an odd-order derivative.

Why do you say necessary?

I saw an example of an inflection point which was not differentiable at all.

for continuous differentiable functions what i said is true. i suppose one could think up some examples where the second derivative doesn't exist.
let me put it this way, if at some point, the second derivative is zero and the next higher non-zero derivative is of odd order, then that point is an inflection point.

by Rishi Tue Apr 30, 2013 11:15 pm

Just tell me if f(x)=x^4 has an inflection point?

by MaxEntropy_Man Tue Apr 30, 2013 11:26 pm

Rishi wrote:Just tell me if f(x)=x^4 has an inflection point?

the only point at which the second derivative is zero is x=0, but all higher order derivatives of odd order are also zero at x=0. so i think no inflection points.

by Rishi Tue Apr 30, 2013 11:39 pm

MaxEntropy_Man wrote:
Rishi wrote:Just tell me if f(x)=x^4 has an inflection point?

the only point at which the second derivative is zero is x=0, but all higher order derivatives of odd order are also zero at x=0. so i think no inflection points.

The professor explained the following way.

f"(x) = 12x^2

f"(x) = 0 for x0=0

But f"(x)=12x^2 > 0 means concave up on both sides of x0

Which means that is not an inflection point.

by MaxEntropy_Man Tue Apr 30, 2013 11:49 pm

Rishi wrote:
MaxEntropy_Man wrote:
Rishi wrote:Just tell me if f(x)=x^4 has an inflection point?

the only point at which the second derivative is zero is x=0, but all higher order derivatives of odd order are also zero at x=0. so i think no inflection points.

The professor explained the following way.

f"(x) = 12x^2

f"(x) = 0 for x0=0

But f"(x)=12x^2 > 0 means concave up on both sides of x0

Which means that is not an inflection point.

for a point (x0, f(x0)) to be an inflection point, in addition to f''(x0) being equal to zero, the second derivative immediately to the right and to the left of x0 must have opposite signs, i.e. f″(x0+δ) and f″(x0-δ), (where δ→0 is an infinitesimally small but positive quantity), must have opposite signs. i.e. the third derivative [f″(x0+δ)- f″(x0-δ)]/(2δ) should be non-zero. that is not the case for x0=0. hence no inflection point.

by Jeremiah Mburuburu Wed May 01, 2013 1:43 am

Rishi wrote:So far I thought (x0,f(x0)) is an inflection point if the second derivative f"(x0) = 0.

It seems that it is a misconception.

1. (xo, y0) being an inflection point does not imply that f"(x0) = 0.

2. Conversely f"(x0)=0 does not imply that x0,f(x0)) is an inflection point.

This I got it from another MOOC calculus course.

Looks like I have to watch all the Calculus videos from different teachers to get the concepts completely correct in my head.

here's a true statement:

A. if (xo, yo) is an inflection point of f, then f ''(xo) = 0.

the converse of A is not true.

f ''(xo) being = 0 is only a necessary condition for (xo, yo) being an inflection point of f, but is a useful condition for finding inflection points.

a statement that helps one to truly understand the necessary condition is the contrapositive of A, which is equivalent to A:

B. if f ''(xo) is not = 0, then (xo, yo) is not an inflection point of f.

B tells one that one need examine only those points in f for which f ''(x) = 0; some of them may be inflection points, and others won't be, but one won't find any outside that set. thus, one can restrict one's attention to those points for which f ''(x) = 0.

one will then need to scrutinize each point for which f ''(x) = 0 for being an inflection point.

here's the test: if f ''(xo) = 0, and f '' changes sign at xo, i.e. from the left of xo to the right of xo, then (xo, yo) is an inflection point, otherwise not.

a change in the sign of f '' signals a change in the concavity of f around xo. a negative f '' indicates an f that is concave down, and a positive f '' indicates an f that is concave up.

one can also perform a first-derivative test for a point of inflection where f ''(x) = 0. you may enjoy developing that test yourself.

by Jeremiah Mburuburu Wed May 01, 2013 2:12 am

rishi,

- what's MOOC?

- i found it odd that you didn't participate in or acknowledge the discussion of your limit problem.

enna vayaththu valiA?

by MaxEntropy_Man Wed May 01, 2013 6:16 am

Jeremiah Mburuburu wrote:
one can also perform a first-derivative test for a point of inflection where f ''(x) = 0. you may enjoy developing that test yourself.

great suggestion.

by Rishi Wed May 01, 2013 7:00 am

Jeremiah Mburuburu wrote:rishi,

- what's MOOC?

- i found it odd that you didn't participate in or acknowledge the discussion of your limit problem.

enna vayaththu valiA?

Oops.

I forgot to thank you, Max and Seva.

Thank you guys.

Your answer is correct. It is -1.

The approach you and Max took is correct. In fact, I checked with a co-worker who majored in Math.

sqrt(x^2) is indeed equal to abs(x).

The problem in effect reduces to sqrt(x^2)/x which is abs(x)/x.

Since x takes on negative value, abs(x)/x indeed is negative when x goes to negative infinity.

The discussion ended up as a long argument with Seva. So I moved on.

But I am curious, I wonder why the text books do not have examples involving limits approaching negative infinity.

MOOC is an acronym for the online courses offered free by the consortium of many schools.

http://en.wikipedia.org/wiki/Massive_open_online_course

by Rishi Wed May 01, 2013 7:02 am

Before I forget, thanks for Max and JM for their inputs on the test for inflection point.

by MaxEntropy_Man Wed May 01, 2013 7:20 am

Rishi wrote:
The discussion ended up as a long argument with Seva. So I moved on.

future episodes will involve less text and this concise rendering of my emotions:
Calculus misconception cleared Bangheadonwall

by Seva Lamberdar Wed May 01, 2013 7:40 am

[quote="Rishi"]

Jeremiah Mburuburu wrote:rishi,

I forgot to thank you, Max and Seva.

Thank you guys.

You are always welcome Rishi.

by Rishi Wed May 01, 2013 8:01 am

MaxEntropy_Man wrote:
Jeremiah Mburuburu wrote:
one can also perform a first-derivative test for a point of inflection where f ''(x) = 0. you may enjoy developing that test yourself.

great suggestion.

f(x) = x^4

f'(x)= 4x^3

f'(0) = 0

Since f'(0) = 0, x=0 should be a maximum or minimum point.

Therefore it cannot be an inflection point

Testing for f' on both sides of x = 0,

f'(1) = 4 and f'(-1) = -4

f' goes from negative to positive. The curve changes from decreasing to increasing. It is a minimum point.

Btw I think if I really explain to myself conceptually from first principles instead of mindlessly applying the tests, I find there is clarity.

by Seva Lamberdar Wed May 01, 2013 8:56 am

Rishi wrote:
f(x) = x^4

f'(x)= 4x^3

f'(0) = 0

Since f'(0) = 0, x=0 should be a maximum or minimum point.

Therefore it cannot be an inflection point

Testing for f' on both sides of x = 0,

f'(1) = 4 and f'(-1) = -4

f' goes from negative to positive. The curve changes from decreasing to increasing. It is a minimum point.

Btw I think if I really explain to myself conceptually from first principles instead of mindlessly applying the tests, I find there is clarity.

What happens (at x=0) if f(x) = x^3?

by Rishi Wed May 01, 2013 9:52 am

oops

Th blanket statement "Since f'(0) = 0, x=0 should be a maximum or minimum point." I made is not totally accurate.

It should also be a critical number you get by solving f'(x) =0 for x and then test for f'(x) > 0 and f'(x) < 0 for x values on both sides of the critical number.

f(x) = 3x^2 fails that test for x=0.

So x=0 is not min or max for y=x^3

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