Third derivative method (1 Viewer)

[buchanan]

Are you sick and tired of testing either side of the point, when verifying inflections? Do you remember that there's a second derivative test for classifying turning points, which is usually faster than testing either side of the point? Well there's also a faster method for verifying inflections. It's called the third derivative method:

If f''(x<SUB>0</SUB>)=0 and f'''(x<SUB>0</SUB>) exists and &ne;0 then (x<SUB>0</SUB>, f(x<SUB>0</SUB>)) is an inflection point.

And woe betide any teacher who punishes their students for using this method.

Example

Find the inflection point for y=x<SUP>3</SUP>-3x<SUP>2</SUP>.

Solution

f''(x)=6x-6=0 implies x=1 and y=-2 and f'''(1)=6&ne;0. Hence (1,-2) is an inflection point.

NO NEED TO WASTE TIME TESTING EITHER SIDE OF THE POINT!

SAVES TIME IN EXAMS!

YEAH! I'M HAPPY!!!!!!

Now try one yourself:

Show f(x)=sin(x) has an inflection at (0,0).

Solution

Spoiler

f''(x)=-sin(x) and f'''(x)=-cos(x).

So f''(0)=0 but f'''(0)=-1&ne;0. Hence (0,0) in an inflection point.

See how much easier and quicker it is than testing either side of the point?

So why does it work?

Proofs

Method 1

I prove it using a contrapositive argument
((IF not q THEN not p) is equivalent to (IF p THEN q)).

Suppose f''(x<SUB>0</SUB>)=0 and f'''(x<SUB>0</SUB>) exists.

IF (x<SUB>0</SUB>, f(x<SUB>0</SUB>)) is not an inflection point THEN either:

• both f''(x<SUB>0</SUB><SUP>+</SUP>) and f''(x<SUB>0</SUB><SUP>-</SUP>) are positive;
• both f''(x<SUB>0</SUB><SUP>+</SUP>) and f''(x<SUB>0</SUB><SUP>-</SUP>) are negative; or
• at least one of f''(x<SUB>0</SUB><SUP>+</SUP>) or f''(x<SUB>0</SUB><SUP>-</SUP>) is zero.

and hence f'''(x<SUB>0</SUB>)=0.

The contrapositive now is

IF f'''(x<SUB>0</SUB>)&ne;0 THEN (x<SUB>0</SUB>, f(x<SUB>0</SUB>)) is an inflection point

Method 2.

Let g(x)=f'(x). Then the second derivative method for y=g(x) is equivalent to the third derivative method for y=f(x).

A local minimum or local maximum for y=g(x) will correspond to an inflection for
y=f(x).

• g'(x₀)=0 and g''(x₀)>0 implies (x₀, g(x₀)) is a local minimum for y=g(x).
  
• g'(x₀)=0 and g''(x₀)<0 implies (x₀, g(x₀)) is a local maximum for y=g(x).

Hence if f''(x₀)=0 and f'''(x₀) exists and &ne;0 then (x₀, f(x₀)) is an inflection point for y=f(x).

The second derivative method is in the 2 unit syllabus, and this second proof shows that the third derivative method is equivalent to it.

Here are some more examples:

Examples from Past Papers

1. (1993 2U Q6a)

Show (0,0) is an inflection point for f(x)=(1/4)x⁴-x³.

Solution
f''(x)=3x²-6x and f'''(x)=6x-6

So f''(0)=0 and f'''(0)=-6&ne;0.

Hence (0,0) is an inflection point.

2. (1990 2U Q5iii)

Show (0,1) is an inflection point for f(x)=1+3x-x³.

Solution

Spoiler

f''(x)=-6x and f'''(x)=-6.

So f''(0)=0 and f'''(0)=-6&ne;0.

Hence (0,1) is an inflection point.

3. (1947 Leaving Certificate Honours I Paper I Q12i)

Show (0,-1) is an inflection point for f(x)=x⁴-2x³+2x-1.

Solution

Spoiler

f''(x)=12x²-12x and f'''(x)=24x-12.

So f''(0)=0 and f'''(0)=-12&ne;0.

Hence (0,-1) is an inflection point.

4. (1997 4U Q3biii)

Show (1,9) is an inflection point for f(x)=3x⁵-10x³+16x.

Solution

Spoiler

f''(x)=60x³-60x and f'''(x)=180x²-60.

So f''(1)=0 and f'''(1)=120&ne;0.

Hence (1,9) is an inflection point.

5. (1997 2U Q8biii)

Show (&pi;, &pi;) is an inflection point for f(t)=t+sin(t).

Solution

Spoiler

f''(t)=-sin(t) and f'''(t)=-cos(t).

So f''(&pi;)=0 and f'''(&pi;)=1&ne;0.

Hence (&pi;, &pi;) is an inflection point.

6. (1991 4U Q4bii)

Show (ln6, 1/2) is an inflection point for g(x)=4e^-x-6e^-2x.

Solution

Spoiler

g''(x)=4e^-x-24e^-2x and g'''(x)=-4e^-x+48e^-2x.

So g''(ln6)=0 and g'''(ln6)=2/3&ne;0.

Hence (ln6, 1/2) is an inflection point.

7. (1946 Leaving Certificate Honours I Paper I Q11)

Show (&pi;/2, 0) is an inflection point for f(x)=sin(x)sin(2x).

Solution

Spoiler

This is made easier by using the product to sum identity
sin(A)sin(B)=(1/2)(cos(A-B)-cos(A+B)) before differentiating.

So f(x)=(1/2)(cos(x)-cos(3x)).

f''(x)=(1/2)(-cos(x)+9cos(3x)) and f'''(x)=(1/2)(sin(x)-27sin(3x)).

So f''(&pi;/2)=0 and f'''(&pi;/2)=14&ne;0.

Hence (&pi;/2, 0) is an inflection point.

Testing either side if the point for all these examples would take a lot more time.

It can also be generalised:

Successive derivative test

If f⁽ⁿ⁾(x₀)=0 for n=2,3,4,...,2k and if f^(2k+1)(x₀) exists and &ne;0, then (x₀, f(x₀)) is an inflection point (i.e., if the order &ge;2 of the first non-zero derivative has odd parity, then it's an inflection.)

If f⁽ⁿ⁾(x₀)=0 for n=1,2,3,4,...,2k-1 and if f^(2k)(x₀)>0, then (x₀, f(x₀)) is a local minimum (i.e., if the parity of the order of the first non-zero derivative is even and the sign of this derivative is positive then it's a local minimum.)

If f⁽ⁿ⁾(x₀)=0 for n=1,2,3,4,...,2k-1 and if f^(2k)(x₀)<0, then (x₀, f(x₀)) is a local maximum (i.e., if the parity of the order of the first non-zero derivative is even and the sign of this derivative is negative then it's a local maximum.)

Note that although the conditions in the second and third derivative tests are sufficient but not necessary, nevertheless the conditions in the successive derivative test are both sufficient and necessary (provided the derivatives exist).

Example 1

Show (0,0) is a local minimum for f(x)=x⁴.

Solution

f(0)=f'(0)=f''(0)=f'''(0)=0 but f''''(0)=24>0, so by the fourth derivative test, (0,0) is a local minimum.

Example 2

Show (0,0) is a horizontal inflection point for f(x)=e^xx⁹⁹.

Solution
f(x)=e^xx⁹⁹=&Sigma;(xⁿ⁺⁹⁹/n!). So f(0)=f'(0)=...=f⁽⁹⁸⁾(0)=0 but f⁽⁹⁹⁾(0)=99!&ne;0. Hence (0,0) is a horizontal inflection point, by the 99-th derivative test.

Example 3

Show (0,0) is an inflection point for f(x)=x⁵.

Solution

f''(0)=f'''(0)=f''''(0)=0 but f'''''(0)=5!=120&ne;0. Hence (0,0) is an inflection point.

More examples

1. Show (5,0) is a local minimum for f(x)=(x-5)⁴.

Solution.

f'(5)=f''(5)=f'''(5)=0 and f''''(5)=24>0.

So the parity of the order of the first non-zero derivative at (5,0) is even and the sign of this derivative is positive therefore it's a local minimum.

(4th derivative method)

2. Show (6,0) is an inflection for f(x)=(6-x)³

Solution.

f''(6)=0 and f'''(6)=-6&ne;0

So the order &ge;2 of the first non-zero derivative at (6,0) has odd parity and therefore it's an inflection.

(3rd derivative method)

3. Show (4,0) is a local maximum for f(x)=-3(x-4)⁶

Solution.

f'(4)=f''(4)=f'''(4)=f⁽⁴⁾(4)=f⁽⁵⁾(4)=0 but f⁽⁶⁾(4)=-3(6!)=-2160<0

So the parity of the order of the first non-zero derivative at (4,0) is even and the sign of this derivative is negative therefore it's a local maximum.

(6th derivative method)

More for you to do

4. Show (3,0) is an inflection for f(x)=(x-3)⁵

Spoiler

f''(3)=f'''(3)=f⁽⁴⁾(3)=0 but f⁽⁵⁾(3)=5!&ne;0

5. Show (2,0) is a local minimum for f(x)=(x-2)⁸.

Spoiler

f'(2)=f''(2)=f'''(2)=f⁽⁴⁾(2)=f⁽⁵⁾(2)=f⁽⁶⁾(2)=f⁽⁷⁾(2)=0 but f⁽⁸⁾(2)=8!>0

6. Show (-1,0) is a local maximum for f(x)=-2(x+1)⁴

Spoiler

f'(-1)=f"(-1)=f'''(-1)=0 but f⁽⁴⁾(-1)=-2*4!= -48<0.

Here's what others have said about it:

Wow. Very handy.

I've used it quite a few times myself in prelim. exams

It's very useful and does save a lot of time

I used it in the exams.

We were taught that method last year in yr11 ext1 maths.

Wow! That's cool!

It seems like a natural extension of the second derivative test.

It's a good method. I don't like wasting time testing either side of the point all the time.

If it's good enough for ngai (2004 TG Room medallist), then it's good enough for the rest of us!

But there have been some objections in the past from some people regarding the use of the third derivative method. This method is sufficient but not necessary and so their objections have been based on the premise that if it doesn't always work, one should never use it. However, anyone who objects to using the third derivative test for inflections should also object to using the second derivative test for turning points. The second derivative test doesn't always work, but it usually does and it usually saves time. The same can be said of the third derivative test. Moreover, their objections are further weakened by the happy fact that although the conditions in the second and third derivative methods are sufficient, but not necessary, nevertheless, the conditions in the successive derivative method are both sufficient and necessary (provided the derivatives exist).

Use it judiciously, like you would the second-derivative test.

This more balanced view is precisely the correct approach.

When not to use this method

The whole point of the method is to save time. So if it doesn't save time, then don't use it. Examples include:

• if the derivatives are hard to find, such as in quotients which are common in HSC exams (like lnx/x). In this case, testing either side of the point is more efficient.
• or less commonly, when the derivatives don't exist (like f(x)=x⁴sin(1/x) for x&ne;0, f(0)=0. f'(0) exists and is 0, but all higher order derivatives at (0,0) do not exist). In these cases other methods, such as symmetry arguments can be used.

In general, do whatever method is the most efficient.

The method didn't just pop out of my head. There are some good references you can check.

References

Ayres, F. and Mendelson, E., Calculus, McGraw-Hill

Dowling, E.T., Mathematical Methods, McGraw-Hill

http://mathworld.wolfram.com/ExtremumTest.html

There have also been some objections regarding whether or not I should be using BOS forums to communicate this method to students.

So I'll let David Hilbert have the last say on the matter, who said in 1930:

We must know.
We will know.

 

[joesmith1975]

So if y''= 6x-6 e.g and x=1 and y=2 for e.g the y''' = 6 so do you write

6 is not = to 0 so its a pt of inflexion and thats it ?

 

[buchanan]

Yep. That's it.

 

[goburnacat]

actually, at some schools its not an accepted thereom- best check with your teachers first
i got a mark taken off for using it

 

[sikeveo]

lol we use this at uni, it would be quite stupid to penalise this method.

 

[buchanan]

i got a mark taken off for using it

This says more about your teachers' lack of mathematical knowledge than about your own mathematical ability. If your solution is correct, you should demand to get the mark. Assessments should be about your mathematical ability and not about either your teachers lack of mathematical knowledge, nor about your ability to regurgitate the syllabus.

(NB. I haven't seen the question, nor your solution, nor your teachers marking of it, so I am assuming your solution is correct. Otherwise your teacher had every right to deduct a mark.)

best check with your teachers first.

No. Best check that your school is employing properly qualified teachers first before you enrol.

at some schools its not an accepted thereom

Some schools are employing poorly trained teachers. I would advise such schools to either employ properly trained teachers or stop teaching maths at all. It's actually better (albeit sadly) for maths to not be taught at all than to have it taught badly.

lol we use this at uni, it would be quite stupid to penalise this method.

Yep. It's quite stupid. I have provided 2 proofs of it above. But there's an even more poignant reason for why it's so stupid. I bet the same teacher who rejects 3rd derivative method for verifying inflections would also accept the 2nd derivative method for classifying turning points, right? Well my second proof actually shows that the 2 methods are equivalent! So if the teacher accepts one, they should accept the other - otherwise they are unwittingly contradicting themselves - and providing me with ammunition with which to publicly pour scorn and derision upon the philistines who dominate mathematical education in NSW.

 

[felixcthecat]

lol...we were taught this method and using test points, but usually i'm just too lazy to think i use test points- but question- in HSC, while using test points do we need to find the exact value or do we just need to find whether it's positive/negative? because our school requires us to find exact value which is realli annoying......

and if it's taught in some schools and not in others, then it'll be accepted in HSC assuming the 'some schools' aren't teaching the wrong material (in this case it's not wrong)- orelse there won't be a point in teaching it

 

[buchanan]

You should definitely learn testing either side of the point as well because sometimes this is more efficient than 2nd or 3rd derivative methods, eg., if it's a quotient it'll be hard to differentiate 2 or 3 times, so if testing either side happens to be a more efficient method for a particular question, then do it this way.

But you just need to set it out clearly for markers in the HSC. Change of sign is sufficient. There's no need for exact values. Change in sign is more important than the magnitude whichever method you use. See for example MANSW's solution to the 2005 Ext. 2 paper, Q8(a)(i), or mine. Neither MANSW's nor mine uses exact values.

 

[ngai]

No. Best check that your school is employing properly qualified teachers first before you enrol.

if 'properly qualified' means that they have to accept things like the 3rd deriv test, then theres probably no school where every math teacher is qualified...and u dont get to pick ur teacher once you're in a school, so dont enrol in any school?

Some schools are employing poorly trained teachers. I would advise such schools to either employ properly trained teachers or stop teaching maths at all. It's actually better (albeit sadly) for maths to not be taught at all than to have it taught badly.

but just how many 'properly trained teachers' are there? enough to fill every maths teaching staff in the country?
just imagine if only the schools that have only 'properly trained teachers' in their maths staff are allowed to teach maths...how many students would leave australia, what would happen to the entry requirements at uni, the standards of our unis, etc
and i dont know how schools choose teachers...but do public or selective schools even get to choose their own teachers?


i understand what youre trying to say, and yea there isnt much reason for rejecting the 3rd deriv test
but i can't agree with advising students "not to check with their teachers if its accepted"...and this goes for any other 'outside syllabus' stuff too
if teachers just wont accept it, then as a student, what can you do?...argue with them?..complain to the principal?..complain to bos?..
as a student, u want to get marks as well...

 

[buchanan]

A mark from a good teacher is worth more than a mark from a bad one.

 

[YBK]

wow, that's cool.

 

[darkliight]

i understand what youre trying to say, and yea there isnt much reason for rejecting the 3rd deriv test
but i can't agree with advising students "not to check with their teachers if its accepted"...and this goes for any other 'outside syllabus' stuff too
if teachers just wont accept it, then as a student, what can you do?...argue with them?..complain to the principal?..complain to bos?..
as a student, u want to get marks as well...

When it comes to the HSC being marked, I don't think the markers can discriminate against your methods for arriving at a correct solution, unless ofcourse they ask you to use a specific method. buchanan will probably have a definitive answer to that, but I can't see how it would even be possible.

No teacher should ever penalise a student for going above and beyond what is required of them, resulting in a more efficient and/or elegant solution to a problem. Stupid would be an understatement.

 

[buchanan]

Yeah. It's not just stupid. It's an abomination.

Most of what is education is encouragement.

How encouraging is it for someone enlightened by the 3rd derivative method only to be penalised for it just because the teacher doesn't know it? Not very.

And therein lies the failure of the assessment as a teaching tool if it has been marked in such an atrocious manner.

Not only should the student have gotten full marks for it, but I'd suggest perhaps also bonus marks for doing it with a better method.

Now that would be more encouraging, wouldn't it? And then the teacher would have been doing their job better, don't you think?

 

[goburnacat]

i dont really think its becuase my teachers are poorly trained, i mean, we're like in the top 5 of the state when it comes to maths. i think theyre just "cautious" when it comes to "new" theorems because they dont want you to risk getting marks off in the hsc for using a theorem out of the syllabus

 

[Raginsheep]

The point is, you don't (or shouldn't) lose marks for using out of syllabus techniques as long as its correct.

 

[buchanan]

Yeah.

Evidently, selective schools are selective of their students, but not of their teachers. goburnacat is obviously a better mathematician than her teacher.

Marking a student incorrect when she is correct will be seen by the intenational mathematical community as idiotic. And on the internet, that's what matters, not the opinions of philistines who say things like "just stick to the syllabus".

It's also important for it to be said that the whole system in NSW should not be tainted with the same brush. The Board of Studies are very interested in maintaining both the integrity of the HSC and also media perceptions thereof.

 

[NightShadow]

well Mr. Ian Woodhouse [the guy who heads and oversees the entire 4 unit marking process] says: if its not in the syllabus, and you want to use it... prove it first... or dont use it.

 

[buchanan]

Theorems should be proved, whether they are in the syllabus or not.

I have proved the third derivative test two ways in my first post in this thread.

But there's no need to continue reproving it every time one uses it.

Actually, once is enough. I only did it two ways to show the different methods of proof.

As for advice on HSC marking, it's available in the Notes from the Marking Centre (including Marking Guidelines) at
http://www.boardofstudies.nsw.edu.au/hsc_exams/
and Standards Packages at
http://arc.boardofstudies.nsw.edu.au/go/hsc/std-packs/#m

 

[NightShadow]

yeh, to use it in the test you have to prove it on the test paper first before you can apply it. thats what he means

 

[buchanan]

To prove a theorem every time one uses it, whether it's in an exam or not, would be quite stupid.

It's a little off-putting isn't it? - By design, I would suggest. I think it might just be his way of trying to force you (or convince you) to just stick to the syllabus and not do anything beyond the syllabus.

Not everyone agrees with this approach. If a teacher doesn't have depth of knowledge beyond the syllabus, then the syllabus is the ceiling. And for good kids, that's not good enough. The syllabus should be the floor, not the ceiling (http://www.theaustralian.news.com.au/story/0,20867,18636129-12332,00.html).

Deducting marks for a valid method outside the syllabus is not a good argument. Saying to students I'll deduct a mark if you use a method I don't like (regardless of the validity of the method) just makes the teacher look rather idiotic and invalidates their own assessments. It doesn't invalidate the method.

Marks should not be deducted if students have done it correctly with a valid method (regardless of where the method came from - the syllabus or elsewhere) and have shown sufficient working. For example, in last year's Ext. 2 paper, the Heaviside method was used without proof for partial fractions, a valid method not in the syllabus, and provided students showed working, they got marks for it. And in the 2004 2 Unit paper, A=rl/2 formula was used without proof by some students to get the area of a sector (which again isn't in the syllabus) and they got marks for it. Essentially the marking process should be fair and reasonable. It wouldn't be fair or reasonable to penalise students for using a valid method just because it's not in the syllabus. It's a decision for the marking teams and they usually make good decisions. I doubt what happenned to goburnacat would happen in the HSC. The way her paper was marked was neither fair nor reasonable, and HSC marking teams try to avoid such unfairness or unreasonableness.

On the other hand, I think a question specifically asking to prove the third derivative method in an exam would be a good question.