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.

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.

 

[shinji]

yep. handy method.

i'm gonna practice this method more ... and then surprise my maths teacher. bahaha

this should get a sticky!!

 

[Riviet]

I understand why it works, and I'm very tempted to use it all the time, so I'll consult my maths teacher about this method.

 

[pLuvia]

Nice work buchanan, but people should consult with their teacher before actually using it

 

[buchanan]

If you consult your teacher, don't forget to educate them by showing them this thread.

If your teacher lets you use the second derivative method, then they should let you use the third derivative method as well. But if they don't let you use the third derivative method then they shouldn't let you use the second derivative method either (contrapositively speaking).

The reason for this is that the two are equivalent.

Suppose 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 shows that the third derivative method is equivalent to it - hence your teacher should let you use both methods.

 

[who_loves_maths]

the moral of this thread should be: do NOT use the 3rd, or nth, derivative test in HSC exams! NEVER.
the testing-of-values with the second derivative is a flawless test for POI 100% of times! use that always.

the use of buchanan's results along with his overtly trivial examples really pre-empt the notion that one would need to already know whether or not a point is a POI (or at least be very confident of it) before using it to "save time". but of course, you can't have that kind of guarantees in the HSC, so again, the second derivative is a much safer approach.
why else do you think HSCers are never taught this otherwise simple general result? because of the possible confusion in its application!

if we are going to talk about the nth-derivative test, then anyone reading this new thread needs to keep in mind that ALL examples buchanan has given thus far in this thread are trivial. and that is why they give the illusion that the nth-derivative test is so simple to apply when it is not:
eg. what if you get a complicated looking rational function or other complex trascendental functions? will you immediately have the confidence to apply your nth-derivative test? who know how many differentiations you will need before you can prove anything!

Originally Posted by buchanan
who_loves_maths' argument for never using the third derivative test is along the lines of: it doesn't always work ∴ don't ever use it.

EXACTLY! it doesn't always work so don't ever use it IN THE HSC!!!

once again buchanan, i advise you not to give irresponsible advice to potential new HSCers perusing these forums. it is because it doesn't always work that you should never use it in an HSC exam where time is of the essence!

and if anyone out there is a smart worker, then you would know that since the nth-derivative result is not apart of the syllabus then the functions you will encounter in any HSC exam will not be nth-derivative friendly!
ie. dont' try to differentiate something like a rational function n times!

Originally Posted by buchanan
Anyone who objects to using the third derivative test for inflections should also object to using the second derivative test for turning points.

agreed. so maybe that is why teachers (well, at least my year 12 maths teacher) usually prefer the testing-of-values technique using the 1st derivative over the use of the second derivative for turning points buchanan?

you 'distrust' testing values because you merely think it is not mathematically 'rigorous' enough.
but if a method is valid and 100% flawless under all circumstances then why not use it?
unless of course one, such as yourself at times, is irrational

 

[buchanan]

If a method is valid and 100% flawless under all circumstances then why not use it?

Because it's not always the most efficient method.

ALL examples buchanan has given thus far in this thread are trivial.

Does that include the 5 HSC examples I used? Are HSC questions trivial? Most HSC students wouldn't think so. And the other examples I used were similar to those anyway.

what if you get a complicated looking rational function

On this, I agree. It's better to use testing either side of the point here because in this case, this is more efficient than third derivative method. I did actually say that before (although what you call rational, I was a bit more general and called them quotients, e.g., lnx/x, which is not rational, but your point is equally true of this example as well). Usually the third derivative method will actually work for these as well, but alas the derivatives are much harder to find, and so won't save you time.

teachers (well, at least my year 12 maths teacher) usually prefer the testing-of-values technique using the 1st derivative over the use of the second derivative for turning points

For classifying turning points, teachers should teach testing either side of the point first, and then the second derivative method.

Likewise for verifying inflection points, teachers should teach testing either side of the point first, and then the third derivative method.

you 'distrust' testing values because you merely think it is not mathematically 'rigorous' enough.

No. I never said this. Testing values either side of the point should definetely be taught. But so should the second and third derivative methods. The second derivative method is in the syllabus and the third derivative method is equivalent to it. Any teacher who doesn't know this equivalence shouldn't be teaching.

 

[Riviet]

Would the average maths teacher that teaches HSC maths know of the 3rd derivative or nth derivative method?

 

[buchanan]

The average maths teacher knows NOTHING! Ha! Ha! And that's why they shouldn't be teaching!

Nevertheless, I am pleased that there are actually some well educated (hence above average) teachers out there, as evidenced in some of the quotes I used in my first post above - who are teaching the third derivative method.

If your teacher is average, educate them by telling them to read this thread and become above average.

And 50% of teachers are below average! (Think about that too! Ha! Ha!)

 

[pLuvia]

It seems useful

 

[Riviet]

And 50% of teachers are below average! (Think about that too! Ha! Ha!)

Well of course there would be, the way you said it previously.
I have a feeling that my teacher wouldn't have a clue how the third derivative would have an application to curves. Hehe, I'm looking forward to asking him...

 

[buchanan]

You're right. I was being too generous. It's probably more like 80% who are below average.

Riviet was talking about average teachers. If one were to talk about median teachers, I'd say the same about them, i.e., they also know nothing.

Also, who_loves_maths's argument is based on the sad fact that the conditions in the second and third derivative methods are sufficient but not necessary.

He seems not to realise the happy fact that the conditions in the successive derivative test are both sufficient and necessary (provided the derivatives exist).

 

[tywebb]

but what about the function defined by

• f(x)=x⁴(1-cos(1/x)) for x &ne;0
• f(x)=0 for x=0

then f'(0) exists and is equal to 0.

obviously (0,0) is a local minimum.

but f''(0), f'''(0), ... and all higher-order derivatives at (0,0) do not exist.

how can you handle such functions?

how can you prove it's a local minimum?

successive derivative test? no way! you can't even get past the first derivative!

testing either side of the point? no way! the behaviour near (0,0) is wild. suppose you test the value of f'(x) at a small distance away from x=0, say x=&plusmn;&epsilon; for some small &epsilon;. then there are infinitely many local minima, local maxima AND inflections between x=-&epsilon; and x=&epsilon;!

 

[icycloud]

how can you prove it's a local minimum?

f(x)=x^4 * (1-cos(1/x)), x&ne;0
f(x)=0, x=0

x^4 >= 0 for all real x
1-cos(1/x) >= 0 for all real x

Therefore, f(x) >= 0 for all real x, x&ne;0
But, f(x)=0 for x=0
Thus, f(x)>=0 for all real x

Thus, x is an absolute minimum when f(x)=0.
By the definition of the function, when x=0, f(x)=0, thus (0,0) is an absolute minimum of f(x).

 

[tywebb]

yeah. but my point is that for my example, both testing either side of the point AND the successive derivative methods fail.

 

[SeDaTeD]

Actually (0,0) isn't an absolute minimum. f(x) = 0 when x = 1/(2kpi), where k is a nonzero integer, thus there are an infinite number of minima, since f(x) => 0 for all x.

 

[icycloud]

Actually (0,0) isn't an absolute minimum. f(x) = 0 when x = 1/(2kpi), where k is a nonzero integer, thus there are an infinite number of minima, since f(x) => 0 for all x.

Then there are infinitely many absolute minima (a.k.a. global minima). It doesn't change the fact that (0,0) is still an absolute minimum (I didn't say the absolute minimum). (Definition of absolute minima: y₀ is the "absolute minimum" of f(x) on I if and only if y₀ <= f(x) for all x on I.)

 

[tywebb]

yep.

but how about this function:

• f(x)=x⁴sin(1/x) for x&ne;0
• f(x)=0 for x=0

then f'(0) exists and is equal to 0.

so (0,0) is a stationary point.

but again, f''(0) and all higher order derivatives at (0,0) don't exist.

likewise to before, if you test the value of f'(x) at a small distance away from x=0, say x=&plusmn;&epsilon; for some small &epsilon; > 0, then there are infinitely many local minima, local maxima AND inflections between x=-&epsilon; and x=&epsilon;.

so again, successive derivative test fails, but so too does testing either side if the point.

so what sort of stationary point is it?

it can't be a local maximum or local minimum because it's an odd function.

could it be a horizontal point of inflection? but aren't we supposed to have f''(0)=0 if it were an inflection? f''(0) doesn't even exist!

if it were a normal inflection, it would change concavity only once in the interval -&epsilon; < x < &epsilon; for sufficiently small &epsilon;. but it's changing concavity infinitely many times in this interval, no matter how small &epsilon; is.

 

[ronaldinho]

so is any1 gonna actually try this method in the upcoming HSC?

 

[chaoscreater]

The third derivative test is not applicaable for all inflection points.

Consider f(x) = x^5
f''(0) = 0
f'''(0) = 0, and since the rule is that if f''' does not equal to zero, then it's a point of inflection. But it doesn't work in this case because (0,0) is a point of inflection.

So how do we test it for cases like this?