Third derivative method (1 Viewer)

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buchanan said:
It can also be generalised:

Successive derivative test

If f<sup>(n)</sup>(x<sub>0</sub>)=0 for n=2,3,4,...,2k and if f<sup>(2k+1)</sup>(x<sub>0</sub>) exists and &ne;0, then (x<sub>0</sub>, f(x<sub>0</sub>)) 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<sup>(n)</sup>(x<sub>0</sub>)=0 for n=1,2,3,4,...,2k-1 and if f<sup>(2k)</sup>(x<sub>0</sub>)>0, then (x<sub>0</sub>, f(x<sub>0</sub>)) 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<sup>(n)</sup>(x<sub>0</sub>)=0 for n=1,2,3,4,...,2k-1 and if f<sup>(2k)</sup>(x<sub>0</sub>)<0, then (x<sub>0</sub>, f(x<sub>0</sub>)) 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).
Quite often the example y=x<sup>4</sup> is used to show the second derivative test is sufficient but not necessary and fails for this example.

However, the successive derivative test still works.

f(0)=f'(0)=f''(0)=f'''(0)=0 but f''''(0)=24 > 0 so (0,0) is a local minimum.

Likewise the third derivative test is sufficient but not necessary and fails for y=x<sup>5</sup>, but the successive derivative test still works.

f''(0)=f'''(0)=f''''(0)=0 but f'''''(0)=120 &ne; 0 so (0,0) is an inflection.
 
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chaoscreater

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Quite often the example y=x<sup>4</sup> is used to show the second derivative test is sufficient but not necessary and fails for this example.

However, the successive derivative test still works.

f(0)=f'(0)=f''(0)=f'''(0)=0 but f''''(0)=24 > 0 so (0,0) is a local minimum.

Likewise the third derivative test is sufficient but not necessary and fails for y=x<sup>5</sup>, but the successive derivative test still works.

f''(0)=f'''(0)=f''''(0)=0 but f'''''(0)=120 &ne; 0 so (0,0) is an inflection.

So if we get zero, do we just keep using the next derivative until we get a non-zero number? What if we're in an exam and get a question that takes ages to find the derivative that gets us a non zero number??

OR, what if there is in fact no inflection point but we keep getting zero, do we keep trying the next derivative then???


Correct me if i'm wrong but from your example i'm guessing that you find whether it's an inflection point from the nth derivative, where n is x^n.

let's say we have y = x^6. So we use the 6th derivative and test it and see whether it's an zero or not, am i right?

So what happens if the answer is zero, do we stop there? Or do we keep trying on and on?
 
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Again, your example y=x<sup>6</sup> is easy by the successive derivative test.

f'(0)=f''(0)=f'''(0)=f''''(0)=f'''''(0)=0 but f''''''(0)=720 > 0 hence (0,0) is a local minimum.

Nevertheless I think the point you are trying to make is what if it's more complicated, like a quotient? The derivatives are more difficult, so even if there is an inflection, and even if the third derivative test works, it probably isn't advisable to use it (or for that matter the successive derivative test either).

For these it's probably easier to just test the sign of f''(x) either side of the point (or points) where f''(x)=0. If it changes sign it's an inflection.
 

arcaox

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regardless of whether it works all the time or not i think the man deserves some sort of praise, his proof was rather interesting but for all your hscers id stick to the normal method that the teachers teach you as you will be required to prove how this 3rd derivative test works if it is true which will consume even more time as it is not in the syllabus and rather a new method you are introducing to the examiner
 

ninetypercent

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it would be a bitch trying to explain to my retarded teachers how to use the third derivative
and most likely, they will not understand and give me a zero on my exam

so nah.. im not gonna use it
 

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