Hi guys!

 

In calculus, we were shown ways of finding maximums and minimums. To do this, we had to differentiate, and then make the differential=0. Then, to make sure whether is was a maximum or minimum, we double differentiated the original equation (or simply differentiate the differential). If the value was positive, the point was a local minimum, and vice versa.

 

The book had stated that if the double differential was equal to 0, then, it was a point of inflexion. However, my teacher stated that this does not apply on some curves, and therefore, not to use it, but find the gradient before and after the point, and decide thereon.

 

Could some of you guys please give me an example of such an occurrance?

 

Thanks alot guys!

i remember seeing an example of this but can't remember what the equation was. i think it was a fifth order polynomial though

oh dear! An example would really help, because I want to try it out for myself. I have a program called omnigraph and I could plot the graph, calculate the derivative and see what my answer leads me to.

It's funny becuase I started this today in maths (yr 12) and asked my teacher the same question without a satisfying response. Though I figured this:

 

[math]f(x) = x^4[/math]

 

[math]f''(x) = 12x^2[/math]

 

[math]f''(0) = 0[/math]

 

Therfore point of inflexion, oh no, razzamatazz, we know this is a minimum from experience. It is in fact a tripple root and is also a double inflexion. Not sure how to actually solve this properly without having to try above and below, though I am guessing that you have to take the quadruple derivative:

 

[math]f''''(x) = 24 > 0[/math]

 

Therfore it is a minimum.

Wow! That's a grat example. I admire the simplicity. Thanks alot for that!

I don't think you need to look at the fourth derivative; the third will suffice. This article at Wikipedia might help to clarify matters.

That's a great link Dave! Thanks for that!

The book had stated that if the double differential was equal to 0, then, it was a point of inflexion. However, my teacher stated that this does not apply on some curves, and therefore, not to use it, but find the gradient before and after the point, and decide thereon.

The fact that the second derivative is zero is a necessary but not a sufficient condition to have a point of inflection. A sufficient condition would be that the second derivative is 0 at the point and changes sign arround the point.

Changes sign around that point? Do you mean that the double differential changes sign in the x value before and after the point?

Indeed.

The method that I tend to use is, instead of using the double derivative, just use the single derivative for the x value before and after the point which is assumed the point of inflexion. If the sign remains constant, you know the the point is a pont of inflexion.

 

Are there any advantages of using the double derivative instead of the single derivative?

The method that I tend to use is' date=' instead of using the double derivative, just use the single derivative for the x value before and after the point which is assumed the point of inflexion. If the sign remains constant, you know the the point is a pont of inflexion.

 

Are there any advantages of using the double derivative instead of the single derivative?[/quote']

You say "the point which is assumed the point of inflexion.", but how did you find this point? With the first or second derivative being 0?

The second derivative being 0 (since that is what the point of inflexion maybe)

In that case you already calculated the 2nd derivative so the 'advantage' isn't that big. You are correct though, a zero second derivative and constant sign of the first derivative is a point of inflection.

The reason I asked is that some people apply your method (looking at the sign of the 1st derivative to see if it's constant) to points where the first derivative is zero. Although this gives points of inflection, it is a sufficient but not a necessary condition so you may not find all points of inflection.

Oh, so its really a matter of opinion!

 

Thanks alot for your help TD!

You're welcome