*i.e.*you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 4.6 : The Shape of a Graph, Part II

15. Determine the minimum degree of a polynomial that has exactly one inflection point.

Show All Steps Hide All Steps

^{nd}derivative that we can have that will guarantee that we have a single inflection point?

First, let’s suppose that the single inflection point occurs at \(x = a\) for some number \(a\). The value of \(a\) is not important, this only allows us to discuss the problem.

Now, if we start with a polynomial, call it \(p\left( x \right)\), then the 2^{nd} derivative must also be a polynomial and we have to have \(p''\left( a \right) = 0\). In addition, we know that the 2^{nd} derivative must change signs at \(x = a\).

The simplest polynomial that we can have that will do this is,

\[\underline {p''\left( x \right) = x - a} \]This clearly has \(p''\left( a \right) = 0\) and it will change sign at \(x = a\). Note as well that we don’t really care which side is concave up and which side is concave down. We only care that the 2^{nd} derivative changes sign at \(x = a\) as it does here.

Okay, saw how to “undo” differentiation in the practice problems of the previous section. We don’t actually need to do that here, but we do need to think about what undoing differentiation will give here.

The 2^{nd} derivative is a 1^{st} degree polynomial and that means the 1^{st} derivative had to be a 2^{nd} degree polynomial. This should make sense to you if you understand how differentiation works.

We know that we have to differentiate the 1^{st} derivative to get the 2^{nd} derivative. Therefore, because the highest power of \(x\) in the 2^{nd} derivative is 1 and we know that differentiation lowers the power by 1 the highest power of \(x\) in the 1^{st} derivative must have been 2.

Okay, we’ve figured out that the 1^{st} derivative must have been a 2^{nd} degree polynomial. This in turn means that the original function must have been a 3^{rd} degree polynomial. Again, differentiation lowers the power of \(x\) by 1 and if the highest power of \(x\) in the 1^{st} derivative is 2 then the highest power of *x *in the original function must have been 3.

So, the minimum degree of a polynomial that has exactly one inflection point must be **three** (*i.e.* a cubic polynomial).

Note that we can have higher degree polynomials with exactly one inflection point. This is simply the minimal degree that will give exactly one inflection point.