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

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

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Hint : What is the simplest possible form of the 2nd derivative that we can have that will guarantee that we have a single inflection point?
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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 2nd derivative must also be a polynomial and we have to have $$p''\left( a \right) = 0$$. In addition, we know that the 2nd 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 2nd derivative changes sign at $$x = a$$ as it does here.

Hint : We saw how to “undo” differentiation in the practice problems in the previous section. Here we simply need to do that twice and note that we don’t actually have to undo the derivatives here, just think about what they would have to look like.
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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 2nd derivative is a 1st degree polynomial and that means the 1st derivative had to be a 2nd degree polynomial. This should make sense to you if you understand how differentiation works.

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

Okay, we’ve figured out that the 1st derivative must have been a 2nd degree polynomial. This in turn means that the original function must have been a 3rd degree polynomial. Again, differentiation lowers the power of $$x$$ by 1 and if the highest power of $$x$$ in the 1st 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.