# Inflection Points of Third Degree Polynomials

## A New Way To Find The Inflection of A Curve Photo by John Moeses Bauan on Unsplash

Math is a subject where even the tiniest thing can be overlooked. Take for example factoring. There are countless ways to factor a polynomial yet we as mathematicians typically only use one. It’s this overlooking that caused me to discover something that has yet to be discovered.

I’m sure everyone that has taken a Calculus course messed with concavity and curve changes. Part of this measurement is finding the inflection point of the curve or the point at which one curve changes to the other curve. Throughout this section of my calculus course, I noticed a trend among every graph. The inflection point was always halfway between the max and the min. In other words it was (Max + Min)/2 = f”(x).

After noticing this trend I set out to prove that it worked. How was it that this worked every single time as if it was the second derivative? The answer lies in the proof. To show this proof, however, I need to show some proofs. The proofs I used to prove this idea were the Quadratic Formula Proof, Inflection Point Proof, and Relative Max and Min Proof.

Why is it important?

But first, why are inflection points important? How could something as silly as inflection points be valuable to the community and better yet how does this idea help or improve our understanding of mathematics?

Inflection points are more important than many realize. It doesn’t just show the point of change but also shows the trend of the data. Take for example the stock market. If we were to map every inflection we could see how the data is trending over the day and use those numbers to predict future trends. So inflection points are very important in the sense that part of our understanding of data and what it communicates is how the data behaves.

Another use of Inflection points is in computer 3D recognition and 3D printing. The computer, to register the object or print the object, has to be able to map the curve of an object. Let’s say a computer is gonna map a shoelace that has 10 different-sized curves. For the computer to process that shoelace and understand what it’s seeing, it has to be able to map the curves on the object.

So how does my idea help or improve these aspects of ideas? The answer is simple: It allows avoidance of unfactorable equations and can be applied to larger polynomials, provided you find the max and min. So in the case of the stock market or the 10 curves shoelace, it allows the process to be less tedious and can measure higher degree polynomials. However, since the proof for higher degree polynomials is still a working progress, I will focus only on cubic polynomials.

Proofs

The first proof we need to prove this idea is something almost every American learns in Math classes. The Quadratic Formula Proof.

So if ax² + bx + c = 0 then,

x= (-b +/– sqrt(b²-4ac )/2a

Proof: Complete The Square Method

ax²+bx+c=0

ax²+bx=-c

x²+(b/a)x=-(c/a)

x²+(b/a)x +(b²/4a²)=(b²/4a²)-(c/a)

(x +(b/2a))²=(b²/4a²)-(c/a)

(x +(b/2a))²=(b²/4a²)-(4ac/4a²)

(x +(b/2a))²=(b²–4ac)/4a²

(x +(b/2a))²=(b²–4ac)/4a²

x+b/2a=b²–4ac/2a

x= (-b +/– sqrt(b²-4ac )/2a

Now that we have the quadratic Formula we will need the traditional method to find inflection points or the inflection point proof (second derivative test).

If xsub0 is a point of inflection of the function f(x), and this function has a second derivative in some neighborhood of xsub0, which is continuous at that point xsub0 itself, then,

f”(x)=0

Proof: Second Derivative Test

Suppose that the second derivative at the inflection point xsub0 is not zero:

f”(xsub0) ≠ 0. Since it is Continuous at xsub0, then there exists a

delta -neighborhood of the point xsub0 where the second derivative preserves its sign,

that is,

f”(xsub0)<0 or f”(xsub0)<0 x(xsub0-delta, xsub0+delta).

In this case, the function is either strictly convex upward (when f”(x) < 0)

or strictly convex downward (when f”(x) > 0). But then the point xsub0 is not an inflection point. Hence, the assumption is wrong and the second derivative of the inflection point must be equal to zero.

Lastly, we will need the relative max and min proof to find the max and min to prove that (max + min)/2=f”(x).

If f(x)has a relative extrema at x=c and f’(c ) exists, then x=c is a critical point of f(x). It will be the critical point such that f’(c )=0.

Fermat’s Theorem: Relative Max and Min Proof

Assume that f(x)has a relative max/min. If we assume that we have a

relative max/min at x=c then we know that f(c) ≥ f(x)for all x that are

sufficiently close to c. in particular for all h that are sufficiently close to zero (positive or negative) we must have,

f(c) ≥ f(c+h)

f(c+h)-f(c)≤0

Now let’s assume that h≥0 and divide both sides by h. Thus,

(f(c+h)-f(c))/h≤0

Since we assume that h≥0then we can take the right-hand limit.

lim(h->0^+)(f(c+h)-f(c))/h ≤lim(h->0^+)0=0

Since we also assume that f’(c ) exists and we recall that the normal limit that exists must be equal to both one sided limits. Then,

f’(c )=lim(h->0^+)(f(c+h)-f(c))/h=lim(h->0^+)(f(c+h)-f(c))/h ≤0

Thus we get,

f’(c )≤0

If we flip the inequality on h and solve for the left-hand limit then we derive,

f’(c )≥0

Since we have both f’(c )≤0 and f’(c )≥0, then the only way for these to be true is if f’(x)=0 which means that x=c must be a critical point.

My Method

In order to show that f”(x)=(max + min)/2, we need to first solve ax³+bx²+cx+d using the traditional method. So,

Given that f(x)=ax³+bx²+cx+d, find the inflection point. Since f”(x)is the

inflection point we get,

f(x)=ax³+bx²+cx+d

f’(x)=3ax²+2bx+c

f”(x)=6ax+2b

Then we solve for x, since f”(x)=0, we get,

x=-b/3a

Now that we have laid the groundwork it’s time to show that f”(x)=(max +min)/2.

If some continuous function f(x)has a max and min or relative max and min, then (max+min)/2=f”(x)or the inflection point.

The first step to prove this is to find the max and min of f(x)=ax³+bx²+cx+d.

So,

f(x)=ax³+bx²+cx+d

f’(x)=3ax²+2bx+c

Now we solve for the x of the first derivative to find the max and min. So we get,

x= (-2b +/- sqrt(4b²–12ac) )/6a

Since (max+min)/2 is commutative inside the parenthesis, it doesn’t matter which we label the max and which we label the min. For this though, I will label the x with the plus as max and the x with the minus as the min. Thus we get,

max = (-2b + sqrt(4b²–12ac ))/6a min = (-2b -sqrt(4b²–12ac ))/6a

Now we must plug these values into the expression (max + min)/2. Thus we produce,

((-2b + sqrt(4b²–12ac ))/6a + (-2b- sqrt(4b²–12ac))/6a)/2

Which reduces to,

-b/3a

So finally we get,

(max+min)/2= -b/3a

And,

f”(x)= -b/3a

Which means by the transitive property we get,

(max+min)/2=f”(x)

Conclusion

So as you can see, this method works. It allows us to solve for the change of concavity in an equation without taking the derivative several times. It’s easier to mess with and has less room for error. Perhaps the coolest thing about this method is that it isn’t constricted by whether or not an equation can be factored. As you go up the chain of higher degree polynomials it becomes harder and harder to find inflection points as well as max and min because of difficulty factoring. This method passes that problem by putting all the focus into the first derivative. Currently, I have only proved this for cubic polynomials but I look forward to proving this for higher degree polynomials. Perhaps there will be a pattern to all of this.

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