## Ask Professor Puzzler

Do you have a question you would like to ask Professor Puzzler? Click here to ask your question!

Phil's question is about the image below, and the blog post that goes with it (click the image to read the other blog post).

Professor Puzzler, in your answer to the "Genius" puzzle you showed a function that gave the right values for 8, 7, 6, and 5, but 3 wasn't in the domain. (I see how you did that, by the way - you multiplied the function by (x - 3), and then divided it by (x - 3) -- that gives you the same function, except with a hole in it!).

But my question is: can you come up with a function that matches all the data points given in the problem, but has a DIFFERENT value for x = 3? I'd like to be able to tell people that the answer is [some number besides 6] and watch them squirm when I show them the function that gave that answer.

Hi Phil, I can *absolutely* give you a function that matches the data points given, and gives you an answer other than 6 when you plug in three!

Of course, you'll need to do a bit of memorizing, because it's kind of an ugly function. Are you ready?

g(x) = x^{4 }- 26x^{3 }+ 252x^{2 }- 1067x + 1680

It's not pretty, is it? When you plug in 3, you get g(3) = 126, which is wonderful for what you want; it's quite startling that you get a result that's so big in comparison to the other data points. Your friends will think you're crazy for coming up with a number so big!

Are you wondering how I came up with this beast of a function? That actually wasn't too difficult. I was helped by the fact that I already had a function which matched the data points: f(x) = x^{2} - x.

In order to create a new function that has all those same data points, I just need to find some function h(x) such that h(8) = 0, h(7) = 0, h(6) = 0, and h(5) = 0. If I have a function like that, I can just add f(x) + h(x), and I'll have my new function that keeps those same data points.

So what function h(x) meets the criteria of having zeroes at 8, 7, 6, and 5? EASY!

h(x) = (x - 8)(x - 7)(x - 6)(x - 5)

Multiply this thing out and you get:

h(x) = x^{4 }- 26x^{3 }+ 251x^{2 }- 1066x + 1680

h(x) + f(x) = **g(x) = x ^{4 }- 26x^{3 }+ 252x^{2 }- 1067x + 1680**

Keep in mind that this is not the *only* such function; there are an infinite number of ways of generating your h(x), and different h functions will produce different values for g(3). For example, we could take our h(x) and multiply it by 1000, before adding f(x), and we'd still have a function that works. Or we could multiply it by some random polynomial like x^{7} + 17x^{3} - 11 (just make sure your polynomial doesn't have (x - 3) as a factor!). By doing this, we haven't changed the fact that the function has zeroes at 8, 7, 6, and 5. However, we've *significantly* changed the value of f(3)!

In fact, we could make g(3) equal anything we want to, just by manipulating our h(x) with a numerical multiplier.

Suppose we want to have g(3) = k. Then our function g(x) will be:

g(x) = (^{(k - 6)}/_{120})h(x) + f(x).

I'll leave it to the reader to figure out how I came up with that, and verify whether or not it works.

Now go annoy your friends, as only a math geek can!