“Lost information” – when NOT to cancel stuff out
I got an interesting email from a customer today which led to a topic that I don’t think I’ve ever addressed on the site, so I thought I’d write a blog post about it since the answer is kind of fun and applies to a lot of situations in Pre-Calc, Calc, and beyond…
The question the customer has is about the last problem in the last video on this page: https://www.thattutorguy.com/calculus/implicit-differentiation/
Here’s the customer’s question:
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Here’s my answer, which deals with the fact that when you cancel variables by dividing, you can “lose information”. It’s not obvious what that means, but check this out:
Hi Fritz,
Wow, you raise an interesting question! Upon initial inspection, I’m stumped because what you did is pretty simple and looks correct. However, on further inspection, I will have to defer to the catch-all answer that my teachers always gave me in situations like this: “When you canceled out those x’s you lost information.” Let me see if I can explain…
Let’s say you have a rational function like f(x) = [(x+1)(x-3)]/(x+1)
It looks like you can cancel out (x+1) from top and bottom and you’d get the same graph, i.e. the same function. It seems like a no-brainer. Yet it’s not the same function anymore because the original graph had a hole at x=-1 (because you’d be dividing by zero) and the new one does not, so it can’t actually be the same function anymore. On the one hand it’s normally totally legal to cancel stuff like that, yet you can’t deny that you “lost” a discontinuity. Something went wrong.
Another example. Let’s say you are asked to solve the equation x^3 – x^2 + 6x = 0. It’s very tempting to divide both sides by x right off the bat so that you can factor the resulting quadratic to (x+2)(x-3)=0 and solve. But if you’re doing it right, you should just factor out the x first and leave it there, so that your factored equation is x(x+2)(x-3)=0. See that extra “x” in there at the beginning? That means that you actually have a third solution: x=0! So you can see from this example that if you jump the gun by dividing both sides by x right away, you’d only get two answers instead of three. Not good!
“Lost information”!
I think your issue with my implicit differentiation example is analogous. If you graph the original equation (by solving for y like you did and canceling out an x from everything), it’s just a line (that happens to have a hole at 0 if you don’t cancel anything). To see if your math matches mine, just grab a couple points off that line and try plugging them into the answer I got for y” and see if my answer gives you 0 like yours did. And yes indeed: my answer gives y”=0 just like yours! Yet my answer for y” is discontinuous at x=0 whereas yours is not. So the “info” that was “lost” is the fact that the actual second derivative is undefined at x=0.
A subtlety to be sure, but I’m glad you raised the question because it’s kind of interesting. Never noticed that about that particular problem before. It’s actually a bad example for implicit differentiation, because whenever they give you an implicit differentiation problem in class it’s going to have enough y’s in there that you won’t be able to simplify it with algebra the way you did.
In general, it’s fine to cancel stuff out in some situations, but you do run this risk of “lost information”. I’m not a math major, so I can’t tell you when it’s okay to cancel and when it’s not from a technical standpoint. My guess (based on experience) would be that it’s okay to cancel things out when finding derivatives, but that you can’t cancel before you take the derivative because that’s when you’re butchering the original function itself. Also, whenever the “simplification” that you want to do involves canceling variables – like if you could reduce a 3rd degree polynomial to 2nd degree by canceling – you’r probably doing something wrong.
cheers,
chris
